TITLE: Prove that the line $PQ$ passes through a fixed point
QUESTION [2 upvotes]: A right isosceles triangle $AOB$ ($O$ being the origin), is such that when $AO$ and $BO$ are extended to points $P$ and $Q$ the relation $2AP.BQ=AB^2$ holds. Prove that the line $PQ$ passes through a fixed point. I tried writing some equations of lines and using parametric equation, tried to get a relation in the distances. But, nothing worked out. How to proceed? Thanks.

REPLY [0 votes]: Hint:
For ease of calculations, let's us assume right angle at the vertex $O$
Let the vertices $A(a, 0)$ on the x-axis & $B(0, a)$ on the y-axis & similarly. let the points $P(h, 0)$ & $Q(0, k)$  Now, we have $$AP=h-a$$   $$BQ=k-a$$ $$AB=\sqrt{a^2+a^2}=a\sqrt 2$$
Now, applying the condition: $2(AP)(BQ)=(AB)^2$
$$2(h-a)(k-a)=(a\sqrt2 )^2$$  $$(h-a)(k-a)=a^2$$ $$a=\frac{hk}{h+k}$$
Now, the equation of the line: PQ $$y-0=\frac{k-0}{0-h}(x-h)$$
$$y=\frac{k}{h}(h-x)$$


TITLE: How to know if $p$ is spanned by the set $\lbrace p1,p2,p3\rbrace$
QUESTION [0 upvotes]: I want to know how to solve the question if you have a set of polynomials
$$
p_1 = 1 + x,
$$
$$
p_2 = 1 + 2x + x^2,
$$
$$
p_3 = 1 + 3x + 3x^2 + x^3,
$$
and if the polynomial $p = 2017 + x^2 + 26x^3$ is spanned by the set
$S = \{p_1,p_2,p_3\}$
Thank you

REPLY [0 votes]: Substitute $x+1=t$
Then $p=26(t-1)^3 +(t-1)^2 +2017=26t^3-77t^2+76t+1992$
Since there exists a constant, $p$ cannot be expressed as a linear combination of
$t, t^2, t^3$


TITLE: $\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$ solve
QUESTION [4 upvotes]: $$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$$
my attempt
$$t = \frac{1}{2}\tan(u)$$
$$dt = \frac{1}{2}\sec^2(u)du\\$$
$$\begin{align}
\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt&=\int_{0}^{1} \frac{\tan^2(u)}{4} \sqrt{1+\tan^2(u)}\frac{1}{2}\sec^2(u)du\\
&=\frac{1}{8}\int_{0}^{1} \tan^2(u)\sec^{3}(u)du\\
&= \frac{1}{8} \int_{0}^{1} (\sec^2(u) - 1)(\sec^{3}(u))du\\
&=\frac{1}{8}\int_{0}^{1} \sec^5(u)du - \frac{1}{8} \int_{0}^{1}\sec^3(u)du
\end{align}$$
what now?

REPLY [0 votes]: $$\begin{align}
\int t^2 \sqrt{(1+4t^2)}dt&=\frac{1}{8}\int{\sinh^2(x)}\sqrt{1+\sinh^2(x)}\cosh(x)dx\\
&=\frac{1}{8}\int{\sinh^2(x)}\cosh^2(x)dx=\frac{1}{8\times 32}\left(\sinh(4x)-4x\right) +C \\
&= \frac{1}{256}\left((4\sinh(x)\cosh^3(x) + 4\sinh^3(x)\cosh(x)) - 4x\right) + C\\
&=\left[\frac{1}{256}\left((4(2t)(\sqrt{1+4t^2})^3 + 4(2t)^3\sqrt{1+4t^2}) - 4\sinh^{-1}(2t)\right)\right]_{0}^{1} \approx 0.6063
\end{align}$$


TITLE: Positivity of operator with variable flip
QUESTION [1 upvotes]: I consider an operator $T: L^2(\mathbb R^2) \to L^2(\mathbb R^2)$ which acts as
$$
  \langle \psi, T \phi \rangle = \int dp dq~ a^2(p,q) \Big( \overline{\widetilde{f \psi}(p,q)} \widetilde{f \phi}(p,q) -  \overline{\widetilde{f \psi}(p,q)} \widetilde{f \phi}(q,p) \Big),
$$
where $a$ is a real-valued function and $f(x,y) = f(y,x)$ is symmetric.
Q: Is it a positive operator or has a spectrum bounded from below?
The operator can be written as $T_1 - T_2$ for the first and second term in the brakets. $T_1$ is obviously positive as it is of the form $T_1 = L^\star L$. $T_2$ is nearly of this form, but has an additional variable flip $T_2 = L^* L F$ if $(F \phi)(x,y) = \phi(y,x)$.
Is there anything known about such a flip operator and the positivity properties of ($1 - F$), that could be used to show positivity (or boundedness from below) for the above operator?
Thinking finite dimensional, a matrix of this form $A = \begin{pmatrix} ~~~1 & -1 \\ -1 & ~~~1 \end{pmatrix}$ would be positive, since $\langle \psi, A \psi \rangle = |\psi_1 - \psi_2|^2 \geq 0$.

REPLY [1 votes]: $T$ is in general not a positive operator!
This is because for some function $g(p,q)$,
$$
  |g(p,q)|^2 - \overline{g(p,q)} g(q,p) \not\geq 0.
$$
So for any $a(p,q)$, we can find a $g(p,q)$ such that the integral
$$
  \int dp dq ~ a^2(p,q) \Big(  |g(p,q)|^2 - \overline{g(p,q)} g(q,p) \Big) < 0.
$$
However, there is a special case (which is relevant in my case):
If $a(p,q) = a(q,p)$, then $T$ is a positive operator!
In this case, one can symmetrize the brakets and $p,q$ and obtain
$$
  |g(p,q)|^2 + |g(q,p)|^2 - \overline{g(p,q)} g(q,p) - \overline{g(q,p)} g(p,q) = |g(p,q) - g(q,p)|^2 \geq 0.
$$
Then, the integral is positive because the integrand is positive.


TITLE: Closed topologist's sine curve is connected
QUESTION [1 upvotes]: I have been given a proof that the closed topologist's sine curve $\{{(t,\text{sin}(1/t)):t\neq 0\}}\cup \{{(0,t):-1\leq t \leq t\}}$ is connected.
First, we assume it is not connected.  Let $I=\{{(0,t):-1\leq t \leq t\}}$ and $G=\{{(t,\text{sin}(1/t)):t\neq 0\}}$.  Then there are disjoint, nonempty, open $U,V$ s.t. $I\cup G=U\cup V$.  Then $I=(I\cap U) \cup (I\cap V)$, but $I\cap U$ and $I\cap V$ are open and $I$ is closed, so we have a contradiction.
I do not know how to see that $I\cap U$ and $I \cap V$ are open.  I thought it is possible that they are closed?  I think this issue is analogous to the unanswered question I have here: (Equivalent definitions of connectedness).

REPLY [2 votes]: An easier way is to use that the closure of any connected set is again connected. The topologist's sine curve is the closure of the graph $\{(t, \sin(1/t)) \mid t > 0  \}$, which is path-connected (hence connected).

By the above argument, both $$\{ (t, \sin(1/t)) \mid t>0 \} \cup \{(0,t) \mid -1 \leq t \leq 1\}$$ and $$\{ (t, \sin(1/t)) \mid t<0 \} \cup \{(0,t) \mid -1 \leq t \leq 1\}$$are connected. Your topologist's sine curve is the union of these two connected sets, which intersection $\{ (0,t) \mid -1 \leq t \leq 1 \}$ is again connected.


TITLE: Does the limit of $(-1)^x$ as $x \to \infty$ exist?
QUESTION [1 upvotes]: Consider the function $f(x)=(-1)^x$, defined with domain as a subset of $\mathbb{R}$ to $\mathbb{R}$. Firstly, how do we find the domain of $f$? I only know that at fractional powers $1/2, 3/2$, or anything with $p/q$ such that $\gcd(p,q)=1$ and $2|q$, the function is not defined. (Of course at the integers the function is defined)
Secondly, I conjecture from the result that $\lim_{n \to \infty}\ (-1)^n$ does not exist that
$$\lim_{x \to \infty}\ (-1)^x$$
also does not exist.
How do I show it?

REPLY [4 votes]: Assuming you want to work in the real numbers, you could say that the domain of the function is the set
$$
\left\{
x\in\mathbb{Q}:x=\frac{r}{s}, s>0, \text{$s$ odd}
\right\}
$$
This is quite a big set and it contains all natural numbers; however
$$
\lim_{n\to\infty}(-1)^{2n}=1,
\qquad
\lim_{n\to\infty}(-1)^{2n+1}=-1,
$$
so what you can say is that, no, the limit doesn't exist.
I'd simply say that the function $x\mapsto a^x$ on the reals is defined only for $a>0$, so the problem is meaningless in this case.


TITLE: $F/E$ a finite Galois extension, the integral closure of $E[X]$ in $F(X)$
QUESTION [2 upvotes]: If $F/E$ is a finite Galois extension, then one can show that $F(X)/E(X)$ is also a finite Galois extension of the same degree (a basis for $F/E$ is also a basis for $F(X)/E(X)$). Since $E[X]$ is a Dedekind domain, what is its integral closure in $F(X)$?  Is it $F[X]$?

The reason I'm asking is I'm trying to solve the following problem:
"Let $h$ be monic and irreducible in $E[X]$, and let $h_1, h_2 \in F[X]$ be monic irreducible divisors of $h$ in $F[X]$.  Then there exists an automorphism $\theta$ of $F[X]$ which leaves $E[X]$ fixed and sends $h_1$ to $h_2$."
If $F[X]$ is in fact the integral closure of $E[X]$, then I can use the result that $(h_1), (h_2)$ are primes of $F[X]$ which lie over $(h)$, so there is a $\theta \in Gal(F(X)/E(X))$ with $\theta(h_1) = (h_2)$ (which is almost what I want).

REPLY [1 votes]: Since $E\subset F$ is algebraic, $E[X]\subset F[X]$ is integral (why?). Moreover, $F[X]$ is integrally closed, so the integral closure of $E[X]$ in $F(X)$ is (as you hoped) $F[X]$.


TITLE: Showing $\frac{1}{2L}\|\nabla f(x)\|^2 \leq f(x)-f(x^*).$
QUESTION [0 upvotes]: If $\nabla f$ has a Lipschitz constant of $L$, then
$$f(y) \leq f(x) + \nabla f(x)^T(y − x) + \frac{L}{2}\|
y − x\|^2. $$
Using this fact I am going to prove
$$\frac{1}{2L}\|\nabla f(x)\|^2 \leq f(x)-f(x^*).$$
I started by writing the following statement,
$$ f(x)-f(x^*)=\arg \max_y[f(x)-f(y)]\geq \arg\max_y[-\nabla f(x)^T(y-x)-\frac{L}{2}\|y-x\|^2].$$
However, I don't know how to find "$\arg\max_y[-\nabla f(x)^T(y-x)-\frac{L}{2}\|y-x\|^2]$. I appreciate any help.

REPLY [1 votes]: This result does not depend on $f$ being convex.
Let $\phi_x(h) = \langle \nabla f(x), h \rangle + {L \over 2} \|h\|^2$
and note that $\phi_x$ is a convex quadratic function and $\nabla \phi_x(h) = \nabla f(x) + L h$, hence
$\phi_x(h^*) \le \phi_x(h)$ where $\nabla f(x) + L h^* = 0$.
In particular, $\phi_x(h^*) = -{1 \over 2L} \| \nabla f(x) \|^2$.
We have
$f(x+h) \le f(x) + \phi_x(h)$ and so
$\inf_{h'} f(x+h') = \inf f \le f(x) + \phi_x(h)$ for all $h$.
Hence $\inf f \le f(x) + \inf_h \phi_x(h) = f(x) - {1 \over 2L} \| \nabla f(x) \|^2$.
Rearranging gives
${1 \over 2L} \| \nabla f(x) \|^2 \le f(x) - \inf f$.


TITLE: Sum of this Geometric Series $\frac{1}{4}, \frac{1}{12}, \frac{1}{36}, \frac{1}{108}, \ldots, \frac{1}{2916}$
QUESTION [2 upvotes]: $$\frac{1}{4}, \frac{1}{12}, \frac{1}{36}, \frac{1}{108}, \ldots,\frac{1}{2916}$$
I am trying to calculate the sum of this geometric series.  Here's what I've got so far.
$a = \frac{1}{4}, r = \frac{1}{3}$ and $n = 7$
So $\sum_{n=7}\frac{1}{4}(\frac{1}{3})^{n}$
Somehow equals $\frac{1093}{2916}$ according to my book?
Here are my questions:

How do I get the sum of this geometric series
Does my work look correct?
How can I find a way to calculate the number of terms in the series that isn't "brute forcing"?  This is quite inelegant.

Thanks!

REPLY [3 votes]: The partial sum of a geometric series is
$$\sum_{k=0}^{n-1}ar^k = a\frac{1-r^n}{1-r}$$
so take $a=1/4, r=1/3, n=7$ to get
$$S = \left(\frac{1}{4}\right)\frac{1-(1/3)^7}{1-(1/3)} = \frac{1093}{2916}.$$


text:
\subsection{Surfactant-covered bubble in steady state}
\noindent Using the exact solutions of \S\ref{sec:validsteady}, the bubble deformation and surfactant concentration at steady state is tested. Given a Capillary number $Q$, the steady state interface position $z(\zeta,T)$ and surfactant concentration $\rho(\zeta,T)$ are known, where $T$ is the time of steady state and $\zeta=e^{i\nu}$ for $\nu\in[0,2\pi)$. An example of how such a steady state may look like can be seen in Figure~\ref{fig:plot12}, where the steady state corresponding to $Q=0.14$ is shown.
\begin{figure}[h!]
	\centering
	\includegraphics[width=0.445\textwidth]{graphics/p1_NEW180607.pdf}
	\includegraphics[width=0.5\textwidth]{graphics/p2.pdf}

    \caption{Interface (left) and surfactant concentration (right) at steady state for Capillary number $Q=0.14$. Elasticity number $E=0.5$, initial surfactant concentration $\rho_0=1$, $Pe_\Gamma=\infty$ and viscosity ratio $\lambda=0$. Steady state was reached at time $T=46.35$.}
    \label{fig:plot12}
\end{figure}
In a simulation, a bubble is said to have reached steady state when the normal velocity, $\mathbf{u}\cdot\mathbf{n}=0$. Practically, here the definition of steady state is that
\begin{align*}
	\left|\mathbf{u}\cdot\mathbf{n}\right|\leq 10^{-8}.
\end{align*}
It should be noted that for large Capillary numbers, $Q$, or high viscosity ratios, $\lambda$, this threshold is slightly too large and should be decreased since the deformation is very slow.

The analytical solutions from Algorithm~\ref{alg:steady} are denoted by $z^{V}(\zeta,T)$ and $\rho^{V}(\zeta,T)$ for interface position and surfactant concentration respectively. Defining the obtained solutions from the numerical method as $z(\alpha,T)$ and $\rho(\alpha,T)$, a comparison between the numerical results and the validation is obtained by studying the point-wise differences in $\alpha\in[0,2\pi)$. Two errors will be studied: in position, $e_z = |z-z^V|$ and surfactant concentration $\rho$, $e_\rho$. In order to compare point-wise in $\alpha\in[0,2\pi)$, a mapping from $\nu$ to the equal-arc length measure $\alpha^V$ is needed. This is defined as
\begin{align}
	\alpha^V(\nu) = \dfrac{S_\nu 2\pi}{L},
	\label{eq:alphaV}
\end{align}
where $S_\nu = \int_0^\nu |z^V_\nu|d\nu$. The solutions $z^V$ and $\rho^V$ are computed at non equidistant discrete points $\alpha_i^V=\alpha^V(\nu_i)$, for $\nu_i\in[0,2\pi)$, $i=1,\hdots,N_V$ for some $N_V$. Finally, to compare the results, $z(\alpha,T)$ and $\rho(\alpha,T)$ need to be interpolated to $\alpha^V_i$. This is done using their Fourier expansions.

\begin{figure}[h!]
	\centering
	\includegraphics[width=0.5\textwidth]{graphics/p3_err.pdf}
	\caption{Point-wise error in $z$ (marker $\circ$) and $\rho$ (marker $\triangle$) vs. $\alpha$ at steady state, for the same parameters as in Figure~\ref{fig:plot12}, grid spacing $\Delta s = 0.0016$. Red line corresponds to time-stepping tolerance set to $10^{-6}$.}
	\label{fig:plot3}
\end{figure}

In Figure~\ref{fig:plot3}, the point-wise absolute error between analytical solution and the BIE simulation can be seen for the case in Figure~\ref{fig:plot12}. The spatial discretisation is adaptive, keeping the spatial distance in arc length similar at all times; here approximately $\Delta s=0.008$. The simulation was run with time-step tolerance $10^{-6}$ and the errors are all below the set tolerance.

\begin{figure}[h!]
	\centering
	\includegraphics[width=0.6\textwidth]{graphics/p4.pdf}
	\caption{Steady state deformation, $D$, vs. Capillary number $Q$ at different elasticity numbers $E$ for inviscid bubbles, $\lambda=0$. Black lines: validation solutions computed through the method in \S\ref{sec:validsteady}. Diamonds: numerical results from the BIE method. Circles show simulations chosen for the examples of deformed bubbles shown.}
	\label{fig:plot4}
\end{figure}

Using the method in \S\ref{sec:validsteady}, one can obtain a graph for how deformation $D$ depends on Capillary number $Q$. Such a graph for two different elasticity numbers, $E=0.5$ and $E=0.9$, is shown in Figure~\ref{fig:plot4}. The black lines represent the validation solutions computed through the method in \S\ref{sec:validsteady} and the blue diamonds show results from simulations using the BIE method of this paper. All simulations have been run from the same initial setting as the previous case, with time-step tolerance $10^{-6}$. The simulations differ only in the choice of Capillary number $Q$. Examples of bubble deformation and surfactant concentration are also shown, corresponding to $Q=0.07$ and $Q=0.13$, shown by circles in the figure.

\begin{figure}[h!]
	\centering
	\includegraphics[width=0.45\textwidth]{graphics/p5.pdf}
	\hspace{1mm}
	\includegraphics[width=0.45\textwidth]{graphics/p6.pdf}

	\caption{$\ell_2$ error in $z$ and $\rho$ vs. (left) deformation $D$ for cases in Figure~\ref{fig:plot4} and (right) vs. viscosity ratio $\lambda$ for deformation $D=0.29$. Solid and dashed lines represent $E=0.5$ and $E=0.9$ respectively. Dots and triangles represent $e_z$ and $e_\rho$ respectively. Red solid line shows time-step tolerance $10^{-6}$.}
	\label{fig:plot56}
\end{figure}

To quantify the results in Figure~\ref{fig:plot4}, the $\ell_2$-error for each deformation $D$ is plotted in Figure~\ref{fig:plot56} (left). For each diamond in Figure~\ref{fig:plot4}, the simulation and validation data is compared and plotted vs. deformation $D$. Note that both cases $E=0.5$ and $E=0.9$ are shown (solid and dashed lines respectively), and that they for the same $D$ were obtained by different $Q$. It is shown that both errors ($e_z$ and $e_\rho$) stay below time-step tolerance.

As mentioned in \S\ref{sec:validsteady}, the viscosity ratio does not affect steady state deformation in the case of $Pe_\Gamma=\infty$. The results of this paper agree with this observation, as is shown in Figure~\ref{fig:plot56} (right). There, droplets of different viscosity ratios ranging from $\lambda = 0$ (corresponding to an inviscid bubble) to $\lambda = 2$ were deformed under Capillary number $Q=0.14$. As can be seen in the figure, both interface position and surfactant concentration coincides with that for bubbles up to time-step tolerance.

config: arxiv


TITLE: Class number parity in pure cubic number fields
QUESTION [20 upvotes]: Consider the family of pure cubic number fields
$K = {\mathbb Q}(\sqrt[3]{m})$ for $m = a^3 \pm 3$.
Proposition. If $4 \mid a$ and $m$ is cubefree, then the
class number of $K$ is even.
Proof. Let $\omega = \sqrt[3]{m}$; the element $\alpha = a - \omega$
   has norm $\pm 3$. Since $3$ is completely ramified, the element
   $\varepsilon = \alpha^3/3$ is a unit.
If $m = a^3 + 3$, then  $\varepsilon = 1 - 3a^2\omega + 3a\omega^2$.
   If $4 \mid m$, then $\varepsilon \equiv 1 \bmod 4$, hence
   $K(\sqrt{\varepsilon}\,)/K$ is an unramified quadratic extension.
Experiments seem to suggest that if $m = a^3+3$ and $a \equiv 2 \bmod 4$,
then $h$ is also even, but there is no explanation, class field theoretic
or otherwise. In fact, the class number is even for all cubefree values
of $m$ for $a = 2, 4, \ldots, 2 \cdot 88$, but is odd for $a = 2 \cdot 89$.
This cannot be an accident; the parity of the class number in the
case $m = a^3 - 3$ for $a \equiv 2 \bmod 4$ shows a more typical
(i.e. more random) behaviour in that the class number is odd quite
often.
Question: How can this behaviour in the case $m = 8a^3+3$ be explained?
My first guess would be that, for fields in this family, there is
a family of ideals ${\mathfrak a}$ such that ${\mathfrak a}^2$ is
principal, but I can't seem to find anything in this direction.
 Edit. Dror's comment made me look at the family of elliptic curves $y^2 = x^3 - m$.
These have rank $\ge 1$, and by the parity conjecture rank $\ge 2$. An inequality due
to Billing now shows that $K$ has even class number. For details, see this
pdf file.
Actually, Paul Monsky stumbled across something similar for pure quartic fields;
see here.

REPLY [5 votes]: Billing (Beiträge zur arithmetischen Theorie der ebenen
kubischen Kurven vom Geschlecht Eins, R. Soc. Scient. Uppsala (4) 11,
Nr. 1. Diss. 165 S. Uppsala 1938; see Ian Connell's Handbook for elliptic curves for
a modern presentation of the result) proved the following result:
Let $f(x) = x^3 + ax^2 + bx + c \in {\mathbb Z}[x]$ be irreducible,
and consider the elliptic curve $E: y^2 = f(x)$. Let $K$ be the
cubic number field generated by a root $\alpha$ of $f$, and let
$E_K$ be its unit group. Write $({\mathcal O}_K: {\mathbb Z}[\alpha]) =: m_f^2$.
Then
$$   r \ \le \ r_2(K) + r_E(K) + 2n_+ + n_-, $$
where $r$ is the Mordell-Weil-rank of $E({\mathbb Q})$,
$r_2(K)$ is the $2$-rank of the ideal class group of $K$,
$r_E(K)$ is the ${\mathbb Z}$-rank of the unit group $E_K$ of $K$,
$n_+$ is the number of primes $p \mid m_f$  that split in $K$,
and  $n_-$ is the number of primes $p \mid m_f$ that decompose
as $p {\mathcal O}_K = {\mathfrak p}{\mathfrak p}'$ or as
$p {\mathcal O}_K = {\mathfrak p}^2 {\mathfrak p}'$.
If ${\mathbb Q}(\sqrt[3]{m})$ is a pure cubic field with
$m \not\equiv \pm 1 \bmod 9$ cubefree, then the index is trivial,
and $n_+ = n_- = 0$ provides us with the bound
$$  r \ \le \ r_2(K) + 1. $$
On the other hand, the parity conjecture (see the article by
Liverance pointed out by Dror) implies that the Mordell-Weil
rank of $E$ is even for squarefree values of $m =  8b^3 + 3$,
and the family of nontorsion points
$$  P_b\Big( \frac{2b^3+1}{b^2}, \frac{3b^3+1}{b^3} \Big) $$
shows that $r \ge 1$. Thus the parity conjecture implies $r \ge 2$,
and Billing's bound finally gives $r_2(K) \ge 1$.


TITLE: Change of Variables Problem. How Does One Know What to Select as $u$ and $v$?
QUESTION [1 upvotes]: I just learned change of variables for double and triple integrals, but I'm having difficulty applying it. When I encounter a problem, I am not sure which variables to set as $u$ and $v$. Unlike single variable substitution ($u$-substitution), there doesn't seem to be any indication of which variables should be selected as $u$ and $v$. In fact, most of the example problems that I have encountered online actually give you the variables $u$ and $v$ to use, rather than having to find them yourself. However, for my studies, I am required to find them myself; this is something I am having trouble with.
Take the following problem I am given:
Evaluate $\iint_D \dfrac{y^4}{x} \ dxdy$ over the region D contained between the parabolas $x = 1 - y^2$ and $x = 4(1 - y^2)$.
The textbook solution says to make $x = v(1 - u^2)$ and $y = u$, but it does not give any indication of how it chose these $u$ and $v$! How did the solution to the above problem find/select $u$ and $v$? As a student, when encountering a problem as above, how can I deduce $u$ and $v$? Is there a proper methodology that is analogous to single variable substitution ($u$ substitution)?
I would greatly appreciate it if people could please take the time to help me understand this methodology.

REPLY [2 votes]: Caveat lector -- this answer is somewhat heuristic.
There is no "methodology". There is a certain mysterious and charming creativity in finding a change-of-variables that solves a particular problem. It involves
playing around with the problem, drawing little pictures, understanding the functions involved, and lots of patience. Moreover, it requires a mental process similar to that which might have led Van-Gogh to choose yellow rather than orange.
True, with time and accumulation of collective experience, people notice that some change of variables are more useful than others. For example, if your function has some spherical symmetry, it makes great sense to use spherical coordinates. Or if your function has some other symmetry, or other properties that concern the geometry of the underyling integration domain, then it makes sense to try and use them.


TITLE: Recursion question for CS
QUESTION [0 upvotes]: Is this correct if not where have i gone wrong?
$$f(x)= 20+f(x+1)$$

REPLY [2 votes]: It is not correct. One thing you need is a starting point (e.g., $f(0) = 20$). This is required if you want to make a recursive function. Additionally, I believe the function you're looking for is the following:
$f(0) = 20, f(x) = f(x-1) + 20 + x $
The rate of change in this function is $20 + x$, so if you were to add it to the last term, you would get the following term. It would help for me to know how you derived your formula in order to tell you where you went wrong.


TITLE: Velocity and acceleration question
QUESTION [1 upvotes]: 1.The magnitude of the total force acting on a ball rolling without
  slipping down a ramp is greater than
  the magnitude of the total force
  acting on the same ball if it slides
  down the ramp without friction. True
  or false?

I chose true for this one, as I figured that if our force has to oppose friction, then it must be greater than that force without friction. However the correct answer is false.

2.The magnitude of the velocity of an object must change if the magnitude of
  its acceleration is a constant. True
  or false?

I chose true for this one too. If acceleration is constant, then velocity is linear. This turned out to be false as well.
Can anyone explain to me why I was wrong for both of these?

REPLY [0 votes]: I am going to comment on the first question only. The sum of the forces is equal to acceleration, so I guess this poorly worded question boils down to which ball is accelerating more: a) With friction and angular acceleration, or b) Without friction and zero angular acceleration.
From an energy perspective you can suspect that with friction some of the work done goes towards rotating the ball and thus will be less left to move it linearly.
The math also supports this argument if you work out all the forces/moments, as well required rotational motion for each scenario.


TITLE: partial derivate and integral equation
QUESTION [2 upvotes]: Good day, I need derivate $f(x)=g(x)+\int_{a}^{x}K(x,y,f(y))dy$ and show that $f{}'(x)=g{}'(x)+K(x,x,f(x))+\int_{a}^{x}\frac{\partial }{\partial x}K(x,y,f(y))dy$.
I think the next but I do not sure: Take x=y=t and $u(x,y)=\int_{a}^{x}K(x,y,f(y))dy$ then $\frac{\partial u}{\partial t}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}$. And $\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\int_{a}^{x}K(x,x,f(x))dx$, $\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}\int_{a}^{x}K(y,y,f(y))dy=K(x,x,f(x))$.
Is correct the processe? Also, $\frac{\partial }{\partial x}\int_{a}^{x}K(x,x,f(x))dx,\int_{a}^{x}\frac{\partial }{\partial x}K(x,x,f(x))dx$ can differ  by constant because a limit of integration is a, or not?

REPLY [0 votes]: It's not correct. You have to apply directly the leibniz integral rule. Consider that in your integral $f(x)=g(x)+\int_{a}^{x}K(x,y,f(y))dy$ the argument and the limits of integration depend on $x$ and the simple partial derivative into the integral doesn't work (further, not important, but $u$ is not function of $y$, $y$ is only a dummy variable to integrate).
$$\require{cancel}f'(x)=g'(x)+K(x,x,f(x))(x)'-K(x,a,f(a))\cancelto{0}{(a)'}+\int_{a}^{x}\frac{\partial }{\partial x}K(x,y,f(y))dy$$
$$f'(x)=g'(x)+K(x,x,f(x))+\int_{a}^{x}\frac{\partial }{\partial x}K(x,y,f(y))dy$$
And it's done.


TITLE: Lower bound for Euler's totient for almost all integers
QUESTION [7 upvotes]: Let $\varphi(n)$ be the Euler's totient function. It is well know that $\liminf_{n \to \infty} \frac{\varphi(n)}{n / \log \log n} = e^{-\gamma}$, so that for $\varepsilon > 0$ it results $\frac{\varphi(n)}{n} \geq \frac{e^{-\gamma}-\varepsilon}{\log \log n}$ for large $n$. Actually, the "local minima" of $\frac{\varphi(n)}{n}$ are attached for $n = p_1 \cdots p_k$ (the product of the first $k$ primes) and the set of primorial is really sparse. I wonder if it is known a lower bound for $\varphi(n)$ like: "$\varphi(n) / n \geq f(n)$ for all $n$ but a set of null asymptotic density", where $f(n)$ is a function bigger then $\frac{e^{-\gamma}-\varepsilon}{\log \log n}$.
Thank you in advance for your help.

REPLY [3 votes]: Your question has been answered by Lucia already, but you might also be interested in looking up the Erdős-Wintner theorem. A special case (proved already by Schoenberg) is that for each $u \geq 0$, the set of $n$ with $\phi(n)/n \leq u$ has an asymptotic density $D(u)$; moreover, $D(u)$ is continuous and increasing on $[0,1]$.
There are also estimates available for the size of $D(u)$ when $u$ is near zero, and of $1-D(u)$ when $u$ is near $1$. For this, see Erdős's paper "Some remarks about additive and multiplicative functions":
http://www.renyi.hu/~p_erdos/1946-11.pdf


TITLE: Derivative of $\|x\|$
QUESTION [3 upvotes]: $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a map $f(x)=\langle x,x\rangle$, $g:\mathbb{R}\rightarrow\mathbb{R}$ is a map $g(x)=\sqrt{x}$, so the composition $h:=g\circ f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a map $x\mapsto \|x\|$, I want the derivative of $h$ at $a$, according to chain rule it should be $D(g(f(a)))\circ D f(a)$,could any one tell me now next steps?

REPLY [0 votes]: Here is a coordinate free  argument:
According to the chain rule
$$dh(a)=dg\bigl(f(a)\bigr)\circ df(a)\ .\tag{1}$$
Now
$$f(a+X)-f(a)=(a+X)\cdot(a+X)-a\cdot a=2a\cdot X+|X|^2\tag{2}$$
and therefore
$$df(a).X=2 a\cdot X\ .$$
Furthermore the formula $$g'(y)={1\over 2\sqrt{y}}\qquad(y>0)$$
can be written in the form
$$dg(y).Y={1\over 2g(y)}Y\ .\tag{3}$$
Letting $y:=f(a)$ and plugging $(2)$ and $(3)$ into $(1)$ we obtain
$$dh(a).X=dg\bigl(f(a)\bigr)\bigl(df(a).X\bigr)={1\over 2g\bigl(f(a)\bigr)}\bigl(2a \cdot X)\bigr)={a\cdot X\over|a|}\ .$$
It follows that $dh(a)$ can be represented by the gradient $$\nabla h(a)={a\over|a|}\qquad(a\ne0)$$ in the form
$$dh(a).X=\nabla h(a)\cdot X\qquad(X\in{\mathbb R}^n)\ .$$


TITLE: Rewrite expressions
QUESTION [0 upvotes]: I have to prove that

$$q\lor(¬q\land(p\lor q))$$

is equal to just $q$.
This is normally done with logical equivalences, but I can't solve this one.
Can somebody please help?
----------EDIT----------
So with some help i now got this with the help of a truth table
\begin{array}{|c|c|c|c|}
\hline p& q & ¬q & p\land q & ¬q\land(p\land q)\\
\hline t& t& f& t&f\\
 t& f& t& f&f\\
 f& t& f& f&f\\
 f& f& t& f&f \\\hline
\end{array}
Hence we have
$$\left[q\lor\left(¬q\land(p\land q)\right)\right] \Leftrightarrow \left[q\lor f\right]\Leftrightarrow q.$$
But the assignment is to prove this with logical equivalences. Look it does not matter on my test friday only the answer does. But i can't stand the fact i could not solve this.
We must use the following paper as a cheat sheet https://www.dropbox.coms/o6pau9a3z7yqi39/Logische_equivalenties.docx (Note, not by OP: dead link.)
---------- EDIT ----------
Everyone thanks for the help got an A on the test.

REPLY [2 votes]: Here are two ways to prove that your expression simplifies to $\;p \lor q\;$ (so not to $\;q\;$).
One way is to use distribution and then simplify:
\begin{align}
& q \lor (\lnot q \land (p \lor q)) \\
\equiv & \qquad \text{"$\;\lor\;$ distributes over $\;\land\;$ -- to bring the $\;q\;$'s together"} \\
& (q \lor \lnot q) \land (q \lor p \lor q) \\
\equiv & \qquad \text{"simplify: excluded middle; $\;q \lor q \equiv q\;$"} \\
& \text{true} \land (p \lor q) \\
\equiv & \qquad \text{"simplify"} \\
& p \lor q \\
\end{align}
Here I've not made explicit the associativity and symmetry of $\;\lor\;$.
The other (quicker) way is as follows:
\begin{align}
& q \lor (\lnot q \land (p \lor q)) \\
\equiv & \qquad \text{"use $\;q \equiv \text{false}\;$ on other side of leftmost $\;\lor\;$"} \\
& q \lor (\lnot\text{false} \land (p \lor \text{false})) \\
\equiv & \qquad \text{"simplify"} \\
& q \lor p \\
\end{align}
Personally I find this rule ('assume one side of $\;\lor\;$ is $\;\text{false}\;$ on other side') often useful, just like its dual ('assume one side of $\;\land\;$ is $\;\text{true}\;$ on other side').


TITLE: What will limit the speed at which one can play the piano?
QUESTION [3 upvotes]: I thought about this as a fun question.  Suppose we provide a piano player that is limited only by the typical relativistic rules (i.e., hands and fingers cannot move faster the speed of light), so choose your favorite speedy superhero or android.
In this scenario, what would limit the speed at which the player could play a typical (i.e., not electronic) piano?  Would it be:

The strength of any of the following parts:

piano keys?
piano wires/strings?
hammer arm?
etc.

The speed of the standing waves on the wires/strings in the piano?
The delay between pressing a key down and the time when the hammer strikes the string?

REPLY [2 votes]: Let the hammer that strikes the string moves a distance $\Delta x$ above the string and then hit the string back due to gravity. Or take some other mechanism like a spring which puts the note back to the unpressed position after pressing that note. The time it takes to get a particular note back to unpressed position after it being pressed is the main factor here. Because for the artist to play something useful he'd need to repress a recently used note.
The problem is not that we can't pluck the string or unpress a note with a speed faster than $c$. The problem is that we can't even play the piano faster than speed of sound in metal. The information of pressing the note reaches the string with the speed of sound in the materials involved. Let's take the distance between the note and the hammer $\Delta y$ then  if we try to press a note twice in a time $\Delta t$ then we have to make $ (\Delta x + \Delta y) v_{sound}< \Delta t$. Otherwise we'd just press the already pressed note. So if a tone involves pressing of a note twice in a time period then we can't make that time period smaller than $\Delta t$.
There might be further restrictions based upon how fast the string vibrates. A vibrating string is actually a standing wave. For a note to be heared the string must at least oscillate once. That is the wave must go from start to the end of wire at least twice. All this depends upon the speed of wave. This speed depends upon the length, weight, meterial and tension in the string. None of these factor involve $c$. Because the speed of wave is way too less than $c$.
If your superhuman piano player also has a superpiano which is made up of super material which can transmit waves in strings with extraordinary speed then in principle the information of the note being pressed can't reach the string faster than $c$. Or if a guitar is being used then the speed of wave on string can't exceed $c$. If the length of the string of guitar is $L$ then you can't pluck a string twice in time $2L/c$


TITLE: If ${x_n}$ converges weakly to $x_0$, then for $T \in B(X,Y)$, $T(x_n)$ converges weakly to $T(x_0)$?
QUESTION [2 upvotes]: Let $X,Y$ be normed spaces.  A bounded operator $T: X \to Y$ is arbitrary. $B(X,Y)$ denotes the set of bounded operators.  How to prove that if  ${x_n}$ converges weakly to $x_0$, then for $T \in B(X,Y)$, $T(x_n)$ converge weakly to $T(x_0)$?
I know it has to be true in the case that weak convergence implies strong convergence. Also, when $T$ is not only bounded but also linear, Riesz representation theorem comes in and the proof becomes trivial. However, if $T$ is only bounded, I have no clue.
This is an exercise from the book. I seem to find counter a example of this though. Can anyone at least show me an example of weak convergence where the limit is not 0?

REPLY [2 votes]: Assume $x_n \xrightarrow{w} x_0$. For any $f \in Y^*$ we have that $f \circ T \in X^*$ so
$$f(Tx_n) = (f \circ T)(x_n) \xrightarrow{n\to\infty} (f \circ T)(x_0) = f(Tx_0)$$
Therefore $Tx_n \xrightarrow{w} Tx_0$.


TITLE: Why values close to critical point in limit are giving different outputs?
QUESTION [0 upvotes]: In evaluation of limits , there's a critical point c at which the output becomes indeterminate, like 0/0 etc. Now I came across some weird examples where when I tried putting values close to that critical value, but I got output different from the answer analytical evaluation gave me.
Example 1: $$f(x)=\frac{\sin\left(\pi {\left(\cos\left(x\right)\right)}^{2}\right)}{{x}^{2}}$$
In this function if we take limit as x-->0 then we should get π. But I tried putting values close to zero in desmos:

Everything goes fine till 10^-5 but after that values start getting deflected from π and after that the value goes to ZERO!!
Example 2: $$f(x)=(\frac{1-5{x}^{2}}{1-3{x}^{2}}{)}^{\frac{1}{{x}^{2}}}$$
In this example if we take limit as x-->0 I should get e^2 which is approx 7.38905609893. But this is what happened in desmos:

At x=10^-7 , the output suddenly became more INACCURATE and more inaccurate at 10^-8 and then goes straight to ONE!!
Main question: Why are the output values different from the expected answer from analytical method and more importantly why are they following this ambiguous pattern?

REPLY [2 votes]: To help illustrate my comment. Look at your second example, though as given in Desmos, not the altered version you reproduced in the question:
$$h(x)=\left(\frac{1+5x^2}{1+3x^2}\right)^{\frac 1{x^2}}$$
The double precision floating point number that is commonly used for computing can store numbers to about 16 digits of accuracy. That is, starting with the leading non-zero digit in the actual number, It can store roughly only the next 16 digits ("roughly" because it actually stores the number in binary, not decimal). Now look what happens when you try to calculate $h(x)$ for $x = 10^{-9})$. First, it calculates

$1 + 5x^2 = 1 + 5\times 10^{-18} = 1.000,000,000,000,000,005$, except that is 19 digits, and it can only store about 16. So it rounds it to the nearest number it can store: $1$.
$1 + 3x^2 = 1.000,000,000,000,000,003$. Again, this is rounded to $1$.

Since the ratio of $1$ to $1$ is $1$, and $1$ raised to any power is $1$, it calculates $h(10^{-9}) = 1$.
At $x=10^{-8}$, it can just barely fit in a bit of extra information (though you need to look at the actual binary representation to see why), but that information is not enough to accurately calculate the value, so you get the bad result. Even at $x = 10^{-7}$, there is not enough precision to get more than a couple digits of accuracy.


TITLE: Is contraction of radical ideals and/or prime ideals surjective in this case?
QUESTION [2 upvotes]: Let $K$ denote a field and $\overline{K}$ denote it's algebraic closure. Then we have an inclusion of polynomial rings $$K[x_1,\ldots,x_n] \subseteq \overline{K}[x_1,\ldots,x_n].$$ Since the preimage of a prime (resp. radical) ideal is prime (resp. radical), the contraction of an ideal $I \subseteq \overline{K}[x_1,\ldots,x_n]$ to an ideal $I^c \subseteq K[x_1,\ldots,x_n]$ defined by $I^c := I \cap K[x_1,\ldots,x_n]$ preserves both radicalness and primeness.
This proceed needn't be injective. For example, let $K = \mathbb{R}$ and $\overline{K} = \mathbb{C}$. Then $(x+i)^c = (x^2+1)$ and $(x-i)^c = (x^2+1)$.
However, I have a hunch it might be surjective on prime ideals and/or radical ideals. If so, it should follow that contraction commutes with radicals $(\sqrt{I})^c = \sqrt{I^c}.$

Question. Is any or all of this true?

REPLY [2 votes]: Yes, this is true.  For prime ideals, this is a special case of the "lying over theorem" for integral extensions.  If $B$ is any commutative ring with a subring $A$ such that $B$ is integral over $A$, then every prime ideal of $A$ is the contraction of some prime ideal of $B$.  In this case, $B=\overline{K}[x_1,\ldots,x_n]$ is integral over $A=K[x_1,\ldots,x_n]$ since it is generated by $\overline{K}$ as an $A$-algebra and every element of $\overline{K}$ is integral over $K$.
As a sketch of a proof, let $P\subset A$ be a prime ideal.  Localizing at $P$, we may assume $P$ is the unique maximal ideal of $A$.  It then suffices to show that $PB$ is a proper ideal of $B$, so it is contained in some maximal ideal of $B$ (which can then only lie over $P$ since $P$ is maximal).  If $PB=B$, then this equation remains true when we replace $B$ with some finitely generated $A$-subalgebra $B_0\subseteq B$.  But then $B_0$ is a finitely generated $A$-module since $B$ is integral over $A$, so $PB_0=B_0$ implies $B_0=0$ by Nakayama.  This is a contradiction, and hence $PB$ must be a proper ideal of $B$, as desired.

Once we have the result for prime ideals, it immediately follows for radical ideals, since radical ideals are just intersections of prime ideals.
Note that the statement $(\sqrt{I})^c = \sqrt{I^c}$ is comparatively trivial and holds for any extension of commutative rings $A\subseteq B$.  Indeed, if $f\in (\sqrt{I})^c$, then $f\in \sqrt{I}$ and so $f^d\in I$ for some $d$.  But then $f^d\in A$ as well and so $f^d\in I^c$ and $f\in \sqrt{I^c}$.  Conversely, if $f\in\sqrt{I^c}$, then $f^d\in I^c$ for some $d$ and so $f^d\in I$ and hence $f\in (\sqrt{I})^c$.


TITLE: Proof of compactness of subset S of a compact set K contained in an open set.
QUESTION [0 upvotes]: Let $A$ be an open subset $\mathbb{R}^{n}$ and $K\subset A$ is compact. Prove there exist an $r\gt0$ such that the set  $$ D=\{y\in \mathbb{R}^{n}: \|y-x\|\leq r \text{ for some }x \in K\}$$ is a compact subset of $A$.

.
Let $y\in cl(D)$ (closure of D), there exist a sequence $y_n$ in $D$ which converges to $y$. This implies that $\|y_n-x_n\|\leq r$, now $x_n$ is a sequence in $K$ and must have a convergent subsequence(due to compactness of K) and point of convergence in $K$, say $x$.
We choose the subsequence of $y_n$  in such way to match with numbering of subsequence of $x_n$.
$\|y-x\|\leq \|y_{n_k}-y\|+\|y_{n_k}-x_{n_k}\|+\|x_{n_k}-x\|\leq 2\epsilon+r$
This implies $y\in D$ and $cl(D)=D$. $\blacksquare$
I have doubt whether I'm  missing something and I haven't made use of the set $A$ at all and also the existence of $r>0$

REPLY [0 votes]: From an advanced standpoint we can use some basic results about compact spaces and metric spaces:
(1). If $K$ is a non-empty compact subset of a space $X$  and if $f:X\to \Bbb R$ is continuous then $\min \{f(x):x\in K\}$ exists.
(2). If $(X,d)$ is a metric space and $V$ is a non-empty subset of $X$ then the function $f(x)=\inf \{d(x,y):y\in V\}$ is continuous from $X$ to $\Bbb R.$
(3). If $(X,d)$ is a  metric space and $K$ is a closed bounded subset of $X$ and $r\in \Bbb R^+$ then $D=  \{z\in X: \exists x\in K\;(d(z,x)\leq r\}$ is closed and bounded.
For the Q, supposing $K\ne \emptyset$ and $A\ne \Bbb R^n$:
Let $f(x)=\inf \{\|x-z\|: z\in \Bbb R^n\setminus A\}$ for $x\in \Bbb R^n.$ Since $A$ is open, $f(x)\ne 0$ for all $ x\in A,$ so $f(x)\ne 0$ for all $x\in K.$
By (1), there exists $x_0\in K$ with $f(x_0)=r=\min \{f(x):x\in K\}.$ And $r\ne 0$ because $x_0\in A.$ So $\forall x\in K \;(f(x)\geq r).$
Let $D= \{z\in \Bbb R^n: \exists x\in K\;(\|z-x\|\leq r/2\}.$  Now $K$ is closed and bounded, so by (3), $D$ is closed and bounded, and in $\Bbb R^n$ this implies compactness.
If $z\in \Bbb R^n \setminus A $ then  for any $x\in K$  we have $\|z-x\|\geq f(x)\geq f(x_0)=r ,$ so $z\not \in D$. Therefore  $D\subset A.$
For the trivial cases: If $K$ is empty or if $A=\Bbb R^n$ let $D=K.$
Remarks: (1) is a corollary of the result that a continuous image of a compact space is compact, so $\{f(x):x\in K\}$ is a non-empty compact sub-space of $\Bbb R,$ so it must have a minimum..... In (3), let $g(z)=\inf \{d(z,x):x\in K\}.$ Since $K$ is closed , we have  $X \setminus D=g^{-1}(r,\infty),$  which is open by (2) so $D$ is closed.


TITLE: Find angle $\theta$ if the following information is known
QUESTION [1 upvotes]: Find $\color{green}\theta$
$\color{blue}{R_b}$ is the radius of the big circle
$R_s$ is the radius of the small circle
$\color{red}{d}$ is known
$x$ is known and is an angle

REPLY [2 votes]: Hint (Sorry for the awful drawing): Parallelity, because otherwise there are infinitely many angles that are possible, then use the law of sines
$$\dfrac{\sin\bar{\theta}}{R_b}=\dfrac{\sin x}{d}$$
to calculate the angle $\bar{\theta}$, then $\theta= 180°-\bar{\theta}$


TITLE: The kernel of a surjective homomorphism $G\rightarrow\mathbb{Z}$ is finitely generated
QUESTION [1 upvotes]: Let $G$ be a finitely generated group. The following is a part 1 of proposition 6.3.10 in Clara Löh's Geometric Group Theory and is left as an exercise, but I can`t figure it out.

Let $G$ be a finitely generated group with polynomial growth that admits a surjective homomorphism $\pi: G \to \mathbb{Z}$. Show that the kernel of $\pi$ is finitely generated.

REPLY [3 votes]: Let $N$ be the kernel of $\pi$ and fix $t\in\pi^{-1}(1)$. Since $G$ is finitely generated and $\mathbf{Z}$ is finitely presented, the kernel is finitely normally generated: there exists a finitely generated subgroup $M$ of $N$ such that $N$ is generated by $\bigcup_{n\in\mathbf{Z}}t^nMt^{-n}$.
If $M$ is both contained in the subgroup generated by $\bigcup_{n\ge 1}t^nMt^{-n}$ and in the one generated by $\bigcup_{n\le -1}t^nMt^{-n}$, then for some $N$ it is both contained in the subgroup generated by $\bigcup_{1\le n\le N}tMt^{-1}$ and in the one generated by $\bigcup_{-N\le n\le -1}t^nMt^{-n}$, and it is easy to deduce that $N$ is finitely generated.
Otherwise, up to change $t$ to $t^{-1}$, $M$ is not contained in the subgroup $N_1=\bigcup_{n\ge 1}t^nMt^{-n}$. Hence, denoting $N_k$ as the subgroup generated by $\bigcup_{n\ge k}t^nMt^{-n}$, we have $N_1=tN_0t^{-1}$ strictly contained in $N_0$, and $N=\bigcup_{k\in\mathbf{Z}} N_k$, so $G$ is a strictly ascending HNN-extension of $N_0$. In particular, by standard arguments, $G$ has exponential growth and contains a free subsemigroup on 2 generators.


TITLE: Why can integrals be written as $I=\int \phi(x) \epsilon$?
QUESTION [1 upvotes]: Carroll's book Spacetime and Geometry gives this new way to think about integrals:
$$I=\int \phi (x) \epsilon\tag{2.98}$$
$\epsilon$ is the Levi - Civita tensor (2.96). I don't see how the RHS equals the integral of $\phi(x)$ though. If we removed the $\int$, then $\phi(x) \epsilon$ remains, which is just a tensor field. Why does it make sense to put a $\int$ before a tensor field? I don't think $\int$ is defined to operate on tensor fields.

REPLY [8 votes]: The Direct Answer.
Here's the short and direct answer. On an oriented $n$-dimensional smooth manifold $M$, we can define the integrals of $n$-forms. So, if $\omega$ is an $n$-form on $M$, then we can define the number $\int_M\omega\in\Bbb{R}$, called the integral of the differential $n$-form over $M$ (ok I'm glossing over some integrablity conditions for the sake of brevity). So, the 'things' we are supposed to integrate on $n$-dimensional oriented smooth manifolds are  differential $n$-forms. You write

which is just a tensor field. Why does it make sense to put a $\int$ before a tensor field? I don't think $\int$ is defined to operate on tensor fields.

and yes, $\phi\cdot\epsilon$ is a tensor field, but it's not some random tensor field; it is an alternating $(0,n)$ tensor field (i.e an $n$-form), and this is what allows us to define its integral.

Brief definition of Integrals of $n$-forms.
The $n$-form $\epsilon$ is what a mathematician would call a volume form (and in this case it's the volume form induced by the Lorentzian metric, and it defines the orientation). So, if you have a function $\phi:M\to\Bbb{R}$, then $\phi\cdot\epsilon$ is again an $n$-form, and this is the 'correct' thing to be integrating on $M$ (because it is oriented).
The topic of integration on manifolds is a basic part of differential geometry, and is covered in any good textbook (e.g Spivak, Lee). The definition of the integral $\int_M\omega$ is slightly subtle, and the details are covered in these textbooks. The key thing is that if you have a coordinate chart $(U,\alpha=(x^1,\dots, x^n))$ which is positively oriented, then you can write $\omega=f\,dx^1\dots\wedge x^n$ for some unique function $f:U\to\Bbb{R}$, and we have
\begin{align}
\int_U\omega&:=\int_{\alpha[U]}(f\circ\alpha^{-1})\cdot d\lambda_n\equiv \int_{\alpha[U]}f\circ\alpha^{-1},
\end{align}
where the symbols on the right are the standard multivariable calculus integrals of functions defined on subsets of $\Bbb{R}^n$ (i.e the usual integral with respect to $n$-dimensional Lebesgue measure... or you can just use Riemann integrals throughout). In words, this is saying the most obvious thing you can do: to integrate an $n$-form in a coordinate patch, write the $n$-form as a function times a wedge of $dx^i$'s and then integrate that chart-dependent function over the corresponding subset of $\Bbb{R}^n$. To then define $\int_M\omega$ from these various $\int_U\omega$, one uses what's called a partition of unity, and the purpose of me emphasizing orientation is so that in this global step, we choose the correct signs... i.e we have to put humpty dumpty back together correctly.

More Context, and Reconciling Various Ideas.
Perhaps the following answer of mine will be helpful in some details. In reality, the correct things to integrate on an arbitrary smooth manifold are scalar densities. However, if your manifold is oriented, then you can construct an isomorphism from the vector space of scalar densities to the vector space of $n$-forms. So, one then, by 'transport of structure', define integrals for $n$-forms.
I think what you're most confused by is that we're integrating complicated types of objects (scalar densities/$n$-forms) as opposed to smooth functions. But this shouldn't be surprising at all! When doing integrals, we don't just integrate functions $f$. We integrate functions with respect to some measure $\mu$ (in $\Bbb{R}$, this is the integral with respect to $1$-dimensional Lebesgue measure, which for continuous functions is equal to the basic vanilla Riemann integral we all know and love). In $\Bbb{R}^n$, we integrate functions $f$ with respect to $n$-dimensional Lebesgue measure: $\int_{\Bbb{R}^n}f\,d\lambda_n$. Or more generally on an abstract measure space $(X,\mathfrak{M},\mu)$, we integrate functions $f:X\to\Bbb{R}$ with respect to the measure $\mu$, to get the number $\int_Xf\,d\mu$. The question now becomes: on an abstract manifold $M$, what is the measure? Is there a natural measure? In the general case the answer is no, but for GR, the answer is yes! The metric tensor $g$ gives rise to a unique measure $\mu_g$, also denoted $\lambda_g$ or $dV_g$, which a mathematician might call the Riemann-Lebesgue volume measure on $M$ (see here for details of the definition and construction). So, we can now talk about integrating functions on smooth manifolds $\int_Mf\,dV_g$.
So, the point is that there are many ways to develop the subject of integration on manifolds:

If you're more analysis-minded, you might go the Lebesgue route and ask 'what measure should I use'? In the case of Riemannian/Lorentzian geometry, you'd end up using the measure $dV_g$. In classical mechanics (particularly the Hamiltonian formulation on phase space, you'd use the measure $dV_{\Omega}$ induced by the symplectic form). Once you have the measure, you can start integrating functions with respect to this measure.
If you were more differential-geometry minded, you'd say 'hmmm...what are the objects I should be integrating? Well, a manifold comes with a bunch of local coordinate charts, so maybe I can use this local information to somehow put together a global definition?' Pursuing this line of thought leads you to the notion of integrating scalar densities.
Going one step further, you'd say scalar densities aren't so nice to work with. Also, from single variable calculus, $f(t)\,dt$ can be thought of as a $1$-form, and differential $n$-forms capture the notion of signed-volumes, and differential forms are very easy to compute with. Can I somehow make a definition for integrating $n$-forms? Pursuing this line of thought, you'd see that if you impose the orientedness condition on $M$, then several pesky minus signs disappear and you now have a notion of integration for $n$-forms (particularly you can now integrate the $n$-form $\phi\cdot\epsilon$).

Finally, you can check that regardless of which route you take, all these approaches give you the same answer given correct hypotheses (oriented $n$-dimensional pseudo-Riemannian manifold). So, the fact that you have to end up considering more complicated objects ($n$-forms) is not really an issue, and is in fact a strength in disguise (they're very computationally flexible, and we have the famous Stokes' theorem).


TITLE: Maximal Interval of Existence Explanation.
QUESTION [1 upvotes]: $\frac{dx}{dt} =x^2$ with $x(0)=x_0$.
$x(t) = \frac{1} {x_0^{−1} −t}$ satisfies the differential equation above (provided that $x_0\neq 0$). Clearly when $t = 0$, we have $x(0) = x_0$.
Our solution has some interesting properties. If $x_0 > 0$ then the denominator is initially positive (at $t = 0$), but decreases as $t$ increases until it reaches zero at time $t = x^{−1}_0$ . This means that the solution, $x(t)$, has become infinite by the time $t = x^{−1}_ 0$; we say that it ‘blows up’ in a finite time.
Things are much nicer, though, if we want to see where our solution came from in the past. We can decrease $t$ (from zero) as much as we like, since as $t$ decreases the denominator becomes larger, and so the solution itself tends to zero as $t →−∞$. So when $x_0 >0$, we can define the solution of the differential equation on the interval $(−∞,x_0^{-1})$, but there is no way to define the solution on an interval that extends further into the future beyond the time $t = x^{−1}_ 0$ . We refer to $(−∞,x_0^{−1})$ maximal interval of existence for the differential equation.
I think this statement is completely wrong "but there is no way to define the solution on an interval that extends further into the future beyond the time $t = x^{−1}_ 0$". I think the solution is also defined even after $t = x^{−1}_ 0$ and here only $t = x^{−1}_ 0$ is discontinuity and it undefined at this point we can still get solution beyond $t = x^{−1}_ 0$.I have graphed the solution using positive $x_0$ and found that after $x^{-1}_0$ there are negative solution. So how am I wrong guys? I don't think book is right.We can go beyond $x_0^{-1}$ without touching $x_0^{-1}$.

REPLY [2 votes]: Look up your definition of solutions of a differential equation.
It will be something like: a differentiable function $x(t):I\to \mathbb{R}$ where $I$ is an interval, satisying the differential equation at each point $t\in I$.
In your case, if you extend the solution past $t=x_0^{-1}$ then either the function will not be differentiable everywhere (it's not in $t=x_0^{-1}$ as it blows up) or the domain of definition is not an interval. Thus it doesn't satisfy the definition.
The reason why we don't like solutions as being defined on disconnected domains is that you lose uniqueness unless you specify new conditions on each disconnected component.
For instance in your case, suppose you want to extend your solution $x(t)=\frac{1}{x_0^{-1}-t}$, defined for $t<x_0^{-1}$, past $t=x_0^{-1}$.
Then you get an infinite family of possible extensions, in the form $x_c(t)=\frac{1}{c-t}$  (for $t>x_0^{-1}$) where $c\in \mathbb{R}$ is completely arbitrary. They don't care at all about what happens before $t=x_0^{-1}$. Unless you specify $x(t)$ for some $t>x_0^{-1}$, you can't narrow it down to a single solution, which is what we expect with IVPs.


TITLE: Finding the determinant of an $n\times n$ matrix.... and the inverse
QUESTION [1 upvotes]: Finding the determinant of a $2\times 2$ matrix is easy and the inverse is even easier. Finding the determinant of a $3\times 3$ matrix and its inverse is a little more difficult but still doable. $4\times 4$ I can use RREF of an augmented identity and deduce the determinant from that. But as the size of the square matrix grows, finding these two becomes a serious PITA. What I was wondering is whether or not there is a  more efficient method to find the determinant of an arbitrarily large $n\times n$ matrix by hand..... just because.

REPLY [1 votes]: The "brute force" method would require $n!$ operations, but if you bring the matrix into upper triangular form by row operations and then read the determinant from th eproduct of the diagonal entries, it only take $O(n^3)$ operations.


TITLE: Hamiltonian System in economics
QUESTION [0 upvotes]: I was wondering if you know about any simple example of a Hamiltonian System that arises in economics or finances. I have been looking on the web, but I could only find one (with variables I didn't really understand). Maybe there is a simpler one (like the hamiltonian for the pendulum). Thank you.

REPLY [1 votes]: There is a collected volume titled The Hamiltonian Approach to Dynamic Economics, edited by David Cass and Karl Shell, published in 1976 by Academic Press.
It collects eight essays originally appeared on the Journal of Economic Theory, vol. 12, n. 1, 1976.


TITLE: Does the power rule come from the generalised binomial theorem or the other way around?
QUESTION [1 upvotes]: Remember how we say $lim \frac{\sin{x}}{x}=1$ doesn't come from the L Hospital's rule, because the differentiation of $sin x$ from first principles uses the fact that $lim \frac{\sin{x}}{x}=1$?
A similar situation is arising for the power rule and the generalised binomial theorem.
The generalised binomial theorem can be proven as the Taylor expansion of $(1+x)^r$. This derivation uses the power rule, as we need the power rule to calculate derivatives of $(1+x)^r$
The derivation of the power rule from first principles would need the generalised binomial theorem, as we'd need to expand the term $(x+h)^r$
So which of these two results is more fundamental? Which one comes from the other and why?

REPLY [0 votes]: The power rule comes from the generalised binomial theorem.
Generalised binomial theorem: ($1665$)
We find in section 1.2 Exponentials and the Binomial Theorem of Analysis by Its History by E. Hairer and G. Wanner a citation from Newton

[Newton:] All this was in the two plague years of $1665$ and $1666$, for in those days I was in the prime of my age for invention, and minded mathematics and philosophy more than at any other time since.
followed by
One of Newton's ideas of these anni mirabiles, inspired by the work of Wallis was to try to interpolate the polynomials $(1+x)^0, (1+x)^1, (1+x)^2,\ldots$ in order to obtain a series for $(1+x)^{\alpha}$ where $\alpha$ is some rational number. This means that we must interpolate the coefficients given in
\begin{align*}
 (1+x)^{n}=1+\frac{n}{1}x+\frac{n(n-1)}{1\cdot 2}x^2
 +\frac{n(n-1)(n-2)}{1\cdot2\cdot3}x^3+\cdots\qquad n=0,1,2,3,\ldots
\end{align*}
Since the latter are polynomials in $n$, it is clear that the result is given by the same expression with $n$ replaced by $a$. We therefore arrive at the general theorem.
Theorem: (Generalized binomial theorem of Newton). For any rational $a$ we have for $|x|<1$
\begin{align*}
 \color{blue}{(1+x)^{a}=1+\frac{a}{1}x+\frac{a(a-1)}{1\cdot 2}x^2
 +\frac{a(a-1)(a-2)}{1\cdot2\cdot3}x^3+\cdots}
 \end{align*}

From todays point of view this argumentation is (of course) not rigorous enough. The authors continue:

Even Newton found that his interpolation argument was dangerous. Euler, in his Introductio ($1748$), stated the general theorem without any further proof or comment. Only Abel, a century later, felt the need for a rigorous proof.

Power rule: ($1670/1671$)

Newton derived the power rule in his Method of Fluxions which was written in $1670/1671$ and published in $1736$.
In a later treatise (german title: Abhandlung über die Quadratur der Kurven) from $1704$ he gave a more mature presentation of the Method of Fluxions where he also explicitly refers to the method of infinite series, i.e. to the generalised binomial theorem to derive the power rule.

A more detailed treatment of this treatise can be found e.g. in Grundlagen der Mathematik by Oskar Becker.
Side note: Limits (ca. $1850$)

The usage of a limit as we use it today was not a theme at that time. It needed the rigor of Weierstrass and his school around $1850$.
D. Hilbert wrote in Über das Unendliche ($1926$):
It is essentially a merit of the scientific activity of Weierstrass that there exists at present in analysis full agreement and certainty concerning the course of such types of reasoning which are based on the concept of irrational number and of limit in general.
We owe it to him that there is unanimity on all results in the most complicated questions concerning the theory of differential and integral equations, despite the most daring and diversified combinations with application of super-, juxta-, and transposition of limits.


TITLE: Need help with a step from exponential functions using graphs.
QUESTION [0 upvotes]: I have a question about a Khan Academy problem for determining an exponential function formula based on information from a graph.
The problem: $h(x)= a \times r^x$
We are given two points. (-2, 4.5) and (0,8)
I'm having a problem with one of the steps in the hints section. It's not all clear to me what they did to get the value. I'll explain:
Firstly we know that the 8 in (0,8) is going to be our "a" in the formula.
The problem I'm having is when we plug the next set of points into the equation.
$4.5 = 8 \times r^{-2}$
$\frac{9}{2} = \frac{8}{r^2} $
I understand how the right side gets converted to $8/r^2$ but I don't understand why 4.5 gets multiplied by 2 when all we did on the right side was convert $r^-2$ to $1/r^2$ and multiply by 8. There's no details associated with the any of the 4 steps that they work out and I've been racking my brain trying to figure out a solution with no luck. Any help is greatly appreciated.

REPLY [1 votes]: As $4.5 = \frac{9}{2}$
So,
$\frac{9}{2} = \frac{8}{r^2}$
Cross multiply terms.
$9 * r^2 = 8 * 2$
$9 * r^2 = 16$
$r^2 = \frac{16}{9}$
$r = \sqrt{\frac{16}{9}}$
$r = \frac{4}{3}$


text:
\begin{document}

\title{Closure properties of solutions to heat inequalities}

\author{Jonathan Bennett}
\author{Neal Bez}
\thanks{Both authors were supported by EPSRC
grant EP/E022340/1. \\ \emph{Acknowledgement}. The first author
would like to thank Tony Carbery and Terence Tao for many
interesting and useful conversations on the context
and perspective of this work. We would also like to thank Dirk Hundertmark for his
contribution to the early stages of this work.}

\address{Jonathan Bennett and Neal Bez\\ School of Mathematics \\
The University of Birmingham \\
The Watson Building \\
Edgbaston \\
Birmingham \\
B15 2TT \\
United Kingdom}\email{J.Bennett@bham.ac.uk\\ N.Bez@bham.ac.uk}

\subjclass[2000]{44A35, 35K99, 52A40}

\date{28th of May 2008}

\begin{abstract} We prove that if $u_1,u_2 : (0,\infty) \times \R^d
\rightarrow (0,\infty)$ are sufficiently well-behaved solutions to
certain heat inequalities on $\R^d$ then the function $u: (0,\infty) \times \R^d
\rightarrow (0,\infty)$ given by $u^{1/p}=u_1^{1/p_1}
* u_2^{1/p_2}$ also satisfies a heat inequality of a similar type
provided $\tfrac{1}{p_1} + \tfrac{1}{p_2} = 1 + \tfrac{1}{p}$. On
iterating, this result leads to an analogous statement concerning
$n$-fold convolutions. As a corollary, we give a direct heat-flow
proof of the sharp $n$-fold Young convolution inequality and its
reverse form.
\end{abstract}

\maketitle

\section{Introduction}
It is known that if $d\in\mathbb{N}$ and
$u_1,u_2:(0,\infty)\times\mathbb{R}^d\rightarrow (0,\infty)$ satisfy
the heat inequality
\begin{equation}\label{first}
\partial_t u\geq\frac{1}{4\pi}\Delta
u,\end{equation} then any geometric mean of $u_1$ and $u_2$
satisfies the same heat inequality; i.e. for $1\leq
p_1,p_2\leq\infty$ satisfying $\tfrac{1}{p_1}+\tfrac{1}{p_2}=1$, the
function $$u:=u_1^{1/p_1}u_2^{1/p_2}$$ also satisfies the heat inequality
\eqref{first}. As a corollary to this, provided that $u_1$ and $u_2$
are sufficiently well-behaved, by the divergence theorem it follows
that the quantity
$$ Q(t):=\int_{\mathbb{R}^d}u_1(t,x)^{1/p_1}
u_2(t,x)^{1/p_2}dx$$ is nondecreasing for all $t>0$. Furthermore, on
insisting that, for $j=1,2$, $u_j$ satisfies \eqref{first} with
equality and sufficiently well-behaved initial data $f_j^{p_j}$, it
follows from this monotonicity that
$$
\int_{\mathbb{R}^d}f_1(x)f_2(x)dx=\lim_{t\rightarrow 0}Q(t)\leq
\lim_{t\rightarrow\infty}Q(t)=\|f_1\|_{p_1}\|f_2\|_{p_2};
$$
that is, we recover the classical H\"older inequality.

This closure property of solutions to heat inequalities may be
generalised considerably. Let $m,d\in\mathbb{N}$,
$d_1,\hdots,d_m\in\mathbb{N}$, $1\leq p_1,\hdots,p_m\leq\infty$ and
for each $1\leq j\leq m$ let
$B_j:\mathbb{R}^d\rightarrow\mathbb{R}^{d_j}$ be such that
$B_j^*B_j$ is a projection and
$$
\sum_{j=1}^m\tfrac{1}{p_j}B_j^*B_j=I_d,$$ where $I_d$ denotes the
identity on $\mathbb{R}^d$. If $u_j:(0,\infty)\times
\mathbb{R}^{d_j}\rightarrow (0,\infty)$ satisfies the heat
inequality \eqref{first} for each $j$, then one may show that the
same is true of the ``geometric mean"
$$u(t,x):=\prod_{j=1}^mu_j(t,B_jx)^{1/p_j}.$$ (Here $\Delta$
acts in the number of variables dictated by context.) Very much as
before, an immediate consequence of this is that $\int u(t,\cdot)$
is nondecreasing for all $t>0$, and that from this monotonicity one
may deduce the inequality
$$
\int_{\mathbb{R}^d}\prod_{j=1}^mf_j(B_jx)dx\leq
\prod_{j=1}^m\|f_j\|_{L^{p_j}(\mathbb{R}^{d_j})}.$$ This is the
celebrated geometric Brascamp--Lieb inequality due to Ball
\cite{Ball} for rank-one projections and Barthe \cite{Barthetrans2}
in the general rank case. Such a heat-flow approach to proving
inequalities, by its nature, generates sharp constants and
guarantees the existence of centred gaussian extremisers. All of
these observations were first made by Carlen, Lieb and Loss
\cite{CLL} for rank-one projections and by Bennett, Carbery, Christ
and Tao \cite{BCCT} in the general rank case. Recently, Barthe and
Huet \cite{BartheHuet} have given a different heat-flow proof of the
geometric Brascamp--Lieb inequality and, moreover, the same line of
argument also led them to a heat-flow proof of Barthe's reverse form
of the geometric Brascamp--Lieb inequality. See \cite{Barthetrans1},
\cite{Barthetrans2} for a statement of this reverse inequality. The
reader is referred to \cite{Barthe2} and the references therein for
further discussion of heat-flow methods in the context of such geometric
inequalities.

Aside from the geometric means above, and the trivial operation of ordinary addition, it is not difficult to verify that harmonic addition also preserves the set of solutions of \eqref{first}; i.e. if $u_1,u_2:(0,\infty)\times\mathbb{R}^d\rightarrow (0,\infty)$ satisfy \eqref{first}, then the function
$u$ given by
$$
\frac{1}{u}=\frac{1}{u_1}+\frac{1}{u_2}$$
also satisfies \eqref{first}. By the divergence theorem, this closure property is easily seen to imply a variant of the triangle inequality for harmonic addition.

All of the above closure properties involve pointwise operations.
The main purpose of this article is to establish closure properties
under rather different operations involving convolution. As a
consequence we provide heat-flow proofs of sharp convolution
inequalities that do not proceed via duality.

\subsection{Main Results}

Let $d \in \N$ and suppose $0<p_1,p_2,p<\infty$ satisfy
\begin{equation} \label{e:scaling}
  \frac{1}{p_1} + \frac{1}{p_2} = 1 + \frac{1}{p}.
\end{equation}

Let $0\leq \sigma_1,\sigma_2<\infty$ satisfy
\begin{equation} \label{e:BL}
  \frac{1}{p_1}\left(1-\frac{1}{p_1}\right)\sigma_2 =
  \frac{1}{p_2}\left(1-\frac{1}{p_2}\right)\sigma_1.
\end{equation}

The main contribution in this paper is captured by the following. We
clarify that the operation $*$ will denote spatial convolution.

\begin{theorem} \label{t:closure} For $j=1,2$ suppose that $u_j:(0,\infty) \times
\mathbb{R}^d \rightarrow (0,\infty)$ is such that
$u_j(t,\cdot)^{1/p_j}$, $\partial_t(u_j(t,\cdot)^{1/p_j})$,
$\nabla(u_j(t,\cdot)^{1/p_j})$, $u_j(t,\cdot)^{1/p_j}|\nabla \log
u_j(t,\cdot)|^2$ and $\Delta(u_j(t,\cdot)^{1/p_j})$ are rapidly
decreasing in space locally uniformly in time $t>0$. Let $u :
(0,\infty) \times \R^d \rightarrow (0,\infty)$ be given by
$$ u^{1/p} := u_1^{1/p_1} * u_2^{1/p_2} $$
and let $\sigma := (\sigma_1p_1 + \sigma_2p_2)/r$. Then
$u(t,\cdot)^{1/p}$, $\partial_t(u(t,\cdot)^{1/p})$,
$\nabla(u(t,\cdot)^{1/p})$, $u(t,\cdot)^{1/p}|\nabla \log u(t,\cdot)
|^2$ and $\Delta(u(t,\cdot)^{1/p})$ are also rapidly decreasing in
space locally uniformly in time $t>0$. Furthermore,
\begin{enumerate}
  \item if $p_j \geq 1$ and
  \begin{equation*}
    \partial_t u_j \geq \frac{\sigma_j}{4\pi}
    \Delta u_j,
  \end{equation*}
  for $j=1,2$, then
  \begin{equation} \label{e:heatinuj}
    \partial_t u \geq
    \frac{\sigma}{4 \pi}\Delta u;
  \end{equation}
\item if $p_j\leq 1$ and
  \begin{equation*}
    \partial_t u_j \leq \frac{\sigma_j}{4\pi} \Delta
    u_j,
  \end{equation*}
  for $j=1,2$, then
\begin{equation} \label{e:heatinu}
    \partial_t u \leq
    \frac{\sigma}{4 \pi}\Delta u.
  \end{equation}
\end{enumerate}
\end{theorem}
An important feature of this closure property is that the
(technical) regularity ingredients are all satisfied when $u_1$ and
$u_2$ are solutions to heat \emph{equations} with sufficiently
well-behaved initial data. Indeed, we shall see that Theorem
\ref{t:closure} implies the following.
\begin{corollary} \label{c:Ymon}
For $j=1,2$ let $u_j$ satisfy the heat equation
\begin{equation*}
    \partial_t u_j = \frac{\sigma_j}{4\pi} \Delta u_j
\end{equation*}
with initial data a compactly supported positive finite Borel
measure. Let $Q : (0,\infty) \rightarrow (0,\infty)$ be given by
\begin{equation*}
  Q(t) := \|u_1(t,\cdot)^{1/p_1} * u_2(t,\cdot)^{1/p_2}\|_{L^p(\R^d)}.
\end{equation*}
If $p_1,p_2 \geq 1$ then $Q(t)$ is nondecreasing for each $t > 0$
and if $p_1,p_2 \leq 1$ then $Q(t)$ is nonincreasing for each $t >
0$.
\end{corollary}
The proofs of Theorem \ref{t:closure} and Corollary \ref{c:Ymon}
appear in Section \ref{section:closure1}.

\begin{remark} \label{remarks:intro} \emph{Under the hypotheses of Corollary
\ref{c:Ymon}, it follows from our proof in Section
\ref{section:closure1} that
\begin{align*}
  Q'(t) = &\frac{\varepsilon}{8\pi Q(t)^{p-1}} \int_{\R^d} \int_{\R^d}
  \int_{\R^d}
    (u_1(t,\cdot)^{1/p_1} * u_2(t,\cdot)^{1/p_2})(x)^{p-2}
  u_1(t,x-y)^{1/p_1}u_2(t,y)^{1/p_2} \times \\ &
  u_1(t,x-z)^{1/p_1}u_2(t,z)^{1/p_2}
  \left|\left(\tfrac{\sigma_1}{p_1}|\tfrac{1}{p_1}-1|\right)^{1/2}\tfrac{\nabla u_1}{u_1}(t,x-y)
+
  \left(\tfrac{\sigma_2}{p_2}|\tfrac{1}{p_2}-1|\right)^{1/2}\tfrac{\nabla u_2}{u_2}(t,y) \right.\\ & \left. \quad   -
  \left(\tfrac{\sigma_1}{p_1}|\tfrac{1}{p_1}-1|\right)^{1/2}\tfrac{\nabla u_1}{u_1}(t,x-z) -
  \left(\tfrac{\sigma_2}{p_2}|\tfrac{1}{p_2}-1|\right)^{1/2}\tfrac{\nabla u_2}{u_2}(t,z)
  \right|^2 \,dxdydz
\end{align*}
for each $t > 0$. Here $\varepsilon$ is defined to be 1 if $p_1,p_2
\geq 1$ and $-1$ if $p_1,p_2 \leq 1$. Consequently, if exactly one
of $p_1$ and $p_2$ is equal to 1 then the corresponding monotonicity
in Corollary \ref{c:Ymon} is strict. In this case and if $p_j$ is
equal to 1, then it is amusing to note that $\sigma_j$ is zero by
\eqref{e:BL}; that is, the heat-flow $u_j$ is constant in time.}
\end{remark}

We now describe the sharp convolution inequalities that follow from
these results. Recall that the sharp form of the Young convolution
inequality on $\R^d$ states that if $p_1,p_2\geq 1$ and
$\tfrac{1}{p_1} + \tfrac{1}{p_2} = 1 + \tfrac{1}{p}$ then
\begin{align} \label{e:sharpYoung}
    \|f_1 * f_2\|_{L^p(\R^d)} \leq \left(\frac{C_{p_1}C_{p_2}}{C_{p}}\right)^d
    \|f_1\|_{L^{p_1}(\R^d)}\|f_2\|_{L^{p_2}(\R^d)}
\end{align}
for any nonnegative functions $f_j$ in $L^{p_j}(\R^d)$, where $C_r
:= \big(r^{1/r}/r'^{1/r'}\big)^{1/2}$. The sharp constant in
\eqref{e:sharpYoung} is due to Beckner \cite{Beckner2},
\cite{Beckner} and Brascamp and Lieb \cite{BL}. The sharp reverse
form of \eqref{e:sharpYoung} states that if $p_1,p_2 \leq 1$ and
$\tfrac{1}{p_1} + \tfrac{1}{p_2} = 1 + \tfrac{1}{p}$ then
 \begin{align} \label{e:reverseYoung}
    \|f_1 * f_2\|_{L^p(\R^d)} \geq \left(\frac{C_{p_1}C_{p_2}}{C_{p}}\right)^d
    \|f_1\|_{L^{p_1}(\R^d)}\|f_2\|_{L^{p_2}(\R^d)}
  \end{align}
for any nonnegative functions $f_j$ in $L^{p_j}(\R^d)$. Leindler
\cite{Leindler} proved \eqref{e:reverseYoung} with a nonsharp
constant and Brascamp and Lieb found the sharp constant in \cite{BL}
(see also Barthe's simpler argument in \cite{Barthe} which proves
both forms with sharp constants). It is easy to see that from
Corollary \ref{c:Ymon} one may recover both \eqref{e:sharpYoung} and
\eqref{e:reverseYoung}. To see this, let $0<p_1,p_2,p<\infty$
satisfy \eqref{e:scaling} and note that it suffices to verify both
inequalities when the $f_j^{p_j}$ are bounded, integrable and
compactly supported functions. For $j=1,2$ let $u_j$ satisfy the
heat equation
\begin{equation} \label{e:heateqn}
    \partial_t u_j = \frac{\sigma_j}{4\pi}\Delta u_j
\end{equation}
with initial data $u_j(0,x) := f_j(x)^{p_j}$. By the dominated
convergence theorem, one can show that,
  \begin{equation*}
    \lim_{t \rightarrow 0} Q(t) = \|f_1 \ast f_2 \|_{L^p(\R^d)}
  \end{equation*}
and, combined with a simple change of variables,
\begin{equation*}
    \lim_{t \rightarrow \infty} Q(t) = \big\|H_{\sigma_1}^{1/p_1} * H_{\sigma_2}^{1/p_2} \big\|_{L^p(\R^d)}
    \|f_1\|_{L^{p_1}(\R^d)}\|f_2\|_{L^{p_2}(\R^d)}.
\end{equation*}
  Here,
  \begin{equation*}
  H_t(x) := (1/t)^{d/2}e^{-\pi|x|^2/t}
  \end{equation*}
  is the appropriate heat kernel at time $t$ and $\sigma_1, \sigma_2$ satisfy \eqref{e:BL}. A direct computation shows that
  \begin{equation*}
   \big\|H_{\sigma_1}^{1/p_1} * H_{\sigma_2}^{1/p_2} \big\|_{L^p(\R^d)} =
   \left(\frac{C_{p_1}C_{p_2}}{C_{p}}\right)^d
  \end{equation*}
  and hence, Corollary \ref{c:Ymon} immediately implies both
  \eqref{e:sharpYoung} and \eqref{e:reverseYoung}. We also remark that if the initial data for $u_1$ and $u_2$
  are extremal then $Q(t)$ is constant in time. It is possible to
  recover the complete characterisation of the extremals
in the Young convolution inequality and its reverse form from the
expression for $Q'(t)$ in Remark \ref{remarks:intro}. We omit the
details of this.

  An alternative perspective on the sharp Young convolution inequality on $\R^d$ is
  to consider the following dual formulation. Suppose $1\leq p_1,p_2,p_3 <\infty$
  satisfy $\tfrac{1}{p_1} + \tfrac{1}{p_2} + \tfrac{1}{p_3} = 2$. The inequality
  \begin{equation} \label{e:dual}
    \int_{\R^d}\int_{\R^d}
    f_1(x)^{1/p_1}f_2(y)^{1/p_2}f_3(x-y)^{1/p_3}dxdy \leq
    \prod_{j'=1}^3 C_{p_{j'}}^d \prod_{j=1}^3
    \|f_{j}\|_{L^1(\R^d)}^{1/p_j}
  \end{equation}
  for all nonnegative integrable functions $f_j$ is equivalent to the Young convolution
  inequality in \eqref{e:sharpYoung}. It is known that if each $f_j$ evolves under an appropriate
  heat-flow $u_j$ then the quantity
  \begin{equation*}
    \int_{\R^d} \int_{\R^d} u_1(t,x)^{1/p_1}u_2(t,y)^{1/p_2}u_3(t,x-y)^{1/p_3} \,dxdy
  \end{equation*}
  is nondecreasing for each $t > 0$ from which the inequality in
  \eqref{e:dual} follows. This type of dualised heat-flow approach to the Young convolution
inequality on $\R^d$ can be found in \cite{CLL} and \cite{BCCT}.
Carlen, Lieb and Loss have also shown that this type of heat-flow
approach can be used to prove certain analogues of the Young
convolution inequality in other settings. See \cite{CLL} and
\cite{CLL2} for analogues on the euclidean sphere and the
permutation group, respectively (see also \cite{BCM}).

It is also worth noting that in our undualised setup when the
exponent $p$ is a natural number and $1\leq p_1, p_2 <\infty$, by
multiplying out the $p$th power of the integral one may deduce the
monotonicity of $Q$ directly from \cite{BCCT} (see also \cite{CLL}).

\subsection{Iterated convolutions}

Naturally Theorem \ref{t:closure} self-improves to a
result involving iterated convolutions, which we now state. Let
$0<p_1,\ldots,p_n,p<\infty$ satisfy
  \begin{equation} \label{e:scalinggen}
    \sum_{j=1}^n \frac{1}{p_j} = n-1+\frac{1}{p}
  \end{equation}
and let $0 \leq \sigma_1,\ldots,\sigma_n < \infty$ satisfy
\begin{equation*}
  \frac{1}{p_j}\left(1-\frac{1}{p_j}\right)\sigma_k =
  \frac{1}{p_k}\left(1-\frac{1}{p_k}\right)\sigma_j
\end{equation*} for each $j,k=1,\ldots,n$. As before these relations
uniquely define $\sigma_1,\hdots,\sigma_n$ up to a common scale factor.

\begin{corollary} \label{c:closure} For $j=1, \ldots, n$ suppose that $u_j:(0,\infty) \times
\mathbb{R}^d \rightarrow (0,\infty)$ is such that
$u_j(t,\cdot)^{1/p_j}$, $\partial_t(u_j(t,\cdot)^{1/p_j})$,
$\nabla(u_j(t,\cdot)^{1/p_j})$, $u_j(t,\cdot)^{1/p_j}|\nabla \log
u_j(t,\cdot)|^2$ and $\Delta(u_j(t,\cdot)^{1/p_j})$ are rapidly
decreasing in space locally uniformly in time for $t>0$. Let $u :
(0,\infty) \times \R^d \rightarrow (0,\infty)$ be given by
$$ u^{1/p} := u_1^{1/p_1} * \cdots * u_n^{1/p_n} $$
and let $\sigma := \tfrac{1}{r}\sum_{j=1}^n \sigma_jp_j$. Then
$u(t,\cdot)^{1/p}$, $\partial_t(u(t,\cdot)^{1/p})$,
$\nabla(u(t,\cdot)^{1/p})$, $u(t,\cdot)^{1/p}$ $|\nabla \log
u(t,\cdot)|^2$ and $\Delta(u(t,\cdot)^{1/p})$ are also rapidly
decreasing in space locally uniformly in time for $t>0$.
Furthermore,
\begin{enumerate}
  \item if $p_j \geq 1$ and
  \begin{equation*}
    \partial_t u_j \geq \frac{\sigma_j}{4\pi}
    \Delta u_j,
  \end{equation*}
  for $j=1,\ldots,n$, then
  \begin{equation*} \label{e:teapot}
    \partial_t u \geq
    \frac{\sigma}{4 \pi}\Delta u;
  \end{equation*}
\item if $p_j\leq 1$ and
  \begin{equation*}
    \partial_t u_j \leq \frac{\sigma_j}{4\pi} \Delta
    u_j,
  \end{equation*}
  for $j=1,\ldots,n$, then
\begin{equation*} \label{e:heatinequality2}
    \partial_t u \leq
    \frac{\sigma}{4 \pi}\Delta u.
  \end{equation*}
\end{enumerate}
\end{corollary}

It is a simple exercise to verify that Corollary \ref{c:closure}
follows from Theorem \ref{t:closure}.

\begin{corollary} \label{cor:iterated}
For $j=1,\ldots,n$ let $u_j$ satisfy the heat equation
\begin{equation*}
    \partial_t u_j = \frac{\sigma_j}{4\pi} \Delta u_j
\end{equation*}
with initial data a compactly supported positive finite Borel
measure. Let $Q : (0,\infty) \rightarrow (0,\infty)$ be given by
\begin{equation*}
  Q(t) := \|u_1(t,\cdot)^{1/p_1} * \cdots * u_n(t,\cdot)^{1/p_n}\|_{L^p(\R^d)}.
\end{equation*}
If $p_1,\ldots,p_n \geq 1$ then $Q(t)$ is nondecreasing for each $t
> 0$ and if $p_1,\ldots,p_n \leq 1$ then $Q(t)$ is nonincreasing for each $t >
0$.
\end{corollary}
We remark that Corollary \ref{cor:iterated} follows from Corollary
\ref{c:closure} in the same way that Corollary \ref{c:Ymon} follows
from Theorem \ref{t:closure}. As one may expect, from Corollary
\ref{cor:iterated} (and its proof) we recover the sharp $n$-fold
Young convolution inequality, its reverse form and a complete
characterisation of extremals. We omit the details of this.

When $p$ is an even integer, this $n$-fold Young convolution
inequality is of course related to the Hausdorff--Young inequality
via Plancherel's theorem. In particular,
\begin{equation} \label{e:FT}
  \big\| \widehat{u(t,\cdot)^{1/p'}}
    \big\|_{L^{p}(\R^d)} = \|u(t,\cdot)^{1/p'} * \cdots *
    u(t,\cdot)^{1/p'}\|_{L^2(\R^d)}^{2/p}
\end{equation}
where \hspace{0.01cm} $\widehat{\;}$ \hspace{0.01cm} denotes the
Fourier transform and the iterated convolution is $p/2$-fold. By
Corollary \ref{cor:iterated} it follows that the above quantity is
nondecreasing for $t > 0$ if $u$ satisfies the heat equation
$\partial_t u = \tfrac{1}{4\pi}\Delta u$ with initial data a
compactly supported finite positive Borel measure. We remark that
the nondecreasingness of the quantity in \eqref{e:FT} also follows
from \cite{BCCT}. We refer the interested reader to \cite{BBC} for
an explicit verification of how this fact follows from \cite{BCCT}
and for counterexamples to the monotonicity of $$\big\|
\widehat{u(t,\cdot)^{1/p'}} \big\|_{L^{p}(\R^d)}$$ whenever $p$ is
not an even integer.

\subsection{Further results}

We now describe some extensions of our results when the scaling
condition \eqref{e:scaling} (or more generally \eqref{e:scalinggen})
is relaxed. Let $0<\alpha_1, \alpha_2\leq 1$ and $1\leq p <\infty$
be such that
\begin{equation*}
  \alpha_1 + \alpha_2 \geq 1 + \frac{1}{p}.
\end{equation*}
Suppose $0\leq \rho_1, \rho_2 \leq 1$ satisfy
\begin{equation} \label{e:weight}
  \rho_1\alpha_1 + \rho_2\alpha_2 = 1+\frac{1}{p}
\end{equation}
and let $0\leq \sigma_1,\sigma_2<\infty$ satisfy
\begin{equation} \label{e:BL2}
  \alpha_1(1-\rho_1\alpha_1)\sigma_2 =
  \alpha_2(1-\rho_2\alpha_2)\sigma_1.
\end{equation}

\begin{theorem} \label{t:closure2}
  For $j=1,2$ suppose that $u_j:(0,\infty) \times
\mathbb{R}^d \rightarrow (0,\infty)$ is such that
$u_j(t,\cdot)^{\A_j}$, $\partial_t(u_j(t,\cdot)^{\A_j})$,
$\nabla(u_j(t,\cdot)^{\A_j})$, $u_j(t,\cdot)^{\A_j}|\nabla \log
u_j(t,\cdot)|^2$ and $\Delta(u_j(t,\cdot)^{\A_j})$ are rapidly
decreasing in space locally uniformly in time for $t>0$. Let $u :
(0,\infty) \times \R^d \rightarrow (0,\infty)$ be given by
$$ u(t,x)^{1/p} := t^{d(\alpha_1+\alpha_2-1-1/p)/2}(u_1(t,\cdot)^{\A_1} *
    u_2(t,\cdot)^{\A_2})(x) $$
and let $\sigma := (\tfrac{\sigma_1}{\A_1} +
  \tfrac{\sigma_2}{\A_2})/p$. Then
$u(t,\cdot)^{1/p}$, $\partial_t(u(t,\cdot)^{1/p})$,
$\nabla(u(t,\cdot)^{1/p})$, $u(t,\cdot)^{1/p}|\nabla \log
u(t,\cdot)|^2$ and $\Delta(u(t,\cdot)^{1/p})$ are also rapidly
decreasing in space locally uniformly in time for $t>0$.

Furthermore, if
\begin{equation} \label{e:he}
  \partial_t u_j \geq \frac{\sigma_j}{4\pi} \Delta u_j
\end{equation}
  and, for each $t>0$,
  \begin{equation} \label{e:lc}
    \sigma_j\emph{div}\left(\frac{\nabla u_j}{u_j}\right)(t,\cdot) \geq -\frac{2d\pi}{t}
  \end{equation}
for $j=1,2$ then
  \begin{equation} \label{e:heforu}
    \partial_t u \geq \frac{\sigma}{4\pi} \Delta u
  \end{equation}
  and, for each $t>0$,
  \begin{equation} \label{e:lcforu}
    \sigma\emph{div}\left(\frac{\nabla u}{u}\right)(t,\cdot) \geq -\frac{2d\pi}{t}.
  \end{equation}
\end{theorem}

\begin{corollary} \label{c:Ymon2}
For $j=1,2$ let $u_j$ satisfy the heat equation
\begin{equation*}
    \partial_t u_j = \frac{\sigma_j}{4\pi} \Delta u_j
\end{equation*}
with initial data a compactly supported positive finite Borel
measure. Let $Q : (0,\infty) \rightarrow (0,\infty)$ be given by
\begin{equation*}
  Q(t) := t^{d(\alpha_1+\alpha_2-1-1/p)/2}\left\| u_1(t,\cdot)^{\A_1} * u_2(t,\cdot)^{\A_2}
  \right\|_{L^p(\R^d)}.
\end{equation*}
Then $Q(t)$ is nondecreasing for each $t > 0$.
\end{corollary}

We remark that the idea behind this extension of Theorem
\ref{t:closure}(1) lies in \cite{BCCT} and involves a certain
log-convexity property for solutions to heat equations. In
particular, if there is equality in \eqref{e:he} and the initial
data for $u_j$ is some finite positive Borel measure then
\eqref{e:lc} is automatic by Corollary 8.7 of \cite{BCCT}.

Theorem \ref{t:closure2} self-improves to a result involving
iterated convolutions, as was the case with Theorem \ref{t:closure}.
Again, we leave the details of this to the interested reader.

Finally we remark that all of our results also hold in the setting
of the torus. This will be clear from our proofs.

\section{Proof of Theorem \ref{t:closure}, Corollary \ref{c:Ymon} and Theorem \ref{t:closure2}} \label{section:closure1}
An elementary but crucial component of the proof of Theorem
\ref{t:closure} and Theorem \ref{t:closure2} is the following.

\begin{lemma} \label{l:main}
  Let $0 < \A_1,\A_2,\Lambda_1,\Lambda_2 < \infty$. For $j=1,2$ suppose that $u_j:(0,\infty) \times
\mathbb{R}^d \rightarrow (0,\infty)$ is such that
$u_j(t,\cdot)^{\A_j}$, $\partial_t(u_j(t,\cdot)^{\A_j})$,
$\nabla(u_j(t,\cdot)^{\A_j})$, $u_j(t,\cdot)^{\A_j}|\nabla \log
u_j(t,\cdot)|^2$ and $\Delta(u_j(t,\cdot)^{\A_j})$ are rapidly
decreasing in space locally uniformly in time for $t>0$.
   Then
  \begin{align*}
    & \Lambda_1(u_1^{\A_1} * u_2^{\A_2})(u_1^{\A_1}|\tfrac{\nabla u_1}{u_1}|^2 * u_2^{\A_2}) +
\Lambda_2(u_1^{\A_1} * u_2^{\A_2})(u_1^{\A_1} * u_2^{\A_2}|\tfrac{\nabla u_2}{u_2}|^2) \\
&  + 2\Lambda_1^{1/2}\Lambda_2^{1/2}(u_1^{\A_1} *
u_2^{\A_2})(u_1^{\A_1}\tfrac{\nabla u_1}{u_1} *
  u_2^{\A_2}\tfrac{\nabla u_2}{u_2})- (\tfrac{\Lambda_1}{\A_1^2} + \tfrac{\Lambda_2}{\A_2^2} +
2\tfrac{\Lambda_1^{1/2}\Lambda_2^{1/2}}{\A_1\A_2})|\nabla(u_1^{\A_1}
* u_2^{\A_2})|^2
  \end{align*}
  evaluated at $(t,x) \in (0,\infty) \times \R^d$ coincides
  with
  \begin{align*}
    \tfrac{1}{2} \int_{\R^d} & \int_{\R^d} u_1(t,x-y)^{\A_1}u_2(t,y)^{\A_2}
    u_1(t,x-z)^{\A_1}u_2(t,z)^{\A_2} \times \\
  & |\Lambda_1^{1/2}\tfrac{\nabla u_1}{u_1}(t,x-y) + \Lambda_2^{1/2}\tfrac{\nabla u_2}{u_2}(t,y) -
  \Lambda_1^{1/2}\tfrac{\nabla u_1}{u_1}(t,x-z) - \Lambda_2^{1/2}\tfrac{\nabla u_2}{u_2}(t,z)|^2 \,dydz.
  \end{align*}
\end{lemma}

\begin{proof} We remark that each convolution term is well-defined by the
regularity hypotheses on $u_1$ and $u_2$. Moreover, since
$\nabla(u_j^{\A_j}) = \A_j u_j^{\A_j} \tfrac{\nabla u_j}{u_j}$ is
rapidly decreasing in space for $j=1,2$ it follows that
$\nabla(u_1^{\alpha_1}
* u_2^{\alpha_2})$ coincides with $\alpha_1(u_1^{\alpha_1}
\tfrac{\nabla u_1}{u_1} * u_2^{\alpha_2})$ and
$\alpha_2(u_1^{\alpha_1} * u_2^{\alpha_2}\tfrac{\nabla u_2}{u_2})$,
depending on whether one applies the gradient to the left or right
of the convolution. Thus, Lemma \ref{l:main} follows upon expanding
the square in the integrand and collecting like terms.
\end{proof}

\subsection{Proof of Theorem \ref{t:closure}}
We begin by justifying the closure of the technical regularity
ingredients in Theorem \ref{t:closure}. For $j=1,2$ let $v_j$ be the
time dependent vector field on $\R^d$ given by $v_j := \frac{\nabla
u_j}{u_j}$.

Since the convolution of two rapidly decreasing functions on $\R^d$
is also rapidly decreasing, it is straightforward to check that
$u^{1/p}$ is rapidly decreasing locally uniformly in time. For the
time derivative, we note that
\begin{equation} \label{e:timederiv}
  \partial_t(u^{1/p}) = \partial_t(u_1^{1/p_1}) * u_2^{1/p_2} + u_1^{1/p_1} *
  \partial_t(u_2^{1/p_2}),
\end{equation}
where the interchange of differentiation and integration is
justified since $u_j^{1/p_j}$ and $\partial_t(u_j^{1/p_j})$ are
rapidly decreasing in space locally uniformly in time for $j=1,2$.
Hence $\partial_t(u^{1/p})$ is also rapidly decreasing in space
locally uniformly in time. Similarly, it follows that
$\nabla(u^{1/p})$ is rapidly decreasing in space locally uniformly
in time and, moreover, we may write
\begin{equation} \label{e:gradu1}
  \nabla(u^{1/p}) = \nabla(u_1^{1/p_1}) * u_2^{1/p_2} =
  \tfrac{1}{p_1} u_1^{1/p_1}v_1 * u_2^{1/p_2}
\end{equation}
or, by symmetry, $\nabla(u^{1/p}) = \tfrac{1}{p_2} u_1^{1/p_1}
* u_2^{1/p_2}v_2$. Next, observe that
\begin{equation*}
  u^{1/p}|\nabla \log u|^2 = p^2
  \frac{|\nabla(u^{1/p})|^2}{u^{1/p}} = \tfrac{p^2}{p_1^2}\frac{|u_1^{1/p_1}v_1 *
  u_2^{1/p_2}|^2}{u^{1/p}} \leq \tfrac{p^2}{p_1^2} u_1^{1/p_1}|v_1|^2
  * u_2^{1/p_2}
\end{equation*}
by \eqref{e:gradu1} and the Cauchy-Schwarz inequality. Since
$u_1^{1/p_1}|v_1|^2$ and $u_2^{1/p_2}$ are rapidly decreasing in
space locally uniformly in time by assumption, it follows that
$u^{1/p}|\nabla \log u|^2$ is also rapidly decreasing locally
uniformly in time.

Finally, we note that $\Delta(u^{1/p})$ is rapidly decreasing in
space locally uniformly in time since \eqref{e:gradu1} and our
hypotheses on $u_1$ and $u_2$ imply that
\begin{equation*}
  \Delta(u^{1/p}) = \Delta(u_1^{1/p_1}) * u_2^{1/p_2}.
\end{equation*}
This concludes our justification of the closure of the regularity
ingredients in Theorem \ref{t:closure}. It is, however, a convenient
opportunity to note that we may also write
\begin{align} \label{e:Deltau1}
    \Delta(u^{1/p}) = \d(\nabla(u_1^{1/p_1}) * u_2^{1/p_2}) = \left\{\begin{array}{llllll}
    \tfrac{1}{p_1^2}u_1^{1/p_1}|v_1|^2 * u_2^{1/p_2} + \tfrac{1}{p_1}u_1^{1/p_1}\d(v_1) *
    u_2^{1/p_2} \\
    \tfrac{1}{p_1p_2} u_1^{1/p_1}v_1 * u_2^{1/p_2}v_2
    \end{array} \right.
  \end{align}
  depending on whether we apply the divergence to the term on the left or right of the convolution.
Since
 \begin{equation*}
   \Delta(u_1^{1/p_1}) = \tfrac{1}{p_1}\d(u_1^{1/p_1}v_1) =
   \tfrac{1}{p_1^2}u_1^{1/p_1}|v_1|^2 + \tfrac{1}{p_1}u_1^{1/p_1}\d(v_1)
 \end{equation*}
 it follows from the regularity of $u_1$ and $u_2$ that each convolution term in \eqref{e:Deltau1} is
well-defined. By symmetry the expression \eqref{e:Deltau1} also holds
with the subscripts 1 and 2 interchanged.

We now turn to proving Theorem \ref{t:closure}(1) where we have $p_j
\geq 1$ and $\partial_t u_j \geq \frac{\sigma_j}{4\pi} \Delta u_j$
for each $j=1,2$. Then,
  \begin{equation*} \label{e:vj}
    \frac{\partial_tu_j}{u_j} \geq \frac{\sigma_j}{4\pi} \frac{\d(\nabla
    u_j)}{u_j} = \frac{\sigma_j}{4\pi} (|v_j|^2 + \d(v_j))
  \end{equation*}
  and therefore, by \eqref{e:timederiv},
  \begin{align*}
    4\pi \frac{\partial_t u}{u^{(p-2)/p}}
   & \geq pu^{1/p}\left(\tfrac{\sigma_1}{p_1}u_1^{1/p_1}|v_1|^2
   * u_2^{1/p_2} + \tfrac{\sigma_2}{p_2}u_1^{1/p_1} * u_2^{1/p_2}|v_2|^2 \right. \\
  & \qquad \qquad \qquad + \left.\tfrac{\sigma_1}{p_1}u_1^{1/p_1}\d(v_1) * u_2^{1/p_2} +
  \tfrac{\sigma_2}{p_2} u_1^{1/p_1} * u_2^{1/p_2}\d(v_2) \right).
 \end{align*}
Furthermore,
\begin{align*}
   \frac{\Delta u}{u^{(p-2)/p}}
   = p(p-1)|\nabla(u^{1/p})|^2 + pu^{1/p}\Delta(u^{1/p})
  \end{align*}
  and therefore,
  \begin{align*}
    -\frac{\sigma_1p_1}{p}\frac{\Delta u}{u^{(p-2)/p}} = -(p-1)\sigma_1p_1|\nabla(u^{1/p})|^2
    + \sigma_1(p-p_1)u^{1/p}\Delta(u^{1/p}) - p \sigma_1u^{1/p}\Delta(u^{1/p}).
  \end{align*}
   Hence, by \eqref{e:Deltau1},
  \begin{align*}
   -\frac{\sigma_1p_1}{p}\frac{\Delta u}{u^{(p-2)/p}} &
   = -(p-1)\sigma_1p_1|\nabla(u^{1/p})|^2 + \tfrac{\sigma_1(p-p_1)}{p_1p_2}u^{1/p} (u_1^{1/p_1}v_1 *
    u_2^{1/p_2}v_2) \\
   & \quad - \tfrac{p\sigma_1}{p_1^2}u^{1/p} (u_1^{1/p_1}|v_1|^2 * u_2^{1/p_2})
   - \tfrac{p\sigma_1}{p_1}u^{1/p} (u_1^{1/p_1}\d(v_1) *
   u_2^{1/p_2}).
  \end{align*}
  By symmetry, it follows that
  \begin{align*}
   -\frac{\sigma_2p_2}{p}\frac{\Delta u}{u^{(p-2)/p}} &
   = -(p-1)\sigma_2p_2|\nabla(u^{1/p})|^2 + \tfrac{\sigma_1(p-p_2)}{p_1p_2}u^{1/p} (u_1^{1/p_1}v_1 *
    u_2^{1/p_2}v_2) \\
   & \quad - \tfrac{p\sigma_2}{p_2^2}u^{1/p} (u_1^{1/p_1} * u_2^{1/p_2}|v_2|^2)
   - \tfrac{p\sigma_2}{p_2}u^{1/p} (u_1^{1/p_1} * u_2^{1/p_2}\d(v_2)).
  \end{align*}
  Thus,
  \begin{align*}
    & \frac{1}{u^{(p-2)/p}}\left[4\pi\partial_t u -
    \tfrac{1}{p}\left(\sigma_1p_1 + \sigma_2p_2\right) \Delta u \right] \\
    & \quad \geq \tfrac{p\sigma_1}{p_1}(1-\tfrac{1}{p_1})u^{1/p}(u_1^{1/p_1}|v_1|^2 *
    u_2^{1/p_2}) + \tfrac{p\sigma_2}{p_2}(1-\tfrac{1}{p_2})u^{1/p}(u_1^{1/p_1} *
    u_2^{1/p_2}|v_2|^2) + \\
    & \quad
    \tfrac{1}{p_1p_2}(\sigma_1(p-p_1) + \sigma_2(p-p_2))
    u^{1/p}(u_1^{1/p_1}v_1 *
    u_2^{1/p_2}v_2) -(p-1)(\sigma_1p_1 + \sigma_2p_2)|\nabla(u^{1/p})|^2.
  \end{align*}
  By Lemma \ref{l:main}, it suffices to verify that
  \begin{equation} \label{e:1}
    p\sigma_1(p_1-1) + p\sigma_2(p_2-1) + 2\left(\tfrac{p\sigma_1}{p_1}(1-\tfrac{1}{p_1})\right)^{1/2}
    \left(\tfrac{p\sigma_2}{p_2}(1-\tfrac{1}{p_2})\right)^{1/2} =
    (p-1)(\sigma_1p_1 + \sigma_2p_2)
  \end{equation}
  and
  \begin{equation} \label{e:2}
   2\left(\tfrac{p\sigma_1}{p_1}(1-\tfrac{1}{p_1})\right)^{1/2}
    \left(\tfrac{p\sigma_2}{p_2}(1-\tfrac{1}{p_2})\right)^{1/2} =
    \frac{1}{p_1p_2}(\sigma_1(p-p_1) + \sigma_2(p-p_2)).
  \end{equation}
  However, \eqref{e:1} and \eqref{e:2} are elementary consequences of the hypotheses
  \eqref{e:scaling} and \eqref{e:BL}. This completes the proof of Theorem
  \ref{t:closure}(1).

  Now suppose that we have $p_j \leq 1$ and $\partial_t u_j \leq \frac{\sigma_j}{4\pi}
\Delta u_j$ for each $j=1,2$. By a very similar argument it follows
that
\begin{align*}
    & \frac{1}{u^{(p-2)/p}}\left[-4\pi\partial_t u + \tfrac{1}{p}\left(\sigma_1p_1 + \sigma_2p_2\right)\Delta u \right] \\
    & \geq \tfrac{p\sigma_1}{p_1}(\tfrac{1}{p_1}-1)u^{1/p}(u_1^{1/p_1}|v_1|^2 *
    u_2^{1/p_2}) + \tfrac{p\sigma_2}{p_2}(\tfrac{1}{p_2}-1)u^{1/p}(u_1^{1/p_1} *
    u_2^{1/p_2}|v_2|^2) \\
    & \quad -
    \tfrac{1}{p_1p_2}(\sigma_1(p-p_1) + \sigma_2(p-p_2))
    u^{1/p}(u_1^{1/p_1}v_1 *
    u_2^{1/p_2}v_2) +(p-1)(\sigma_1p_1 + \sigma_2p_2)|\nabla(u^{1/p})|^2,
\end{align*}
which is nonnegative by \eqref{e:scaling}, \eqref{e:BL} and Lemma
\ref{l:main}. This completes the proof of Theorem
\ref{t:closure}(2).

\subsection{Proof of Corollary \ref{c:Ymon}} Firstly, we verify that
$u_1$ and $u_2$ are sufficiently regular to apply Theorem
\ref{t:closure}. For $j=1,2$ suppose $u_j(0,\cdot) = d\mu_j$ and
that $d\mu_j$ is supported in the euclidean ball of radius $M$
centred at the origin. From the explicit formula for the solution
\begin{equation*}
  u_j(t,x) = \frac{1}{t^{d/2}}\int_{\R^d} e^{-\pi|x-y|^2/t}
  \,d\mu_j(y)
\end{equation*}
it is clear that $u_j(t,x) \leq t^{-d/2}e^{-\pi|x|^2/4t}$ for $|x|
> 2M$ and all $t > 0$. It follows that $u_j^{1/p_j}$ is rapidly
decreasing in space locally uniformly in time. Furthermore, by
interchanging differentiation and integration, it is easy to see
that
\begin{equation} \label{e:vjpoly}
|\nabla u_j(t,x)| \leq 2\pi t^{-d/2}(|x|+M)u_j(t,x).
\end{equation}
Consequently, $\nabla(u_j^{1/p_j})$ and $u_j^{1/p_j}|\nabla \log
u_j|^2$ are rapidly decreasing in space locally uniformly in time.
Similar considerations show that the quantities
$\partial_t(u_j^{1/p_j})$ and $\Delta(u_j^{1/p_j})$ are rapidly
decreasing in space locally uniformly in time.

To complete the proof of Corollary \ref{c:Ymon} it suffices, by the
divergence theorem, to show that for each $t > 0$,
$\int_{R\mathbb{S}^{d-1}} |\nabla u(t,\cdot)|$ tends to zero as $R$
tends to infinity. To see this, note that
$$\nabla u = p\nabla(u^{1/p})u^{1-1/p} =
p(u_1^{1/p_1}v_1 * u_2^{1/p_2})u^{1-1/p}$$ where, as before, $v_1 =
\tfrac{\nabla u_1}{u_1}$. However, if $r$ is chosen such that $1 >
1/r > 1-p$ then, by H\"{o}lder's inequality,
\begin{equation*}
  |\nabla u| \leq p (u_1^{\varepsilon_1
  r'}|v_1|^{r'} * u_2^{\varepsilon_2})^{1/r'} u^{1-1/p + 1/rp}
\end{equation*}
where $\varepsilon_j$ satisfies $(1/p_j - \varepsilon_j)r = 1/p_j$
for $j=1,2$. Any nonnegative power of $u_1$ or $u_2$ is rapidly
decreasing in space and, by \eqref{e:vjpoly}, $|v_1|$ has at most
linear growth in space. Hence, for each $t > 0$, $|\nabla
u(t,\cdot)|$ is rapidly decreasing in space which is clearly
sufficient to see that $\int_{R\mathbb{S}^{d-1}} |\nabla
u(t,\cdot)|$ tends to zero as $R$ tends to infinity.

\subsection{Proof of Theorem \ref{t:closure2}} The verification of the closure of the
regularity properties follows in the same way as in Theorem
\ref{t:closure}. We also remark that, as before, these ingredients
are sufficient to justify all convergence issues related to the
integrals which appear in the proof below.

Let $\beta := d(\A_1+\A_2-1-1/p)/2$ and for $j=1,2$ let $v_j$ denote
the time dependent vector field on $\R^d$ given by $v_j :=
\tfrac{\nabla u_j}{u_j}$. Thus,
  \begin{align*}
    4\pi \frac{\partial_t u}{t^{\beta p}(u_1^{\A_1} * u_2^{\A_2})^{p-2}}
    & \geq p(u_1^{\A_1} * u_2^{\A_2})\left[\A_1\sigma_1u_1^{\A_1}|v_1|^2 * u_2^{\A_2}
  + \A_2\sigma_2u_1^{\A_1} * u_2^{\A_2}|v_2|^2 \right. \\
  & \quad  + \left.\A_1\sigma_1u_1^{\A_1}\d(v_1) * u_2^{\A_2} +
  \A_2\sigma_2 u_1^{\A_1} * u_2^{\A_2}\d(v_2) \right] + \tfrac{4\pi \beta p}{t}(u_1^{\A_1} * u_2^{\A_2})^2.
  \end{align*}
The proof of \eqref{e:heforu} proceeds in a similar way to the proof
of Theorem \ref{t:closure}(1). The difference is that we use some of
the divergence terms on the right hand side of the above inequality
to kill off the factor $\tfrac{4\pi \beta p}{t}(u_1^{\A_1} *
u_2^{\A_2})^2$. In particular, by \eqref{e:lc},
\begin{align*}
  & (1-\rho_1)\A_1\sigma_1(u_1^{\A_1}\d(v_1) * u_2^{\A_2})
  + (1-\rho_2)\A_2\sigma_2(u_1^{\A_1} * u_2^{\A_2}\d(v_2)) + \tfrac{4 \pi \beta}{t}(u_1^{\A_1} *
  u_2^{\A_2})\\
  & \quad \geq \tfrac{2d\pi}{t}(\rho_1 \A_1 + \rho_2 \A_2 - 1 -
  \tfrac{1}{p})(u_1^{\A_1} * u_2^{\A_2})
\end{align*}
and the right hand side of the above inequality vanishes by
\eqref{e:weight}. Now, guided by the proof of Theorem
\ref{t:closure}(1), it follows that
  \begin{align*}
    & \frac{1}{t^{\beta p}(u_1^{\A_1} * u_2^{\A_2})^{p-2}}
    \left[4\pi\partial_t u -
    \tfrac{1}{p}\left(\tfrac{\sigma_1}{\A_1} + \tfrac{\sigma_2}{\A_2}\right)
    \Delta u \right] \\
    & \geq p\A_1(1-\rho_1\A_1)\sigma_1(u_1^{\A_1} * u_2^{\A_2})(u_1^{\A_1}|v_1|^2 *
    u_2^{\A_2}) + p\A_2(1-\rho_2\A_2)\sigma_2(u_1^{\A_1} * u_2^{\A_2})(u_1^{\A_1} *
    u_2^{\A_2}|v_2|^2) \\
    & \quad +
    \A_1\A_2(\sigma_1(p\rho_1-\tfrac{1}{\A_1}) + \sigma_2(p\rho_2 -
    \tfrac{1}{\A_2}))
    (u_1^{\A_1} * u_2^{\A_2})(u_1^{\A_1}v_1 *
    u_2^{\A_2}v_2) \\
    & \quad -(p-1)(\tfrac{\sigma_1}{\A_1} + \tfrac{\sigma_2}{\A_2})|\nabla(u_1^{\A_1} * u_2^{\A_2})|^2.
  \end{align*}
  By \eqref{e:weight}, \eqref{e:BL2} and Lemma \ref{l:main}
  the right hand side of the above inequality is nonnegative. This completes the proof of \eqref{e:heforu}.

  To prove \eqref{e:lcforu} we let $v$ be the time dependent
  vector field on $\R^d$ given by $v := \tfrac{\nabla u}{u}$. Thus,
  \begin{equation*}
    \d(v) = \frac{p}{(u_1^{\A_1} * u_2^{\A_2})^2}\left[\d(\nabla(u_1^{\A_1}
    * u_2^{\A_2}))(u_1^{\A_1} * u_2^{\A_2}) - |\nabla(u_1^{\A_1}
    * u_2^{\A_2})|^2\right].
  \end{equation*}
For $j=1,2$ let
\begin{equation} \label{e:ldefn}
  \lambda_j := \left(\frac{1-\rho_j\alpha_j}{2-\rho_1\A_1 - \rho_2\A_2}\right)^2
\end{equation}
and write
\begin{align*}
  \d(\nabla(u_1^{\A_1} * u_2^{\A_2})) & = \tfrac{\alpha_1}{2}\d(u_1^{\A_1}v_1 *
  u_2^{\A_2}) + \tfrac{\alpha_2}{2}\d(u_1^{\A_1} * u_2^{\A_2}v_2)
  \\
  & = \lambda_1\A_1u_1^{\A_1}\d(v_1) * u_2^{\A_2} +
  \lambda_2\A_2u_1^{\A_1} * u_2^{\A_2}\d(v_2) \\
  & \quad + \lambda_1\A_1^2u_1^{\A_1}|v_1|^2 * u_2^{\A_2} + \lambda_2\A_2^2u_1^{\A_1} *
  u_2^{\A_2}|v_2|^2 \\
  & \quad + \A_1\A_2(1-\lambda_1-\lambda_2)(u_1^{\A_1}v_1 *
  u_2^{\A_2}v_2).
\end{align*}
We shall assume that neither $\sigma_1$ nor $\sigma_2$ is equal to
zero; a simple modification of the argument will handle the
degenerate cases. By \eqref{e:lc},
\begin{align*}
 \lambda_1\A_1u_1^{\A_1}\d(v_1) * u_2^{\A_2} + \lambda_2\A_2u_1^{\A_1} *
 u_2^{\A_2}\d(v_2) & \geq - (\tfrac{\lambda_1 \alpha_1}{\sigma_1} +
 \tfrac{\lambda_1\alpha_1}{\sigma_2})\tfrac{2d\pi}{t}(u_1^{\alpha_1} *
 u_2^{\alpha_2}).
\end{align*}
Moreover, by \eqref{e:BL2} and \eqref{e:ldefn}, it is easy to check
that
\begin{equation*}
  r\sigma(\tfrac{\lambda_1 \alpha_1}{\sigma_1} +
 \tfrac{\lambda_1\alpha_1}{\sigma_2}) = \lambda_1^{1/2} +
 \lambda_2^{1/2} = 1
\end{equation*}
and therefore,
\begin{align*}
\d(v) \geq -\tfrac{2d\pi}{\sigma t} & + \tfrac{r}{(u_1^{\A_1} *
u_2^{\A_2})^2} \left[\lambda_1\A_1^2(u_1^{\A_1} *
u_2^{\A_2})(u_1^{\A_1}|v_1|^2
* u_2^{\A_2}) + \lambda_2\A_2^2(u_1^{\A_1} * u_2^{\A_2})(u_1^{\A_1}
* u_2^{\A_2}|v_2|^2) \right.\\
& + \left. \A_1\A_2(1-\lambda_1-\lambda_2)(u_1^{\A_1} *
u_2^{\A_2})(u_1^{\A_1}v_1 *
  u_2^{\A_2}v_2) - |\nabla(u_1^{\A_1} * u_2^{\A_2})|^2 \right]
\end{align*}
Since $\lambda_1 + \lambda_2 + 2\lambda_1^{1/2}\lambda_2^{1/2} =
(\lambda_1^{1/2} + \lambda_2^{1/2})^2 = 1$ it follows immediately
from Lemma \ref{l:main} that the term in square brackets in the
above inequality is nonnegative. This completes the proof of Theorem
\ref{t:closure2}.

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