TITLE: Unbounded operator with no bounded extension.
QUESTION [1 upvotes]: Let $X$ be a measurable subset of $\mathbb{R^d}$, $g \in L^\infty(X)$, and $p \in [1,\infty]$. Define the operator $T_g:L^p(X) \rightarrow L^p(X)$ by $f \mapsto fg$. Suppose that $T_g$ is injective but not surjective and show that $T_g^{-1}: \text{range}(T_g) \rightarrow L^p(X)$ is not bounded and that there is no bounded extension of $T_g^{-1}$ to $L^p(X)$.
I know that we can prove the first bit by showing that the $\text{range}(T_g)$ is a dense and proper subset of $L^p(X)$ and this will show that $T_g^{-1}$ is not bounded. However, I am not sure how to show that it is dense. I have looked at questions similar to this part on here but none seem to use the argument I am going for here. The proper subset is given to us by assumption. As for the existence of a bounded extension, how does one usually go about these arguments? Thanks in advance!

REPLY [1 votes]: Let $X,Y$ be Banach spaces and $T\in \mathcal{B}(X,Y)$. Let $T^\star$ be the adjoint of $T$.
Then

(1) $T$ is an isomorphism into $Y$ if and only if $T^\star$ is onto.
(2) $T^\star$ is an isomorphism into $X^\star$ if and only if $T$ is onto.
(3) $T$ is an isomorphism onto if and only if $T^\star$ is an isomorphism onto.
For a proof, see, for example, Banach Space Theory Exercise 2.49 on p. 78 (note that (1)+(2) imply (3)).

If $1\leq p<+\infty$. Suppose that $T_g^{-1}:gf\in T_g(L^p(X))  \mapsto f\in L^p(X)$ is bounded. Then
$T_g$ is an isomorphism from $L^p(X)$ into $T_g(L^p(X))$. According to step 1-(1),
$T_g^\star$ is onto, but
$$T_g^\star : h\in L^q(X)\mapsto gh\in L^q(X),$$
where $q$ is the conjugate
exponent of $p$.
It follows that $\forall f\in L^q(X)$, $\exists h\in L^q(X)$ such that $f=gh$. Note that
$g\neq 0$ a.e. (Indeed, if $\mu(\{g=0\})>0$, since $\mathbb{R}^d$ is $\sigma$-finite,
we can find a set $A\subset \{g=0\}$ with $0<\mu(A)<+\infty$, let $f= 1_A\in L^q(X)$, then
there doesn't exist an $h\in L^q$ such that $f=gh$, a contradiction.) Hence,
For any $f\in L^q(X)$, $h= \frac{f}{g}$ is unique. Hence, $T_g^\star$ is bijective.
According to step 1-(3), $T_g$ is surjective, a contradiction.

If $p=+\infty$.
Define $$\widetilde{T}_g: f\in L^1(X)\mapsto gf\in L^1(X).$$
Then $\widetilde{T}_g^\star =T_g$. The proof is entirely analogous.

Another solution:
$T_g^{-1}$ is continuous iff there is a constant $c>0$ such that $\|T_gf\|_{L^p(X)} \geq c \|f\|_{L^p(X)}$ for all $f\in L^p(X)$,
to put it another way,
$T_g^{-1}$ is not continuous iff there is a sequence $\{f_n\}_{n\geq 1}\subset L^p(X)$ with $\|f_n\|_{L^p(X)}=1$ such that
$T_gf_n = gf_n \to 0$ in $L^p(X)$. So, our purpose is to find such a sequence $\{f_n\}_{n\geq 1}$.
Since $T_g$ is injective, we obtain that $g\neq 0$ a.e. on $X$.
Note that $\frac{1}{g}\notin L^\infty(X)$, otherwise $T_g$ is bijective. That means, there doesn't exist a constant $c>0$ such that
$\left|\frac{1}{g}\right|\leq c$ (or $|g|\geq \frac{1}{c}$) a.e. on $X$. Hence, for each $n$, we can find
a measurable subset $S_n \subset X$ with
$\mu(S_n)>0$ such that $|g|\leq \frac{1}{n}$ a.e. on $S_n$. Since $\mathbb{R}^d$ is $\sigma$-finite,
we can assume that $\mu(S_n)<+\infty$.
For each $n$, we can find a measurable function $f_n \in L^p(S_n)$ such that $\|f_n\|_{L^p(S_n)}=1$ (this is possible since we can use those simple functions).
Extend $f_n$  by $0$ outside $S_n$. Also,
$$\|T_gf_n\|_{L^p(X)}= \|gf_n\|_{L^p(X)}= \|gf_n\|_{L^p(S_n)}\leq \frac{1}{n} \cdot \|f_n\|_{L^p(S_n)}=\frac1n\to 0 \quad \text{as $n\to\infty$}.$$


TITLE: Boundaries on double integral help
QUESTION [0 upvotes]: New to double integrals I'm not understanding how to apply the domain on this so I can solve.
$\iint_D \frac{x}{\sqrt{x^2+y^2}} \ln(x^2+y^2)\ dA$
$D=\{(x,y)\ |\ x>0, y>0, \text{ and } 1<x^2+y^2<2\}$

REPLY [0 votes]: \begin{gather*}
\iint \frac{x}{\sqrt{x^{2} +y^{2}}}\ln\left( x^{2} +y^{2}\right) dxdy\\
Convert\ ot\ polar\ coordinates,\ \\
x=r\cos \theta \\
y=r\sin \theta \\
The\ given\ intgeral\ then\ becomes\\
I=\iint \frac{r\cos \theta }{r^{2}} \cdotp \ln\left( r^{2}\right) \cdotp rdrd\theta \\
where\ 1< r< \sqrt{2}\\
0< \theta < \frac{\pi }{2}\\
I=\left(\int ^{\sqrt{2}}_{1} 2\ln rdr\right)\left(\int ^{\frac{\pi }{2}}_{0}\cos \theta d\theta \right)
\end{gather*}
Can you solve these two integrals and complete the answer?


TITLE: Process quicker than Fourier for squares of polynomials
QUESTION [5 upvotes]: FFT is a quick algorithm for multiplying two polynomials, but given it's a square (i.e. multiplying the polynomial with itself) can we find something better?

REPLY [5 votes]: As requested, here is an answer.
$$2(AB + BA) = (A+B)^2 - (A-B)^2$$
Is an identity that holds in rings in general. When the ring multiplication is commutative and one can divide by 4 nicely, this gives a means of multiplying $A$ and $B$ in terms of adding, subtracting, and two squaring operations.  So any fast routine for squaring in the right kind of ring leads to a fast multiplication algorithm.  It should be clear how fast multiplication leads to fast squaring.
Back in another lifetime, I looked at implementing the above using programmable logic chips to produce a fast multiplier. (The basic idea was that minimizing the logic for squaring binary integers might lead to a smaller circuit depth.) The result did not gain much for 16 bit integers. Nowadays 64 bit arithmetic is "kept under the hood" meaning multiplication is seen as fast enough, so this approach of reducing circuit depth by squaring is unlikely to be practical.
In the case that your domain is restricted, so that you are squaring a small class of polynomials, you might find a more optimal circuit/method, especially if the polynomials are multiplicatively generated, or if you have a fast transform like a logarithm/exponential pair available.  At present, I am not aware of anything that in practice is much faster than FFT for squaring.
Gerhard "It's All About The Speed" Paseman, 2019.08.24.


TITLE: Simplification of nasty expression
QUESTION [0 upvotes]: I have the following equation which I am trying to solve,
$$  \frac{x^2}{y^2}- \frac{x^2}{y^4} -\frac{1}{2} \leq 0$$
Can anyone think of a way of simplifying the above, I don't think this is a form of quadratic equation but I want to simplify it and have a proper relationship between $x$ and $y$

REPLY [1 votes]: Not sure if this is what you want, but you can write for $ y \ne 0$:$$\frac{x^2}{y^2}-\frac{x^2}{y^4}\le \frac12 \iff 2x^2(y^2-1 ) \le y^4 \implies \begin{cases} 2x^2 \le \dfrac{y^4}{y^2-1} && y^2 > 1 \\x \in \mathbb R&& y^2 \le 1\end{cases}$$
The first case can also be written as $|x| \le \dfrac{y^2}{\sqrt{2(y^2-1)}}$, for $|y| > 1$.