text:
\section{The Hexagon Property}\label{sec: hexagon}

Let $G$ be a group with an involution $\phi$, let $X\le G$ be a subgroup. Recall that $G$ has the \emph{hexagon property} with respect to $X,\phi$ if for all $\xi,\xi'\in X$ and $z\in G$: $\xi ^z = \phi((\xi')^z)$ implies $\xi'=\xi^{\pm 1}$.

\subsection{Hexagon condition for HNN extensions}

\begin{lemma}\label{lem: hexagon property under HNN}
    Let $A$ be a group with an involution $\phi$, $X\le A$ a subgroup. Let $C \le X$ and $C'\le A$ such that $C,C',D=\phi(C),D'=\phi(C')$ satisfy the conditions of Lemma \ref{lem: 2 acylindericity}. Set $G=\gen{A,s,t \;|\; C^s=C',D^t=D}$. Extend $\phi$ to an involution of $G$ by setting $\phi(s)=t$. If $A$ satisfies the hexagon property with respect to $X,\phi$, then so does $G$.
\end{lemma}

\begin{proof}
    Assume $\xi ^z = \phi({\xi'}^z)$, for some $\xi,\xi'\in X$ and $z\in G$.

    Write $z$ in normal form as $z=a_0 x_1 a_1 \ldots x_n a_n\in G$, where $a_i\in A, x_i \in \{s,s\ii,t,t\ii\}$. Without loss of generality, assume that $z$ has the minimal $n$ among all that satisfy $\xi ^z = \phi({\xi'}^z)$.

    By the assumption on $A$, $z\notin A$. Hence, $n\ge 1$. The word $z \phi (z)\ii$ is reduced in the HNN extension.
    By Lemma \ref{lem: 2 acylindericity}, the extension $G$ is 2-acylindrical. It follows that $n\le 1$.

    Write $z=axb$ where $a,b\in A, x\in \{s,s\ii,t,t\ii\}$.
    The relation $\xi ^z = \phi({\xi'}^{z})$ becomes
    \[ b\ii x\ii a\ii \xi  a x b \quad \phi(b\ii x\ii a\ii {\xi'}\ii a x b)  =1 .\]

    By symmetry, there are two cases to consider:

    Case 1: $x=s\ii$.
    Here the relation becomes
    \[ b\ii  \;\; \overbrace{ s \;\; \underbrace{a\ii \xi  a}_{\in A} \;\; s\ii}^{\heartsuit} \;\; \underbrace{b  \phi(b)\ii}_{\in A} \;\; \overbrace{t \;\; \underbrace{\phi(a \ii{\xi'} \ii a)}_{\in A} \;\; t\ii }^{\heartsuit}\;\;\phi(b)  =1 .\]

    By Britton's Lemma, the word must be non-reduced at both expressions marked with $\heartsuit$.

    After reducing and rearranging we get $(c)^b =   (d) ^{\phi(b)}$
    where $c=s a\ii \xi  a s\ii \in C$ and $d=t \phi(a\ii \xi' a) t\ii \in D=\phi(C)$. Since $c\in C\le X$ and $d=\phi(c')$ for some $c'\in C \le X$ we can apply the hexagon condition of $A$ to deduce that $c'=c^{\pm 1}$. Tracing back the definition of $c,c'$, it follows that $\xi'= \xi^{\pm 1}$, as desired.

    Case 2: $x=s$.
    Applying the same argument we get
    $(c')^b =   (d') ^{\phi(b)}$
    for some $c'\in C', d'\in D'$.
    However, this contradicts the assumption that $g C' g\ii \cap D'=1$ for all $g\in A$.
\end{proof}

\subsection{Hexagon property for small cancellation quotients}

\begin{lemma}\label{lem: hexagon property under small cancellation}
    Let $G$ be a torsion-free hyperbolic group with an involution $\phi$, let $X\le G$ be a quasiconvex subgroup. For all $\mathcal{R}$ such that $\phi(\mathcal{R})=\mathcal{R}$ with small enough cancellation and small enough overlap with $X$, if $G$ has the hexagon property with respect to $X,\phi$ then so does $G/\ngen{\mathcal{R}}$.
\end{lemma}

\begin{proof}
Assume for contradiction that there exist $\xi,\xi'\in X,z\in G$ such that \[\xi ^z = \phi((\xi')^z) \in G/\ngen{\mathcal{R}}\]
but $\xi'\ne\xi^\pm 1$.
Let us assume that $\xi,\xi'$ are $(\lambda,c)$-quasigeodesics in $G$, and that $z$ is a geodesic in $G/\ngen{\mathcal{R}}$.
The word $q:=z\ii \xi z \phi(z)\ii \phi(\xi')\ii \phi(z)$ is trivial in $G'$ but is not trivial in $G$ since $G$ is assumed to satisfy the hexagon property.
We would like to apply Greendlinger's Lemma to the path $q$.
However, even though the path $q$ is a concatenation of 6 quasigeodesic paths in $G$, it might not be a quasigeodesic because of ``backtracking''.
However, one can fix this by trimming the backtracking corners as described in Lemma \ref{lem: corner trimming}. There exist (possibly empty) subwords $z_1,z_2,z_3,z_4$ of $z$ and subwords $\eta,\eta'$ of $\xi,\xi'$ respectively, and words $v_1,\ldots,v_6$ of length $\le \delta'$  such that the path
\[p:= z_1\ii v_1 \eta v_2 z_2 v_3 \phi(z_3)\ii v_4 \phi(\eta')\ii v_5 \phi(z_4) v_6\]
is a conjugate of $q$ in $G$, and the path $p$ is a $(\lambda',c')$-quasigeodesic, where $\delta',\lambda',c'$ depend only on $\lambda,c$ and $G$. See Figure \ref{fig: hexagon}.
Moreover, by symmetry of $z\phi(z)\ii$ we may assume that $z_2$ and $z_3$ end at the same place in $z$ (i.e, $z=z' z_2 u = z'' z_3 u$ as words, for some $z',z',u$). A similar statement holds for $z_4, z_1$. By replacing $\xi,\xi'$ with large enough powers, we may assume that $\eta$ and $\eta'$ are arbitrarily long, and in particular non-empty.

Since $p$ and $q$ are conjugates, we have that $p=1\in G/\ngen{\mathcal{R}}$ while $p\ne 1$ in $G$.
By Greendlinger's Lemma there exists a cell labeled $r\in\mathcal{R}$ with contiguity degree $>(1-13\mu)$ assuming $\mathcal{R}$ satisfies small enough cancellation.
Let us denote by $r',p'$ the subwords of $r,p$ respectively which label the opposite sides of the contiguity subdiagram.
As in the proof of Item \ref{lem: summary of properties of sc quotients - quasiconvex} of Lemma \ref{lem: summary of properties of sc quotients} let $r'',p''$ be the $\delta''$-fellow-travelling subpaths of $r',p'$ of length $\|r''\| > \|r'\| - \epsilon', \|p''\| > \|p'\| - \epsilon'$ provided by Lemma \ref{lem: fellow-traveling}, and let $r''=r_{z1}r_{\eta}\ldots r_{z4}$, where $r_{z1}, r_{\eta}\ldots, r_{z4}$ are the (possibly empty) subwords of $r''$ which correspond to the paths that $\delta''$-fellow-travel with $z_1\ii,\eta,\ldots,\phi(z_4)$ respectively. Since $\|r''\|>\|r'\|-\epsilon'$,
\[ \| r_{z1}\| + \|r_{\eta}\| + \ldots + \|r_{z4}\| > (1-13\mu)\|r\| - \epsilon'=:\omega\]

We now divide into cases:

\begin{figure}
  \centering
  \def\svgwidth{\textwidth}
  \input{hexagon.pdf_tex}
  \label{fig: hexagon}
  \caption{The trimmed hexagon, and the 3 cases of the contiguous cell in the proof of Lemma \ref{lem: hexagon property under small cancellation}}
\end{figure}

Case 1. $\|r_\eta\| > \mu\| r\|=:\omega_1$ or $\|r_{\eta'}\| > \mu \| r\|$. This is impossible when $\mathcal{R}$ has small enough overlap with $X$ since $\eta,\eta'\in X$.

Case 2. $\min \{ \|r_{z2} \|, \| r_{z3}\|\} > \lambda\ii(\mu\|r\|+2\delta''+c)+2\delta''=:\omega_2$ and the path $p''$ contains $v_3$.
In this case, let $p''_{z2} = p''\cap z_2$, $p''_{v3} = p''\cap v_3$ and $p''_{z3} = p''\cap \phi(z_3)\ii$.
Since $\|r\|$ is a $(\lambda,c)$-quasigeodesic and $z_2$ and $\phi(z_3)\ii$ are geodesics we get that $\min\{\|p''_{z2}\|,\|p''_{z3}\|\} > \mu \|r\|+2\delta''$.
Recall that $z_2$ and $z_3$ end at the same place in $z$, thus there is a subword $V$ of $z$ of length $\|V\|>\mu\|r\|+2\delta''$ such that $V$ is in $p''_{z2}$ and $\phi(V)\ii$ is in $p''_{z3}$.
Let $U$ and $U'$ be the subwords of $r$ that  $\delta''$-fellow-travel with $V$ and $\phi(V)$. $\|U\|\ge \|V\| - 2\delta'' > \mu \|r\|$ and similarly $\|U'\|  > \mu\|r\|$.
Since $\phi(r)\in \mathcal{R}$ we get that $r$ has a $2\delta''$-piece (and hence an $\epsilon$-piece) with $\phi(r)$ of length $>\mu \|r\|$ which is impossible if $\mathcal{R}$ has $\torsionsccK{X}$.

Similarly one proves the case $\min \{ \|r_{z1} \|, \| r_{z4}\|\} > \lambda\ii(\mu\|r\|+2\delta''+c)$ and the path $p''$ contains $v_6$.

Case 3. $\|r_{z4}\|>\omega - \omega_1 - \omega_2=:\omega_3$ (and similarly for $r_{z1}$, $r_{z2}$ and $r_{z3}$). For small enough $\mu$ and large enough $\rho$ we can assume that $\omega_3/\|r\|$ is arbitrarily close to $1$, and thus we can assume $(\|r\|-\omega_3)+2\delta'' < \lambda \omega_3 - c$.
However, as in Case 2 in the proof of Item \ref{lem: summary of properties of sc quotients - quasiconvex} of Lemma \ref{lem: summary of properties of sc quotients}, there exists a shortcut to $z$ (in $G'$), contradicting the assumption that $z$ is a geodesic.

\end{proof}


TITLE: A wrong corollary of Kolmogorov's $0-1$ Law
QUESTION [4 upvotes]: I either broke Probability Theory, or what is more likely, I am confused.
Let $X_{n}$ be a sequence of i.i.d random variables.
Then by Kolmogorov's $0-1$ Law, the probability that $S_{n}=\sum_{k=1}^n X_{k}$ converges is $\in \{0,1\}$.
Furthermore, the random variables $\lim \sup S_{n}$ and $\lim \inf S_{n}$ are measurable with respect to the terminal sigma-field generated by the $X_{n}$ and hence a.s. constant (a consequence of the 0-1 law) - is this true?
Now if $S_{n}$ converges a.s. then $\lim \sup S_{n} = \lim \inf S_{n}$ a.s.
Does this mean that if $S_{n}$ converges it does so to a constant?
It can't...

REPLY [2 votes]: If $S_n$ converges a.s. then $\limsup S_n = \liminf S_n$ a.s.

So far, so good.

Does this mean that if $S_n$ converges it does so to a constant?

That the answer is "yes" seems surprising, until you remember that a conditional statement "If $P$, then $Q$" is vacuously true when $P$ is false.
In your example, $S_n$ does not converge unless each of the $X_n = 0$ a.s., in which case, $S_n = 0$ a.s. for all $n$ as well and therefore we trivially have $S_n \to 0$ a.s.
If $P(X_n = 0) \neq 1$, then since $S_n$ does not converge, the "if" part of "if $S_n$ converges it does so to a constant" is false, and so the conditional statement is vacuously true.


TITLE: Variance of conditional geometric random variable
QUESTION [1 upvotes]: Given a coin which comes up Heads with probability $p$, let $N_p$ denote the number of times the coin is tossed until the first Heads. Argue that $$N_p = \begin{cases} 1 & \text{with probability } p\\ 1+\tilde{N}_p & \text{with probability } 1-p \end{cases}$$
where $\tilde{N}_p$ is an independent random variable with the same distribution as $N_p$.
Using this, show that $\mathbb{E}[N_p] = 1/p$. Also use a similar argument to find $Var(N_p)$.

I find the expectation fairly straightforward:
By rule of iterated expressions, we have:
$$E[E[N_p|\tilde{N_p}]] = E[N_p] = E[p\times 1 + (1-p)\times (1+\tilde{N_p})]$$
$$= p\times p + (1-p) +  (1-p)\times E[\tilde{N_p}] = 1 + \tfrac 1p - 1 = \tfrac 1p$$
However, I am having trouble with finding $V(N_p)$. There is an identity where
$$V(N_p) = E[V(N_p|\tilde{N_p})] + V(E[N_p|\tilde{N_p}])$$
but I do not know how to handle $V(N_p|\tilde{N_p})$. By definition this should be
$$V(Y|X=x)=\sum_{x\in X}(y-E(Y|X=x)^2P(Y|X=x)$$
What is little $y$ here? Is it simply $1$ and $1+N_p$? What does $E(Y|X=x)$ look like? This is a number, whereas $E(Y|X)$ is a random variable whose actualizations is $E(Y|X=x)$, but I am not sure what these actualizations look like.How do I find $V(Y|X)$ with $V(Y|X=x)$?

REPLY [1 votes]: You should condition on the outcome of the first toss (call this $Y$), not on $\tilde{N}_p$. Then,
$$V\left(N_p\,|\,Y\right) = \left\{
\begin{array}{c c}
0 & Y=1\\
V\left(N_p\right) & Y=0
\end{array}
\right.$$
and, similarly,
$$\mathbb{E}\left(N_p\,|\,Y\right) = \left\{
\begin{array}{c c}
1 & Y=1\\
1+\mathbb{E}\left(N_p\right) & Y=0
\end{array}
\right.$$
Combining these (and having solved for $\mathbb{E}\left(N_p\right)$ before) should give you an equation that has $V\left(N_p\right)$ on both sides, but you can rearrange terms and solve for it.


TITLE: Trivial question about limit of vectors to zero divided by its norm related to proof of 2nd order Optimality Condition
QUESTION [1 upvotes]: Let $x,y\in \mathbb{R}^{n}$, I want to ask how to know whether the following limit exists or not.
$$\lim\limits_{y \to x}\frac{y-x}{||y-x||}$$

I am not sure about my answer because when I see for each component, for example when I have $x,y\in \mathbb{R}^{2}$ with $x=(x_1,y_1)$ and $y=(y_1,y_2)$, the following statement
$$\lim\limits_{(y_1,y_2) \to (x_1,x_2)}\frac{y_1-x_1}{\sqrt{(y_1-x_1)^{2}+(y_2-x_2)^{2}}}$$ simply does not exist.
My initial problem is that I want to prove the necessity condition for 2nd order optimality condition.
I have followed the steps up until this point:
Let $x$ be a local minimizer point of a $C^2$ function $f:S\to\mathbb{R}$ in a neighborhood $B\subset \mathbb{R}^{n}$ ($S \subset \mathbb{R}$ is the domain of $f$). I want to prove that the Hessian matrix of $f$ denoted by $H$ is semi-positive definite.
By Taylor Expansion and 1st order optimality necessary condition, we have $\forall y \in B$,
$$f(y)-f(x)\geq0$$
$$f(y) = f(x) + \frac{1}{2}(y-x).(H(x)(y-x)) + o(||y-x||^3)$$
$$f(y) - f(x) = \frac{1}{2}(y-x).(H(x)(y-x)) + o(||y-x||^3) \geq 0$$
Now, in the prove that I read, it says that by multiplying both sides of equation by $\frac{2z}{||y-x||^{2}}$ for any arbitrary $z \in \mathbb{R}^{n}$.
But then, I will get the following inequality
$$\frac{2(f(y)-f(x))z}{||y-x||^{2}}= \frac{(y-x).(H(x)(y-x))}{||y-x||^{2}}z+o(||y-x||)z\geq 0_{\mathbb{R}^{n}}$$
Now, if I take the limit $y\to x$, I have the initial problem. How can I show that
$$\lim\limits{y\to x}\frac{(y-x).(H(x)(y-x))}{||y-x||^{2}}z \geq 0_{\mathbb{R}^{n}} \implies z.(H(x)z)\geq 0$$
Of course I know this property : given a matrix $A_{n\times n}$, $\forall z \in \mathbb{R}^{n} Az\geq 0_{\mathbb{R}^{n}}\implies z.(H(x)z)\geq 0$
I appreciate the help!

REPLY [1 votes]: One way to prove that the Hessian is positive semi-definite at point $x$ is to look at the all feasible directions $d = y-x$. Now write whatever you have for $f(td+x)$ and $f(x)$, for $t\in[0,1]$. If you write it down this way, then you will have $2(f(td+x)-f(x))$ on the left and $t^2 d^T H d$ plus some term which is of order $t^3$ on the right-hand side. The left hand side is non-negative for the any $t\in[0,1]$. And the $O(t^3)$ term cannot dominate $t^2 d^T H d$ when $t\rightarrow 0$. So $d^T H d\geq 0$, if the right-hand side wants to be non-negative too for any small value of $t$. And this has to be true for any feasible direction at d. If $x$ is inside the interior of the feasible set, then this means that $H$ is positive semi-definite (all directions are feasible locally). If not, then the positive semi-definiteness is only valid in the feasible directions.


TITLE: Help gaining a basic understanding of sequences in $\mathbb{R}$
QUESTION [2 upvotes]: So I can state the definition:
A sequence of real numbers is said to converge to a Real number $a$ iif for every $\epsilon \gt 0$ there is an $N \in \mathbb{N}$ such that.
$n\geq\mathbb{N}$ implies $\vert x_n - a \vert \lt \epsilon$
I don't understand what $n \geq \mathbb{N}$ means and it's connection to $\vert x_n - a \vert$.
Given a sequence I am supposed to show that the limit exists by using the archimedian principle which I'm not sure how to do with this.
For example:
$x_n = 2-\frac{1}{n}$
$x_n \to 2$ as $n \to \infty$
$\vert 2 - \frac{1}{n} - 2 \vert \lt \epsilon$
$\vert -\frac{1}{n} \vert \lt \epsilon$
Which implies $\frac{1}{n} \lt \epsilon$?
$\frac{1}{\epsilon} \lt n$
I have no idea if this is even remotely close. Can someone explain in simple terms what it is I am exactly supposed to do and why. I am bad with technical definitions until I have some foundation.

REPLY [1 votes]: If I may, I would like to summarize the idea:
A sequence $x_{n}$ converges to $a$ provided that given an $\epsilon > 0$, there exists a positive integer $N$ such that whenever $n \geq N$ it follows that $|x_{n} - a| < \epsilon$.
This means that if a sequence $x_{n}$ converges, then regardless of what is happening to the terms initially, eventually all of the terms must reside in an open interval centered at $a$ of radius $\epsilon$. That is what is meant by "there exists a positive integer $N$ for which whenever $n \geq N$, it follows that $|x_{n} - a| < \epsilon$."
So, for example, consider the sequence $x_{n} = \frac{\sin n}{n}$, which converges to $0$ (though not monotonically). We can show that it converges to $a = 0$ by being given an arbitrary $\epsilon > 0$, and then proceed to determine a positive integer $N$ that works:
$$ \left| \frac{\sin n}{n} - 0 \right| = \left| \frac{\sin n}{n} \right|  < \frac{1}{N} < \epsilon; $$
and so, if we take $N$ to be any integer greater than $1/\epsilon$, it follows that whenever $n > N$, we have
$$\left| \frac{\sin n}{n} \right| < \frac{1}{N} < \frac{1}{\frac{1}{\epsilon}} = \epsilon. $$


TITLE: Mittag-Leffler Expansion of gamma function
QUESTION [3 upvotes]: Mittag-Leffler's expansion theorem states that
$$f(z)=f(0)+\sum_{n=1}^\infty \text{Res}_{z=z_k}\left\{\frac{1}{z-z_k}+\frac{1}{z_k}\right\}$$
For gamma function $\Gamma(z)$
$$\Gamma(z)+\frac{1}{z}=1+\sum_{n=1}^\infty \frac{(-1)^n}{n!}\left(\frac{1}{z+n}-\frac{1}{n}\right)=1+\sum_{n=0}^\infty\frac{(-1)^n}{n!(z+n)}-\sum_{n=1}^\infty \frac{(-1)^n}{n(n)!}$$
But this doesn't simplifies to what I have to prove. That is
$$\Gamma(z)=\sum_{n=0}^\infty\frac{(-1)^n}{n!(z+n)}+\int_1^\infty t^{z-1}e^{-t}dt$$
I get the first term right!, but I'm not sure how does this integral shows up here. Please help me with this

REPLY [3 votes]: I think there's no point in trying to invoke the general Mittag-Loeffler result. Just break the Euler integral into $\int_0^1$ and $\int_1^\infty$. In the $\int_0^1$, expand $e^{-t}$ in the usual power series. Interchange integral and sum, and you get the result you're aiming for.


TITLE: Isomorphism of projective varieties and coordinate rings
QUESTION [0 upvotes]: Let $K$ be a field.
Let $f: \mathbb{P}^1(K) \rightarrow \mathbb{P}^2(K)$ be a morphism given by $$f([x_1:x_2])=[x_1^2:x_1x_2:x_2^2].$$
(a) Are $\mathbb{P}^1(K)$ and $f(\mathbb{P}^1(K))$ isomorphic as projective varieties?
(b) Are the rings $\dfrac{K[X,Y]}{I_P(\mathbb{P}^1(K))}$ and $\dfrac{K[X,Y,Z]}{I_P(f(\mathbb{P}^1(K)))}$ isomorphic?
My attempt:
So for the first question, we want to show that $f$ is a morphism that has an inverse.
$f$ is a morphism since for $P=[x:y] \in \mathbb{P}^1(K)$, we can write $f(P)=(f_0(P), f_1(P), f_2(P))$ where $f_0=X^2, f_1=XY, f_2=Y^2 \in K[\mathbb{P}^1(K)]$ are homogeneous. Moreover, $\mathbb{P}^1(K) \subset \mbox{dom } f$.
Now, I want to find an inverse. I'm not really sure how to compute it, but my guess is for any $[x:y:z] \in \mathbb{P}^2(K)$,with $x \neq 0$, $f^{-1}([x:y:z])=[x:y]$. It will follow that if $x_1 \neq 0$, $f^{-1}([x_1^2:x_1x_2:x_2^2])=[x_1^2:x_1x_2]=[x_1:x_2]$. If $x=0$, it will follow that $y=0$. We have the point $[0:0:1]$ so $f^{-1}([0:0:1])=[0:1]$. I am not sure for the other cases.
For the second question, I observed that $f(\mathbb{P}^1(K)) =V(Y^2-XZ)$ where $V(Y^2-XZ)$ is the zero locus of $g=Y^2-XZ$. We can show that $g$ is irreducible in $K[X,Y,Z]$ so
$$\dfrac{K[X,Y,Z]}{I_P(f(\mathbb{P}^1(K)))}=\dfrac{K[X,Y,Z]}{I_P(V(Y^2-XZ))}=\dfrac{K[X,Y,Z]}{(Y^2-XZ)}$$. Meanwhile,
$\dfrac{K[X,Y]}{I_P(\mathbb{P}^1(K))}=  \dfrac{K[X,Y]}{\{0\}} \cong K[X,Y].$
Thus, this problem is reduced to showing if $$K[X,Y] \cong \dfrac{K[X,Y,Z]}{(Y^2-XZ)}.$$ But these two rings are not isomorphic since the latter is not a UFD.
Are my thoughts correct? Any help would be appreciated. Thank you!

REPLY [0 votes]: Your work in the post is correct. One easier way to write down an inverse morphism (which has the benefit of being more obviously a morphism) is as follows: on the affine patch in $\Bbb P^2$ given by $X\neq 0$, the inverse morphism can be defined by $[1:y:y^2]\mapsto [1:y]$, and on the affine patch given by $Z\neq 0$, the inverse morphism can be given as $[x^2:x:1]\mapsto [x:1]$. These maps are clearly regular and agree everywhere they're both defined, and their domain of definition is the whole of the image of $\Bbb P^1$.


TITLE: Induction Inequality with Summation
QUESTION [0 upvotes]: I can't seem to figure out this problem. Do you have any ideas?

For an integer $n > 1$, show that
  $$
  \sum_{k=1}^n {1\over \sqrt{{n^2}+k}} > {{\sqrt{1+{1\over n}}}\over 2}
$$

REPLY [0 votes]: Third proof: $f(x) = \dfrac{n}{\sqrt{x+n^2}}$ has $f''(x) = \dfrac{3n(x+n^2)^{-\frac{5}{2}}}{4}> 0$, thus $f$ is convex. Apply Jensen's inequality:
$\displaystyle \sum_{k=1}^n \dfrac{n}{\sqrt{k+n^2}}=\displaystyle \sum_{k=1}^n f(k)\geq nf\left(\dfrac{1+2+\cdots+n}{n}\right)=n\cdot \dfrac{n}{\sqrt{n^2+\dfrac{1+2+\cdots + n}{n}}}=\dfrac{n^2}{\sqrt{n^2+\dfrac{n(n+1)}{2n}}} > \dfrac{\sqrt{n^2+n}}{2}\iff 6n^3>3n^2+2n+1$, and  this is true by induction on $n$.


text:
\section{Maximal One-to-One Theta Relation}

\subsection{One-to-one theta relation}\label{0407}
We can regard $\Theta$ as a mapping given by
\[
\Theta\colon (\bfG,\bfG')\mapsto \Theta_{\bfG,\bfG'}\subset\cale(G)\times\cale(G').
\]
If $\vartheta$ is another mapping $(\bfG,\bfG')\mapsto\vartheta_{\bfG,\bfG'}\subset\cale(G)\times\cale(G')$ such that
$\vartheta_{\bfG,\bfG'}$ is a subset of $\Theta_{\bfG,\bfG'}$ for each dual pair $(\bfG,\bfG')$,
then $\vartheta$ is called a \emph{sub-relation} of $\Theta$.
For a sub-relation $\vartheta$ and $\rho\in\cale(G)$,
we define
\[
\vartheta_{\bfG'}(\rho)=\{\,\rho'\in\cale(G')\mid(\rho,\rho')\in\vartheta_{\bfG,\bfG'}\,\}.
\]
Among the set of all sub-relations of $\Theta$,
we can give a partial ordering by inclusion.
Moreover precisely, for two sub-relations $\vartheta^1,\vartheta^2$ of $\Theta$ we say that
$\vartheta^1\subseteq\vartheta^2$ if for each dual pair $(\bfG,\bfG')$,
we have $\vartheta^1_{\bfG,\bfG'}\subseteq\vartheta^2_{\bfG,\bfG'}$ as subsets of $\cale(G)\times\cale(G')$.

Now we have several definitions:
\begin{itemize}
\item A sub-relation $\vartheta$ of $\Theta$ is called \emph{semi-persistent} (on unipotent characters)
if it satisfies the following conditions:
\begin{enumerate}
\item if either
\begin{enumerate}
\item $(\bfG,\bfG')=(\Sp_{2n},\rmO^+_{2n'})$ and $\Lambda\in\cals_{\bfG}$ with ${\rm def}(\Lambda)=4d+1$; or

\item $(\bfG,\bfG')=(\rmO^+_{2n'},\Sp_{2n})$ and $\Lambda\in\cals_{\bfG}$ with ${\rm def}(\Lambda)=4d$,
\end{enumerate}
then $\vartheta_{\bfG'}(\rho_\Lambda)\neq\emptyset$ for any $n'\geq n-2d$;

\item if either
\begin{enumerate}
\item $(\bfG,\bfG')=(\Sp_{2n},\rmO^-_{2n'})$ and $\Lambda\in\cals_{\bfG}$ with ${\rm def}(\Lambda)=4d+1$; or

\item $(\bfG,\bfG')=(\rmO^-_{2n'},\Sp_{2n})$ and $\Lambda\in\cals_{\bfG}$ with ${\rm def}(\Lambda)=4d+2$,
\end{enumerate}
then $\vartheta_{\bfG'}(\rho_\Lambda)\neq\emptyset$ for any $n'\geq n+2d+1$.
\end{enumerate}

\item A sub-relation $\vartheta$ of $\Theta$ is called \emph{symmetric} if
for each dual pair $(\bfG,\bfG')$, any $\rho\in\cale(G)$ and $\rho'\in\cale(G')$,
we have $\rho'\in\vartheta_{\bfG'}(\rho)$ if and only if $\rho\in\vartheta_{\bfG}(\rho')$.

\item A sub-relation $\vartheta$ of $\Theta$ is said to be \emph{compatible with the Lusztig correspondence},
if for each dual pair $(\bfG,\bfG')$, any $\rho\in\cale(G)_s$ and $\rho'\in\cale(G')_{s'}$
for some semisimple elements $s,s'$,
there exist choices of Lusztig correspondences $\Xi_s,\Xi_{s'}$ such that the following diagram
\[
\begin{CD}
\rho @> \vartheta_{\bfG'} >> \rho' \\
@V \Xi_s VV @VV \Xi_{s'} V \\
\rho^{(1)}\otimes\rho^{(2)}\otimes\rho^{(3)} @> {\rm id}\otimes{\rm id}\otimes\vartheta_{\bfG'^{(3)}} >> \rho'^{(1)}\otimes\rho'^{(2)}\otimes\rho'^{(3)}.
\end{CD}
\]
is commutative.
\end{itemize}

A sub-relation $\vartheta$ of $\Theta$ is called a \emph{theta-relation} if it is
semi-persistent, symmetric and compatible with the Lusztig correspondence.

A sub-relation $\vartheta$ of $\Theta$ is called \emph{one-to-one} if for each dual pair $(\bfG,\bfG')$
and each $\rho\in\cale(G)$,
there exists at most one $\rho'\in\cale(G')$ such that $(\rho,\rho')\in\vartheta_{\bfG,\bfG'}$.
If $\vartheta$ is a one-to-one sub-relation of $\Theta$, then for a dual pair $(\bfG,\bfG')$ and $\rho\in\cale(G)$
we know that either $\vartheta_{\bfG'}(\rho)=\emptyset$ or $\vartheta_{\bfG'}(\rho)=\{\rho'\}$ for some
$\rho'\in\cale(G')$.
For the latter case, we may just abuse the notation a little by writing $\vartheta_{\bfG'}(\rho)=\rho'$.

\begin{prop}\label{0604}
Both $\underline\theta$ and $\overline\theta$ are one-to-one theta-relations.
\end{prop}
\begin{proof}
By definition, it is clear that both $\underline\theta$ and $\overline\theta$ are sub-relations of $\Theta$.
When $(\bfG,\bfG')$ consists of a symplectic group and an even orthogonal group,
both $\underline\theta$ and $\overline\theta$ are clearly semi-persistent from the definitions in
Subsection~\ref{0428}.
We know that both $\underline\theta$ and $\overline\theta$ are symmetric by Lemma~\ref{0406},
and they are compatible with the Lusztig correspondence by the definition in (\ref{0402}).
Therefore, both $\underline\theta$ and $\overline\theta$ are theta-relations.
Finally, both $\underline\theta$ and $\overline\theta$ are one-to-one by Lemma~\ref{0408}.
\end{proof}

\subsection{Maximal one-to-one theta-relation}
The $\eta$-correspondence is only defined for dual pairs in stable range.
Now we know that both $\underline\theta$ and $\overline\theta$ are
one-to-one theta-relations which extend $\eta$ to general dual pairs.
The following proposition means that both $\underline\theta$ and $\overline\theta$
can not be extended any more if we require the extension to be a one-to-one theta-relation.

\begin{prop}\label{0605}
No one-to-one theta-relation can be contained properly in another one-to-one theta-relation.
\end{prop}
\begin{proof}
Let $\vartheta$ be a one-to-one theta-relation.
Suppose that $\vartheta'$ is a theta-relation which properly contains $\vartheta$.
So we need to show that $\vartheta'$ is not one-to-one.
Let $(\bfG,\bfG')$ be a dual pair, $\rho\in\cale(G)$ and $\rho'\in\cale(G')$
such that $(\rho,\rho')\in\vartheta'_{\bfG,\bfG'}$ and $(\rho,\rho')\not\in\vartheta_{\bfG,\bfG'}$.
If $\vartheta_{\bfG'}(\rho)$ is defined, i.e., $(\rho,\rho'')\in\vartheta_{\bfG,\bfG'}\subset\vartheta_{\bfG,\bfG'}'$ for some $\rho''\in\cale(G')$,
then $\vartheta'$ is not one-to-one.
So we may assume that $\vartheta_{\bfG'}(\rho)$ is not defined.
Since both $\vartheta'$ and $\vartheta$ are compatible with Lusztig correspondence,
we have a commutative diagram
\[
\begin{CD}
\rho @> \vartheta'_{\bfG'} >> \rho' \\
@V \Xi_s VV @VV \Xi_{s'} V \\
\rho^{(1)}\otimes\rho^{(2)}\otimes\rho^{(3)} @> {\rm id}\otimes{\rm id}\otimes\vartheta'_{\bfG'^{(3)}} >> \rho'^{(1)}\otimes\rho'^{(2)}\otimes\rho'^{(3)}.
\end{CD}
\]
This means that $\vartheta'_{\bfG'^{(3)}}(\rho^{(3)})=\rho'^{(3)}$ and $\vartheta_{\bfG'^{(3)}}(\rho^{(3)})$ is undefined.
Now $(\bfG^{(3)},\bfG'^{(3)})$ is a dual pair of one symplectic group and one even orthogonal group,
and both $\rho^{(3)},\rho'^{(3)}$ are unipotent.
So we write $\rho^{(3)}=\rho_\Lambda$ and $\rho'^{(3)}=\rho_{\Lambda'}$ for some
$\Lambda\in\cals_{\bfG^{(3)}}$ and $\Lambda'\in\cals_{\bfG'^{(3)}}$.

Suppose that $(\bfG^{(3)},\bfG'^{(3)})=(\Sp_{2n},\rmO^+_{2n'})$ for some $n,n'$,
$\Lambda\in\cals_{\bfG^{(3)}}$ with ${\rm def}(\Lambda)=4d+1$ for some $d\in\bbZ$.
Then ${\rm def}(\Lambda')=-4d=4(-d)$ by Proposition~\ref{0204}.
Because now $\vartheta$ is semi-persistent and $\vartheta_{\bfG'^{(3)}}(\rho_\Lambda)$ is not defined,
we must have $n'<n-2d$.
So we have $n>n'-(-2d)$, and this means that $\vartheta_{\bfG^{(3)}}(\rho_{\Lambda'})$ is defined.
Let $\Lambda''=\vartheta_{\bfG^{(3)}}(\Lambda')\in\cals_{\bfG^{(3)}}$.
Now $(\rho_{\Lambda'},\rho_{\Lambda''})\in\vartheta_{\bfG'^{(3)},\bfG^{(3)}}\subseteq\vartheta'_{\bfG'^{(3)},\bfG^{(3)}}$ implies
that $(\rho_{\Lambda''},\rho_{\Lambda'})\in\vartheta'_{\bfG^{(3)},\bfG'^{(3)}}$ since $\vartheta'$ is symmetric.
Moreover, $(\rho_\Lambda,\rho_{\Lambda'})$ is in $\vartheta'_{\bfG^{(3)},\bfG'^{(3)}}$ by our assumption.
However, $\Lambda$ and $\Lambda''$ are not equal
because $\vartheta_{\bfG'^{(3)}}(\Lambda)$ is not defined and $\vartheta_{\bfG'^{(3)}}(\Lambda'')=\Lambda'$
by the symmetricity of $\vartheta$.
So we conclude that $\vartheta'$ is not one-to-one.

The proof for other cases (i.e., $(\bfG^{(3)},\bfG'^{(3)})=(\Sp_{2n},\rmO^-_{2n'})$, $(\rmO^+_{2n},\Sp_{2n'})$, or $(\rmO^-_{2n},\Sp_{2n'})$)
are similar.
\end{proof}

\begin{cor}\label{0701}
Both $\underline\theta$ and $\overline\theta$ are maximal one-to-one theta-relation, i.e.,
they are not properly contained in any other one-to-one theta-relation.
\end{cor}
\begin{proof}
This follows from Proposition~\ref{0604} and Proposition~\ref{0605} immediately.
\end{proof}

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