17 - Convexity

Concavity of a Function lemma

Let $g:\mathbb{R}\to \mathbb{R}$ be a differentiable function then

If $[a,b]$ is an interval where $g^{\prime \prime }(x)<0$ then function $g$ is concave on $[a,b]$ if

For all $x<y\in [a,b]$ then

$$g(tx+(1-t)y)\ge tg(x)+(1-t)g(t)\forall t\in [0m1]$$

In other words the chords between any two points on the graph lies below the graph itself

Proof

For $x,y\in [a,b]$ and $t\in [0,1]$

Let $\xi =tx+(1-t)y$ be a point in interval between $x,y$

Slope of chord between $(x,g(x))$ and $(\xi ,g(\xi ))$ by MVT is equal to

$$g^{\prime }(s_1)\text{ where }{s}_1\text{ lies between }x\text{ and }\xi$$

whilst slope of chord between $(\xi ,g(\xi ))$ and $(y,g(y)$ is equal to

$$g^{\prime }(s_2)\text{ where }{s}_2\text{ lies between }\xi \text{ and }y$$

If $g(\xi )<tg(x)+(1-t)g(y)$ then

$$g^{\prime }(s_1)<0\text{ and }{g}^{\prime }(s_2)>0$$

So by MVT for $g^{\prime }(x)$ applied to $s_1$ and $s_2$ then exists $s\in (S_1,s_2)$ with

$$g^{\prime \prime }(s)=\frac{g^{\prime }(s_2)-g^{\prime }(s_1)}{s_2-s_1}>0$$

which contradicts assumption that $g^{\prime \prime }(x)$ is negative on $(a,b)$

Jordon's Lemma lemma

let $f$ be a continuous function on ${\gamma }_R^{\ast }$ where

$${\gamma }_R(t)=R{e}^{it}\text{ where }t\in [0,\pi ]$$

Then for all positive $\alpha$

$$\left|{\int }_{{\gamma }_R}f(z)e^{i\alpha z}dz\right|\le \frac{\pi }{\alpha }{M}_R\text{ where }{M}_R=\underset{t\in [0,\pi ]}{\max }|f({\mathrm{Re}}^{it}])|$$

In particular, suppose $f$ is holomorphic on $\mathbb{H}\setminus S$ where

$$\mathbb{H}=\{z\in \mathbb{C}:\mathrm{Im}(z)>0\}\text{ is the upper half-plane}$$

and $S$ be the finite set of isolated singularities

Suppose $f(z)\to 0$ as $z\to \infty$ in $\mathbb{H}$ then

$${\int }_{{\gamma }_R}f(z)e^{i\alpha z}dz\to 0\text{ as }R\to \infty \text{ for all }\alpha >0$$

Proof

Apply Concavity of a function to function $g(t)=\sin (t)$ with $x=0$ and $y=\frac{\pi }{2}$
So

$$\sin (t)\ge \frac{2}{\pi }t\text{ for }t\in \left[0,\frac{\pi }{2}\right]$$

and similarly

$$\sin (\pi -t)\ge \frac{2(\pi -t)}{\pi }\text{ for }t\in \left[\frac{\pi }{2},\pi \right]$$

Thus

$$|e^{i\alpha z}|\le e^{-\alpha R\sin (t)}\le \{\begin{array}{ll}{e}^{-2\alpha Rt/\pi }& \text{ if }t\in \left[0,\frac{\pi }{2}\right]\\ & \\ e^{-2\alpha R(\pi -t)/\pi }& \text{ if }t\in \left[\frac{\pi }{2},\pi \right]\end{array}$$

But then

$$\left|{\int }_{{\gamma }_R}f(z)e^{i\alpha z}\right|\le 2R{M}_R{\int }_0^{\frac{\pi }{2}}{e}^{-2\alpha Rt/\pi }dt=M_R\frac{\pi }{\alpha }(1-e^{-\alpha R})<M_R\frac{\pi }{\alpha }$$

Fix $ϵ>0$
Since $f\to 0$ as $z\to \infty$ then as $S$ is finite
There exists $R_0$ such that $M_R<ϵ$ for all $R>R_0$ hence

$$\left|{\int }_{{\gamma }_R}f(z)e^{i\alpha z}dz\right|\le ϵ\frac{\pi }{\alpha }\text{ for }R>R_0$$

Contour Bound for $\cot (\pi z)$

Let $f(z)=\cot (\pi z)$

Let ${\Gamma }_N$ denote the square path with vertices $\left(N+\frac{1}{2}\right)(\pm i\pm i)$

Then there is constant $C$ independent of $N$ such that

$$|f(z)|\le C\text{ for all }z\in {\Gamma }_N^{\ast }$$

Proof

Consider horizontal and vertical sides of square separately

Note

$$\cot (\pi z)=\frac{e^{i\pi z}+e^{-i\pi z}}{e^{i\pi z}-e^{-i\pi z}}$$

So on horizontal sides of ${\Gamma }_N$ where $z=x\pm \left(N+\frac{i}{2}\right)i$ for $-\left(N+\frac{1}{2}\right)\le x\le \left(N+\frac{1}{2}\right)$

$$\begin{array}{rl}|\cot (\pi z)|& =\left|\frac{e^{i\pi \left(x\pm \left(N+\frac{i}{2}\right)i\right)}+e^{-\pi \left(x\pm \left(N+\frac{i}{2}\right)i\right)}}{e^{i\pi \left(x\pm \left(N+\frac{i}{2}\right)i\right)}-e^{-i\pi \left(x\pm \left(N+\frac{i}{2}\right)i\right)}}\right|\\ & \le \frac{e^{\pi (N+1/2)}+e^{-\pi (N+1/2)}}{e^{\pi (N+1/2)}-e^{-\pi (N+1/2)}}\\ & =\coth \left(\pi \left(N+\frac{1}{2}\right)\right)\end{array}$$

As $\coth (x)$ is a decreasing function for $x\ge 0$ then on horizontal sides of ${\Gamma }_n$ then

$$|\cot (\pi z)|\le \coth \left(\frac{3}{2}\pi \right)$$

On vertical sides then $x=\pm \left(N+\frac{1}{2}\right)+iy$ where $-N-\frac{1}{2}\le y\le N+\frac{1}{2}$
As $\cot (z+N\pi )=\cot (z)$ for any integer $N$ and $\cot \left(z+\frac{\pi }{2}\right)=-\tan (z)$ then

If $z=\pm \left(N+\frac{1}{2}\right)+iy$ for any $y\in \mathbb{R}$ then

$$|\cot (\pi z)|=|-\tan (iy)|=|-\tanh (y)|{\le }_1$$

So set

$$C=\max \left\{1,\coth \left(\frac{3\pi }{2}\right)\right\}$$