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I have the following question. I have to find a $\delta>0$ such that for all complex numbers $x,y$ the following holds true - \begin{equation} \frac{1}{2\pi}\int_0^{2\pi}|x+e^{it}y|\,dt \ge (|x|^2+\delta|y|^2)^{1/2}. \end{equation}

I have proceeded in the following way. Clearly, if $x=0$, then the requirement on $\delta$ is that $\delta\le 1$. So assume that $x\ne 0$. Then we write $x=re^{i\varphi}$ and $y=se^{i\tau}$ \begin{eqnarray} LHS &=& \frac{1}{2\pi}\int_0^{2\pi}|x+e^{it}y|\,dt \\ &=& \frac{r}{2\pi}\int_{\tau-\varphi}^{2\pi+\tau-\varphi}|1+e^{it}r^{-1}s|\,dt \\ &=& \frac{r}{2\pi}\int_{\tau-\varphi}^{2\pi+\tau-\varphi}(|1+e^{it}r^{-1}s|^2)^{1/2}\,dt \\ &=& \frac{r}{2\pi}\int_{\tau-\varphi}^{2\pi+\tau-\varphi}(1+r^{-2}s^2+2r^{-1}s\cos t)^{1/2}\,dt \end{eqnarray} I am stuck at this point. What inequality should I use to get the desired inequality.

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This question is in the context of 2-uniform PL convexity. Here the task is to prove that $\mathbb{C}$ is 2-uniform PL convex. – Nirakar Neo Mar 4 '13 at 12:42
up vote 1 down vote accepted

I think the inequality is not correct if we are looking at 2-uniform PL convex (see page 750) .

The case if a power of 2 should be on the integrad and no square root on the RHS.

Note if $y=0$ the equality holds for all $\delta \in \mathbb{R}$. Hence let $y \neq 0$ and let

\begin{align*} x &=re^{i\theta} \\ y &=Re^{i\phi} \end{align*}

First note

\begin{align*} \left| x+e^{it}y \right|^2 &= \left( x+e^{it}y \right)\overline{\left( x+e^{it}y \right)} \\ &= \left| x \right|+2\text{Re} \left(\overline{x}ye^{it} \right) +\left| y \right| \end{align*}

and \begin{align*} \text{Re}\left(\overline{x}ye^{it}\right) &= \text{Re}\left(rR\; e^{i\left(t+\phi-\theta \right)} \right) \\ &= rR \cos{ \left( t+\phi-\theta \right)} \end{align*}


\begin{align*} \frac{1}{2 \pi} \int_{0}^{2\pi}\!{\left|x+e^{it}y \right|^2}\,\text{d}t &= \frac{1}{2 \pi} \int_{0}^{2\pi} \! {\left|x\right|^2+ 2rR \cos{ \left( t+\phi-\theta \right)}+\left|y\right|^2}\,\text{d}t \\ &= \left|x\right|^2+\left|y\right|^2 + \frac{rR}{\pi}\int_{0}^{2\pi}\! {\cos{ \left( t+\phi-\theta \right)}} \,\text{d}t\\ &= \left|x\right|^2+\left|y\right|^2 \\ \end{align*}

Hence \begin{align*} \left|x\right|^2+\left|y\right|^2 \geq \left|x\right|^2+ \delta \left|y\right|^2 \; \iff \; \delta \leq 1 \end{align*}

The case if I am wrong.

An idea for a way to proceed:

\begin{align*} \text{Re}\left(\overline{x}ye^{it}\right) &= \text{Re}\left(rR\; e^{i\left(t+\phi-\theta \right)} \right) \\ &= rR \cos{ \left( t+\phi-\theta \right)} \\ &\geq -rR \end{align*} Since $-1 \leq \cos{ \left( t+\phi-\theta \right)} \leq 1$.

Hence \begin{align*} \frac{1}{2 \pi} \int_{0}^{2\pi}\!{\left|x+e^{it}y \right|}\,\text{d}t &= \frac{1}{2 \pi} \int_{0}^{2\pi} \! \sqrt{\left|x\right|^2+ 2rR \cos{ \left( t+\phi-\theta \right)}+\left|y\right|^2}\,\text{d}t \\ &\geq \frac{1}{2 \pi} \int_{0}^{2\pi} \!\sqrt{\left|x\right|^2 - 2rR + \left|y\right|^2}\,\text{d}t \\ &=\frac{1}{2 \pi} \int_{0}^{2\pi} \!\sqrt{\left(\left|x\right| - \left|y\right| \right)^2}\,\text{d}t \\ &= \left|x\right|-\left|y\right| \end{align*}

Hence \begin{align*} \left|x\right|-\left|y\right| & \geq \left( \left|x\right|^2+ \delta \left|y\right|^2 \right)^{1/2} \end{align*} But this will lead to a contradiction if $\delta>0$. So I must of lost too much information in $\text{Re}\left(\overline{x}ye^{it}\right)\geq -rR$.

Alternatively you can try to expand the square root form of the integrand by the Binomial expansion to some order and then see where integration leads you to.

Visualise the problem and think of the vector $y$ added to $x$ giving a vector $x+y$. Now the introduction of $e^{it}$ turns the $y$ vector such that the point $z=x+y$ rotates about a circle of radius $|y|$ centred at $x$. Hence if you plot $|z|$ against $t$ you will get a periodic graph of period $2\pi$. So the integral will be the area of this graph between $0$ and $2\pi$ ("averaged" by $1/2\pi$). Consequently, this should be bigger than the area represented by the RHS. According to your inequality.

Hope this helps.

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