I've tried to prove/disprove the statement that $\left(\sum_{m=1}^M t_m\right)^2$ is convex for $\forall t_m\geq 0$, but the proving seems difficult to me. Below is part of my proof.
\begin{align} \left(\sum_{m=1}^M t_m\right)^2 = \underbrace{\sum_{m=1}^M t_m^2}_{term 1} + 2 \underbrace{\sum_{i=1}^M \sum_{j=1}^{i-1} t_i t_j}_{term 2} \end{align}
I find that term 1 is convex, thus taking term 2 under consideration. Let's define $f=t_i t_j$ with $i\neq j$. My aim is to prove or disprove $f$ is convex! Firt, I calculated the Hessian matrix of $f$ as \begin{align} A &= \left[ \begin{array}{cccc} 0 & 1 \\ 1 & 0 \end{array} \right]. \end{align} Second, I found that there are 2 ways to proceed:
- (case 1): if $\textbf{z}^H=[z_1^* ~ z_2^*]$ is a complex vector, then $\textbf{z}^H A \textbf{z} = 2 Re\left\{z_1z_2^*\right\}$. This result implies that $f=t_it_j$ is non-convex.
- (case 2): if if $\textbf{z}^H=[z_1^* ~ z_2^*] = [z_1~z_2]$ is a real vector, then $\textbf{z}^H A \textbf{z} = 2 z_1z_2$. This result implies that $f=t_it_j$ is convex with $z_1\geq 0$ and $z_2\geq 0$.
I get stuck at the above step. Could you help me to improve/give the proof of the statement? Thanks in advance.
Your approach does not work, because $f=t_i t_j$ is not convex. You can see that if you plot the areas in $\mathbb R^2$ where $f(x,y)=xy \geq0$.
Hint: Your function is a composition of a linear function and a convex function. It can be shown that these types of functions are convex again.