Is this kind of function convex ?
$f\left( {{w_1},{w_2},..,{w_n}} \right) = {a_1}{\left\| {h_1^T{w_1}} \right\|^2} + {a_2}{\left\| {h_2^T{w_2}} \right\|^2} + .... + {a_n}{\left\| {h_n^T{w_n}} \right\|^2} + b$
where ${h_1},{h_2}...{h_n}$ are vector that contain real number.
Also, we have that ${a_1},{a_2},..,{a_n} > 0$ and $b>0$
My guess is it kinda is but I dont know how to proof it ?
Knowing that $$f_i(\omega_i) := a_i \Vert h_i^\intercal \omega_i \Vert^2$$ is convex for each $i \in \{1,\dots,n\}$(proven easily by using properties of norms), it suffices to show, that a finite sum of convex functions is convex.
Let $g,h$ be convex functions. So we have for each $\tau \in (0,1)$, and for each $x,y$ in the domain of $g,h$ that $$ (i) \quad g(\tau x + (1-\tau)y) \leq \tau g(x)+(1-\tau) g(y) $$ $$(ii) \quad h(\tau x + (1-\tau)y) \leq \tau h(x)+(1-\tau) h(y)$$ Adding the inequalities we get: $$(g+h)(\tau x + (1-\tau y)) \leq \tau (g+h)(x) + (1-\tau) (g+h)(y) $$ Inductively it follows that finite sums of convex functions are convex.
In fact, have you noticed that if one includes all the $a_k$s into the norms which is possible because you assume them $>0$, and drop $b$ which is unimportant, i.e., if we work on :
$$f\left( {{w_1},{w_2},..,{w_n}} \right) = \left\| {h_1^T{w_1}} \right\|^2 + \left\| {h_2^T{w_2}} \right\|^2 + \cdots + \left\| {h_n^T{w_n}} \right\|^2$$
we can write $f$ in a compact way:
$$\underbrace{f\left( {{w_1},{w_2},..,{w_n}} \right)}_{f(W)}=\|diag(H^TW)\|^2\tag{1}$$
$H$, resp. $W$, being the matrix whose rows are the $h_k$s, resp. the $w_k$s, and "diag" is the "diagonal extraction operator".
Out of expression $(1)$, the proof of convexity is a matter of calculation on a matrix space, this time.
