I have been stuck on this problem for a while.
After I use the definition of convexity and some algebra, I end with something like this:
$$ \lambda f(x^{(1)}) + (1-\lambda)f(x^{(2)}) + \lambda(1-\lambda) (x^{(2)T}Mx^{(1)} + x^{(1)T}Mx^{(2)} - f(x^{(1)}) - f(x^{(2)})) $$
And I need to proof that this is indeed $\ge$ than $\lambda f(x^{(1)}) + (1-\lambda)f(x^{(2)})$ however I can't see that it is trivial.
As an alternative, I have done the algebra a little differently, and ended up with the inequality:
$$ \lambda(1-\lambda)(f(x^{(1)}) + f(x^{(2)}) - x^{(2)T}Mx^{(1)} - x^{(1)T}Mx^{(2)}) \ge 0 $$
Which I am not entirely sure it is true.
If you guys need more specification, I will go into more detail. Thanks for your help.
I assume you mean $M$ is a SPD matrix here. If $M = -1$ and $f(x) = -x^2$, we cannot obtain the conclusion.
If $M$ is a SPD matrix, we can write its Cholesky factorization $M=LL^T$.
$f(x) = x^TMx = ||L^Tx||_2^2$ is actually a norm on $\mathbb{R}^n$ and therefore is a convex function.
More directly, we have $\nabla f = 2Mx$ and hence
$$(y - x)^T M (y - x) \geq 0$$ $$y^TMy \geq x^TMx + 2x^TM^T(y-x)$$ $$f(y) \geq f(x) + \nabla f^T(y-x)$$
for $x,y \in \mathbb{R}^n$, which is a N&S condition for a differentiable function to be convex.