f is convex if and only if $f(x):=\left \langle Mx,x\right \rangle \ge 0$

Question

I have to show for $M \in \mathbb{R}^{(N \times n)\times (N \times n)}$ symmetric Matrix and $f:\mathbb{R}^{N\times n} \longrightarrow \mathbb{R}$ with $f(x):= \langle Mx,x\rangle$ that:

f is convex if and only if $f(x) \ge 0.$

where $\langle\cdot,\cdot\rangle$ is the standard inner product on $\mathbb{R}^{N\times n}$.

I don't know how to proof that. Perhaps someone has a hint.

Thanks in advance!!!

Answer 1

$f$ is convex $\iff$ the Hessian of $f$ is positive semi-definite. The Hessian of $f$ is the martix $2M.$

Hence: $f$ is convex $ \iff M$ is positive semi-definite $ \iff \left \langle Mx,x\right \rangle \ge 0$ for all $x$.

Answer 2

That does not seem right to me. Let $ \mathbb{R}^{{n \cdot (n + 1)}/{2}}$ be the space of the $n\times n$ symmetric matrices . Not every convex function $ f: \mathbb{R}^{{n \cdot (n + 1)}/{2}}\to \mathbb{R} $ is of the form $ f (X) = \langle M \cdot X, X \rangle $. The function $$ \varphi(X)=\|X\|_{\infty}= \max\left\{|X_{ij}|: X=(X_{ij})_{n\times n}\in \mathbb{R}^{{n \cdot (n + 1)}/{2}}\right\} $$ is convex because that every norm in a finite dimensional space is a convex fuction. It is a consequence of triangular inequality: for all $t\in[0,1]$ we have $$ \|t\cdot X+(1-t)Y\|_\infty\leq \|t\cdot X\|_{\infty}+\|(1-t)\cdot Y\|_\infty\leq t\cdot\|X\|_{\infty}+(1-t)\cdot \|Y\|_\infty $$ It is easy to see that there is no constant matrix $M\in \mathbb{R}^{({n \cdot (n + 1)}/{2})\times ({n \cdot (n + 1)}/{2}) }$ such that $$ \langle M\cdot X, X\rangle_{\mathbb{R}^{{n \cdot (n + 1)}/{2}}}=\|X\|_\infty=\max_{ij}|X_{ij}| $$