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Suppose that $H$ is a real Hilbert space and that $f:H \to \mathbb{R}$ is differentiable in the Frechet sense. Then we can think of the derivative as a function $f': H \to H^* = H$. Suppose that this function also has a continuous Frechet derivative $f'': H \to B(H)$. My question is now: is it true that $f$ is convex if and only if $f''(x)$ is a positive operator for all $x \in H$, i.e., that $\langle f''(x)y, y \rangle \geq 0$ for all $x,y \in H$?

The question is motivated by the analogous case for $C^2$ functions $f: \mathbb{R} \to \mathbb{R}$, in which case $f$ is convex if and only if $f''(x) \geq 0$ for all $x \in \mathbb{R}$.

Thanks in advance.

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For the real case, $f''(x) \geq 0$ (ie, strictness is not necessary) is equivalent to convexity. – copper.hat May 16 '12 at 8:05
@copper.hat Right, that's TEXing laziness on my part. Thanks also for your answer. – user12014 May 16 '12 at 8:18
up vote 1 down vote accepted

It is true. Convexity is a 'one dimensional' characteristic, so you only need to examine it on one dimensional (affine) spaces.

You need only look at the function $\phi_{x,h}(\lambda) = f(x+\lambda h)$. The function $f$ is convex iff $\phi_{x,h}$ is convex ($\forall x,h$, of course).

Since $\phi_{x,h}''(\lambda) = \langle f''(x+\lambda h)h, h \rangle$, we see that $f$ is convex iff $f''(x) \geq 0 $ (ie, positive semi-definite), $\forall x$.

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