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Call the differentiable function $f: \mathbb K \to \mathbb R$ on some compact convex $K \subset \mathbb R^n$strictly convex to mean $f(\lambda x + (1- \lambda)y) < \lambda f(x) + (1-\lambda)f(y)$ for all $x,y \in K$ and $\lambda \in (0,1)$.

For $n = 1$ strict convexity is equivalent to the derivative $f'$ being strictly increasing. In that case we can show $\min f$ occurs at the minimiser of $|f'|$.

Consider first the case that $f'(a) = 0$ for some $a \in K$. By strict convexity there is only one such $a$. Clearly $a$ is the unique minimiser of $|f'|$. But $f'(a) = 0$ also means $a$ is a local minimum, and then convexity implies $a$ is the global minimum.

Otherwise $f'$ is strictly positive or negative negative. First assume the former. Without loss of generality we have $K =[0,1]$. Then since $f'$ is strictly increasing $|f'(x)| = f'(x)$ has unique minimum at $0$. Also by writing $f(x) = f(0) + \int_0^x f'(y) \, dy$ as the integral of a strictly positive function we see $f$ is also strictly increasing hence also has unique minimum at $0$. A symmetric argument shows for $f'$ strictly negative both functions have unique minimum at $1$.

Of course for $n>1$ the derivative of $f$ is the gradient vector $\nabla f$ of partial derivatives. Since there is no notion of a vector being positive the proof fails to generalise. Also the higher-dimensional analogue of $f(x) = \int_0^x f'(y) \, dy$ is Stoke's theorem which does not recover values of $f$ at a point. Rather the left-hand-side becomes the integral over the boundary of whatever region we're integrating over on the right.

Nevertheless my intuition says some variant of the result above should hold in higher dimensions. Does anyone have any ideas or a good reference?

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  • $\begingroup$ Your definition of strictly convex is wrong (consider $\lambda=0$). Strict convexity is not equivalent to an increasing derivative, since a convex function may not have a derivative. $\endgroup$ – LinAlg Dec 1 '18 at 19:54
  • $\begingroup$ You're right. Changed to $\lambda \in (0,1)$. We are assuming all functions are differentiable. $\endgroup$ – Daron Dec 2 '18 at 13:17
  • $\begingroup$ if $f$ is vector valued, you can look at $g(t) = f(x+td)$. $\endgroup$ – LinAlg Dec 2 '18 at 14:56
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No, this statement is not true in general, here is a counter-example.

Let $f(x,y) := \frac{1}{2}x^2 + y^2$ which is a strongly convex differentiable function on $\mathbb{R}^2$ (in particular it is strictly convex). Let's take a point $(\bar x, \bar y) = (1,1)$, and consider the following closed convex constraint $K:=\{(x,y) \ | \ y \geq \frac{-1}{2} x + \frac{3}{2} \}$.

First, we check that $(\bar x,\bar y)$ is the unique minimum of $f$ restricted to $K$. For this, we compute $\nabla f(\bar x, \bar y)=(1,2)$, and observe that $y = \frac{-1}{2} x + \frac{3}{2}$ is the equation of the line passing trough $(\bar x,\bar y)$, and being orthogonal to $\nabla f(\bar x, \bar y)$. Since the gradient of a differentiable function is always orthogonal to the sublevel set, we deduce that this line is the tangent to the sublevel set of $f$ at $(\bar x,\bar y)$. This implies that $(\bar x,\bar y)$ is a minimum of $f$ restricted to $K$. Its uniqueness follows from the strong convexity of $f$.

Now, let's look at the minimum of $\Vert \nabla f(x,y) \Vert$ on $K$ (here the norm is the Euclidean one). At $(\bar x,\bar y) \in K$ we have $\Vert \nabla f(\bar x, \bar y)\Vert =\sqrt{5}$. But if you take for instance $(x,y) = (\frac{3}{2},\frac{3}{4}) \in K$, you can see that $\Vert \nabla f(x, y)\Vert =\sqrt{\frac{9}{2}}$ which is strictly less than $\sqrt{5}$, and this contradicts your claim. (Remark: Actually, it is not hard to verify that $(\frac{3}{2},\frac{3}{4})$ minimizes $\Vert \nabla f(x,y) \Vert$ over $K$).

Note that here the constraint $K$ is not compact. If you want a compact counter-example, just take any compact convex subset of $K$ containing both $(1,2)$ and $(\frac{3}{2},\frac{3}{4})$.

EDIT: If you want to properly handle the constraint, you cannot avoid making use of the tangent/normal cones to the constraint. If $N_K(x)$ denotes the normal cone to $K$ at $x$, a well-known first order condition for $x$ to be a minimizer of $f$ over $K$ is:

$$0 \in N_K(x) + \{\nabla f(x) \}.$$

The above condition can be rewritten as $\nabla f(x) + \text{proj}_{N_K(x)}(-\nabla f(x))=0$, where $\text{proj}_{N_K(x)}(-\nabla f(x))$ is the projection of $-\nabla f(x)$ onto the cone $N_K(x)$. From this it can be directly seen that the minimizers of $f$ coincide with the ones of $x \mapsto \Vert \nabla f(x) + \text{proj}_{N_K(x)}(-\nabla f(x)) \Vert$. If we look at this function in your one-dimensional case, with $K=[-1,1]$, we obtain

$$\Vert \nabla f(x) + \text{proj}_{N_K(x)}(-\nabla f(x)) \Vert= \begin{cases} 0 & \text{ if } x \text{ minimizes $f$ on $K$,}\\ \vert f'(x) \vert & \text{ else.} \end{cases}$$

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