This question is migrated from MathStackExchange where it seemed to be too hard. I wonder does anyone here have any ideas?

Suppose $f: K \to \mathbb R$ is $\mathcal C^2$ and strictly convex on some compact convex $K \subset \mathbb R^n$. That means $f(a x + by) < a f(x) + bf(y)$ for all $x,y \in K$ and $a,b \in (0,1)$ with $a+b=1$.

Strict convexity implies $f$ has a unique minimum over $K$. In all examples I have found, this minimum also coincides with the minimum of $\|\nabla f\|$. I wonder if it is always true but am unable to find or invent a proof unless $n=1$.

For $n = 1$ the gradient $\nabla f(x)=f'(x)$ is just the derivative and strict convexity is equivalent to $f'$ 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 $n>1$ there is no notion of the gradient vector being *positive* and 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.

One fact I imagine is useful is the function $F(x) = \|\nabla f(x)\|^2$ is differentiable with gradient $\nabla F(x) = H(x) \nabla f(x)$ where $H(x)$ is the Hessian of $f$ at $x$. By strict convexity the Hessian is positive definite.

Now $F(x)$ achieves its minimum at some $a \in K$. Without loss of generality $a=0$. Unless $F(a) = 0$ we know the gradient vector $\nabla F(a)$ is normal to $K$. That means $K$ is contained in the halfspace $\{x \in \mathbb R^n: H(x) \nabla f(x) \cdot x\ge 0\}$. That tells us moving in the direction $\nabla f$ a small amount will increase $F$ since the directional derivative is $H(x) \nabla f(x) \cdot \nabla f(x) = \nabla f(x) ^T H(x) \nabla f(x) > 0$ by positive definiteness.

If $H(x) \nabla f(x) $ was parallel to $\nabla f(x) $ we would be done but in general this doesn't occur. For example consider the function $f(x) = x^2 + 2y^2$ over $K=[1,2]^2$. Clearly $f$ is minimised at $(1,1)$ where the gradient is $\nabla f(1) = (2,4)$. We can compute $F(x) = (2x)^2 + (4y)^2 = 4x^2 + 16y^2$. This is also minimised at $(1,1)$ where the gradient is $\nabla F(1) =(8,32)$. In this case the gradients are not parallel but they are both normal to $K$.

**Note:** There is something perverse about the statement that $f$ has the same minimiser as $\|\nabla f \|$ because $f$ is coordinate-independent (obviously) but $\|\nabla f \|$ is not -- it depends on some choice of inner-product to define. However I don't think this should be a problem since changing the inner product should only amount to *skewing* the domain. That transform takes straight lines to straight lines so it should preserve convexity and take the points we're interested to onto each other.