If we consider a $\mathcal C^1$ function $f:\mathbb R^d\rightarrow \mathbb R$ that is convex, is it true that $f$ is strictly convex $\iff$ $\nabla f$ is injective?
This result is clearly true for $d=1$ since strict convexity is equivalent to having a strictly increasing hence injective derivative. Therefore, does this one-dimensional property generalize to a function of $\mathbb R^d$?
On
If $f$ is not strictly convex, then there would exist $x,y\in\mathbb{R}^n$ such that for all $t\in[0,1]$, $$ (1-t)f(x) + tf(y) = f((1-t)x+ty). $$ Differentiating with respect to $t$ yields, $$ f(y) - f(x) = \nabla f((1-t)x+ty)\cdot (y-x) $$ for all $t\in [0,1]$. Namely, $f(y) = f(x) +\nabla f(x)\cdot (y-x)$. As $f\in C^1$ and convex, $\ell(z) = f(x) +\nabla f(x)\cdot (z-x)$ is the unique supporting hyperplane at $x$. As $(y, f(y))$ is contained on this hyperplane, then this is also the unique supporting hyperplane at $y$, i.e. $\nabla f(x) = \nabla f(y)$. Therefore $\nabla f$ is not injective.
Consider $g(t)=f(tx_1+(1-t)x_2).$ Then $g$ is strictly convex on $[0,1 ]$ like $f$. As a consequence, from the one dimensional case. $g'(0)=(x_1-x_2)^Tf'(x_2)<g'(1)=(x_1-x_2)^Tf'(x_1)$, which implies $f'(x_1)\neq f'(x_2).$