# Is the gradient of a strictly convex, continuously differentiable function a homeomorphism?

Let $$X\subseteq\mathbb{R}^n$$ be a convex set. Let $$f:X\to\mathbb{R}$$ be a strictly convex function that is differentiable on the (non-empty) relative interior of $$X$$.

$$\nabla f$$ is a bijection, but is it a homeomorphism? This question is in the context of information geometry and Bregman divergences, where $$\nabla f$$ induces a change of coordinates.

• If ∇𝜙 takes the same value at two distinct points P, Q ∈ 𝑋, then consider the restriction f = 𝜙 | [P, Q] of 𝜙 to the line segment [P, Q] connecting P and Q. This reduces the situation to the 1-dimensional case. Commented Jan 25 at 20:03
• Note that every continuous injective map $F:\Omega\to\mathbb R^n$ on an open subset $\Omega$ of $\mathbb R^n$ is a homeomorphism with its image $F(\Omega)$, which is an open subset of $\mathbb R^n$, by the “theorem of invariance of domain” Commented Jan 26 at 9:58
• @PietroMajer : Good point. I felt that something like this theorem should be true, but did not know of it. The answer below still seems to make sense, being elementary, in contrast to that theorem. Commented Jan 26 at 13:35
• Yes I completely agree Commented Jan 26 at 13:59

$$\newcommand{\na}{\nabla}\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$$Let $$Y$$ denote the interior of $$X$$. Then indeed $$g:=\na f$$ is a homeomorphism of $$Y$$ onto $$g(Y)$$.

You already know that $$g$$ is bijective. So, it remains to show that $$g$$ and $$g^{-1}$$ are continuous.

That $$g$$ is continuous follows e.g. from the last sentence of Theorem 24.5.

It remains to show that $$g^{-1}$$ is continuous. By easy manipulations, this reduces to the following:

Claim: Suppose that $$0\in Y$$, $$f(0)=0$$, $$g(0)=0$$, and $$Y\ni h\to0$$. Then $$g^{-1}(h)\to0$$.

Proof: Take any real $$\de>0$$ such that $$x\in Y$$ whenever $$|x|\le\de$$, where $$|\cdot|$$ denotes the Euclidean norm. Since $$f$$ is strictly convex (and hence continuous on $$Y$$), there is some real $$\ep(\de)>0$$ such that $$f(x)\ge\ep(\de)$$ whenever $$|x|=\de$$, so that for $$\eta(\de):=\frac{\ep(\de)}\de$$ we have $$\begin{equation*} x\in X\ \&\ |x|\ge\de\implies f(x)\ge\eta(\de)\,|x|. \tag{1}\label{1} \end{equation*}$$

Let $$g_h(x):=f(x)-h\cdot x$$, where $$|h|<\eta(\de)$$. Then, by \eqref{1}, $$\begin{equation*} g_h(x)\ge\Big(\eta(\de)-|h|\Big)|x|>0=g_h(0) \end{equation*}$$ if $$x\in X$$ and $$|x|\ge\de$$. So, the continuous strictly convex function $$g_h$$ attains its minimum on $$X$$ at a unique point $$x_h\in X$$ such that $$|x_h|<\de$$, so that for all $$x\in X$$ we have $$g_h(x)\ge g_h(x_h)$$ or, equivalently, $$f(x)\ge f(x_h)+h\cdot(x-x_h)$$, so that $$h=g(x_h)$$ and $$x_h=g^{-1}(h)$$. Thus, $$|g^{-1}(h)|<\de$$ if $$|h|<\eta(\de)$$. $$\quad\Box$$