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.

  • $\begingroup$ 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. $\endgroup$ Commented Jan 25 at 20:03
  • 2
    $\begingroup$ 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” $\endgroup$ Commented Jan 26 at 9:58
  • $\begingroup$ @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. $\endgroup$ Commented Jan 26 at 13:35
  • $\begingroup$ Yes I completely agree $\endgroup$ Commented Jan 26 at 13:59

1 Answer 1


$\newcommand{\na}{\nabla}\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$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 $|h|<\eta(\de)$. $\quad\Box$


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