If $f:\mathbb R^n\to\mathbb R$ is strictly convex, i.e. its hessian is everywhere positive definite, does that imply its gradient is bijective? To ensure the well-definedness of the Légendre transform, which is usually used on s.c. functions, I need the gradient to be bijective, so as to be sure that for any $p\in\mathbb R^n$ there exists one and only one $x\in\mathbb R^n$ for which $p=\nabla f(x)$...
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$\begingroup$ you might want to check out this question: math.stackexchange.com/questions/1372692/…. Obviously injective does not implies surjetive... $\endgroup$– user251257Sep 30, 2015 at 15:34
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$\begingroup$ Partial duplicate, indeed. But it would be nice to have things all in one place, wouldn't it? $\endgroup$– MickGSep 30, 2015 at 15:35
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1$\begingroup$ if $a = b$ then $h^T a = h^T b$ for any $h$. By negation you get $h^T a \ne h^T b$ implies $a \ne b$. As for your question, consider $f(x) = \exp(x)$. $\endgroup$– user251257Sep 30, 2015 at 15:38
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$\begingroup$ Your example seems to say surjectivity is not implied, but then the transform cannot be defined on all $\mathbb R^n$! Is that so? So is the transform perhaps defined only on the codomain of the gradient? It must be so. $\endgroup$– MickGSep 30, 2015 at 15:41
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$\begingroup$ Legendre transformation is always defined for convex functions. I have no idea where you need strict convexity. Also notice, that strict convexity does not imply positive definite hessian (only the reverse direction is correct) $\endgroup$– user251257Sep 30, 2015 at 15:45
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2 Answers
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Consider $f$ of class $C^1$ and strictly convex: $f(\lambda x+(1-\lambda) y)<\lambda f(x)+(1-\lambda)f(y)$ when $x\neq y$ and $\lambda \in (0,1)$.
Injectivity: look at $t \mapsto f(tx_1+(1-t)x_2)$ for $t \in [0,1]$. This function is strictly convex. Writing $f'(0)<f'(1)$ gives that $\nabla f(x_1)\neq \nabla f(x_2)$.
Surjectivity: This does not hold in general. It is enough to have a sublinear/linear growth in one direction. The exponential is an easy example. $x \mapsto \sqrt{1+x^2}$ is another example.
On the other hand, if the function $f$ has super-linear growth towards infinity, $f(x)/|x|\to \infty$, then by definition, $$ \min_{x \in \Bbb{R}^d}f(x)-x\cdot p$$ has a unique solution $x(p)$ for a given $p$ (minimization of a strictly convex function which is infinity at infinity). This solution verifies $\nabla f(x(p)) = p$. Since this can be done for every $p$, the gradient is surjective in this case.
answered Dec 19, 2023 at 22:21
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Since nobody closed or answered this, I will self-answer and self-accept, so as to make it answered.
That the Hessian be everywhere p.d. implies the gradient is injective, as proved here. Basically, the proof uses Taylor expansion with integral form for the remainder, then using p.d.-ness to prove the remainder is always strictly positive.
$f(x)=e^x$ has everywhere p.d. Hessian (i.e. everywhere strictly positive second derivative) but is surely not surjective.
