Let $f \colon \mathbb R^+ \to \mathbb R^+$ be a strictly increasing, positive convex function and let $f^{-1} \colon \mathbb R^+ \to \mathbb R^+$ be its left inverse. Define $$ v(h) = f^{-1} \left( \int f(h(x)) \, dx \right). $$ I claim that $v$ is convex. For example, when we restrict the domain of $f$ to the non-negatives and set $f(x) = x^p$ for some $p \ge 1,$ this becomes the statement that $L^p$ norms are convex on positive functions. When $f(x) = e^x,$ this says that the "partition function" $\int e^{- H(x)} \, dx$ is log-convex with respect to its energy $H$.

These two special cases are well-known. My question is: is my statement true in general?

The best that I came up with is the following. Let $f'$ give a subdifferential for $f.$ Then, if $v$ were convex, taking a variational derivative of $v$ at $g$ would give the lower bound $$ v(h) \ge v(g) + \frac{1}{f'(v(g))} \int f'(g(x)) (h(x) - g(x)) \, dx $$ for any other function $h.$ In fact, because the RHS equals $v(h)$ when $g = h,$ bounds of this form would tell us that the restriction of $v$ to the domain where $v < +\infty$ is a supremum of affine functions, which would prove that that $v$ is convex over its whole domain.

We can rearrange our bound to read $$ f'(v(g)) (v(h) - v(g)) \ge \int f'(g(x))(h(x) - g(x)) \, dx. $$ For example, in the case of $f(x) = e^x,$ this would read $$ \left( \int e^{g(x)} \, dx \right) \left( \ln \int e^{h(x)} \, dx - \ln \int e^{g(x)} \, dx \right) \ge \int e^{g(x)}(h(x) - g(x)) \, dx $$ In the case of $f(x) = x^2,$ this would read $$ \lVert g \rVert (\lVert h \rVert - \lVert g \rVert) \ge \langle g, h - g \rangle $$ in terms of the $L^2$ inner product.

However, here I'm stuck. Since $f'(g(x))$ is a subdifferential, we could bound the integrand on the RHS as $$ f'(g(x))(h(x) - g(x)) \le f(h(x)) - f(g(x)), $$ which upper bounds the RHS by $$ \int f(h(x)) - f(g(x)) \, dx = f(v(h)) - f(v(g)). $$ However, our desired quantity of $f'(v(g))(v(h) - v(g))$ is in fact a lower bound for this last expression. Is my statement even true?


1 Answer 1


In fact the statement I am trying to prove is not true.

Let's reduce it to a simpler case. Let our measure space be the counting measure on a set with two elements, in which case we can write $$ v(x, y) = f^{-1}(f(x) + f(y)). $$ Now consider $\gamma(t) = v(t, y)$ for some fixed $y$. For $v$ to be convex, $\gamma$ needs to be convex. Supposing that $f$ is differentiable, we can compute $$ \frac{d}{dt} \gamma(t) = \frac{f'(t)}{f'(\gamma(t))}. $$ The restriction $\gamma$ is convex only when this function is non-decreasing, meaning informally that $f'(t)$ increases more quickly than $f'(\gamma(t)).$ However, it is easy to construct a function where this does not happen. For example, let $$ f(x) = \begin{cases} x & : x \le 1 \\ 2 (x - 1) + 1 & : \text{otherwise}. \end{cases} $$ Then $\gamma(t) = f^{-1}(f(t) + f(1/2))$ is not convex, because its derivative jumps from $1$ to $1/2$ as $t$ passes through $1/2.$

On the other hand, I'm still interested in knowing under what conditions my claim remains true.


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