# "Conjugation" of integration by an increasing convex function is convex

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?

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 $$1/2.$$