# Is 'Jensen type inquality' true in the following form $h\circ(g*f)\leq g* (h\circ f)$

Given a convex function $$h$$ and some other integrable functions $$g$$ and $$f$$, is it true that $$h\circ(g*f)\leq g* (h\circ f)\;?$$

Note that $$\circ$$ denotes composition, while $$*$$ denotes convolution.

My idea:

$$h\bigg(\int_{0}^{t}g(t-s)f(s)ds\bigg)=h\bigg(\int_{0}^{t}f(s)\mu_{g}(ds)\bigg),$$

where $$\mu_{g}$$ is the measure with density $$g(s)$$ with respect to the Lebesgue measure. I assume now we can apply Jensen, and then we are done?

• Jensen's inequality requires $\mu_g$ to be probability measure. Jul 7, 2022 at 23:16

As geetha290krm points out in the comment, Jensen's inequality requires you to have a probability measure. For your case, suppose $$g$$ is non-negative and $$\int_{\mathbb R} g(s) ds = 1$$ Then we can define a probability measure on $$\mathbb{R}$$ by $$\mu_{g;t}(A) = \int_{\mathbb R} \mathbb{1}_{A}(s)g(t-s) ds = (\mathbb{1}_A \star g)(t)$$ It $$h$$ is convex, then Jensen's inequality implies your inequality: $$h\circ (g \star f)(t) = h\left(\int_{\mathbb{R}} f(s)g(t-s) ds\right) = h\left(\int_{\mathbb R} f(s) \mu_{g;t}(ds)\right) \le \int_{\mathbb R} h(f(s)) \mu_{g;t}(ds) = ((h\circ f)\star g)(t)$$
If $$g(s)$$ is non-negative but $$\int_{\mathbb{R}} g(s) ds = C \ne 1$$, the best we can do is $$h\circ(f\star g)(t)= h\left(\int_{\mathbb{R}} C f(s)\frac{g(t-s) ds}C\right) \le \int_{\mathbb R} h(C f(s)) \frac{g(t-s) ds}C = \frac1C (h(C f(t)) \star g(t))$$
Your inequality does not hold in general if $$\int_{\mathbb{R}} g = C \ne 1$$ or $$g$$ is negative on a set of positive measure. Some counterexamples: If $$C>1$$, let $$f(x) = 1$$ and $$h(x) = x^2$$: Then $$h\circ(f\star g)(t) = h(C) = C^2 > C = (h\circ f)\star g$$ If $$C<1$$, then let $$h(x) = 1$$. Then $$h\circ(f\star g)(t) = 1 > C = (h\circ f)\star g$$ If $$C = 1$$ and $$g(x) <0$$ on some set with positive measure: Let $$A$$ be some set such that $$\int_A g < 0$$ and let $$f(x) = 1-\mathbb{1}_{A}(-x)$$ and let $$h(x) = x^2$$. Then $$\begin{eqnarray} h(g\star f)(0) &=& \left(\int_{\mathbb{R}} (1-1_A(-s))g(-s)ds\right)^2 = \left(1-\int_{A} g\right)^2\\ &>& 1-\int_{A} g = \int_{\mathbb R} (1-\mathbb{1}_{A}(-s))g(-s)ds \\&=& (h\circ f)\star g (0) \end{eqnarray}$$