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 $$h\circ(g*f)\leq g* (h\circ f)\;?$$
If not, which additional assumption would lead to the statement holding?
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?
 A: 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}
