Inequality like Minkowski's for integrals I'm reading an article and the author write something that i can't understand: "it is easy to see that"
$$\int_0^t\int_{\mathbb{R}^2}K(t-s,x-y)u^p(s,y)dyds\geq \int_0^t\left(\int_{\mathbb{R}^2}K(t-s,x-y)u(s,y)dy\right)^pds, p>1$$ where
$$K(t,x_1,x_2)=\dfrac{\sqrt{2}}{2\pi t^{3/2}}e^{-\frac{x_1^2}{2t}-\frac{x_2^2}{t^2}},$$
and $u$ is a function non-negative, continuous and bounded.
I can't see the "jump". I'm trying something with the Minkowski's inequalities for integrals but i didn't get anything so far. Can anyone help me, please?
 A: This is use of Jensen's inequality.
You can easily check that
$K$ is probability density function of 2-dimensional normal distribution,
and function $f(x) = |x|^p$ is convex for $p>1$.
Inner integral on the LHS is actually expected value  $\mathbb E [u(Z_1, Z_2)^p]$ for some appropriate normal variables, and we have
$$
\mathbb E [u(Z_1, Z_2)^p] \geq (\mathbb E [u(Z_1, Z_2)])^p.
$$
A: Define a measure $\mu$ by $$ d \mu = K(t-s,x-y) dy .  $$ Then, by Hölder's inequality, \begin{align*} 
 \bigg ( \int_{\mathbb R^2} u(s,y)K(t-s,x-y) \, d y \bigg )^p  &= \bigg ( \int_{\mathbb R^2} u(s,y) \, d \mu \bigg )^p \, \\
&\leqslant \mu \big ( \mathbb R^2\big )^{1-1/p} \int_{\mathbb R^2} u^p(s,y) \, d \mu \\
&= \mu \big ( \mathbb R^2\big )^{1-1/p} \int_{\mathbb R^2} u^p(s,y) K(t-s,x-y) dy.
\end{align*} Now letting $\tau =s-t$ (for simplicity) and making the change of variables $z=x-y$, we obtain \begin{align*}
\mu (\mathbb R^2) &= \int_{\mathbb R^2} K(\tau,x-y) \, d y \\
&= \int_{\mathbb R^2} K(\tau,z) \, d z \\
&= \frac{\sqrt 2}{2\pi \tau^{3/2}} \int_{\mathbb R^2} e^{-\frac{z_1^2}{2\tau}-\frac{z_2^2}{\tau^2}} \, d z \\
&= \frac{\sqrt 2}{2\pi \tau^{3/2}} \bigg (  \int_{-\infty}^{\infty} e^{-\frac{x_1^2}{2\tau}} \, d z_1 \bigg ) \bigg ( \int_{-\infty}^\infty e^{-\frac{z_2^2}{\tau^2}} \, d z_2 \bigg ) \\
&= 1
\end{align*} where I used that $$\int_{-\infty}^\infty e^{-\frac{x^2}{2\sigma^2}} = \sigma \sqrt{2\pi } $$ provided that $\sigma >0$. This implies the result.
