Swapping $L^p$ norm and integral

Question

Is it true that $$\left\|{\int_0^T}f(x,t)dt\right\|_{L^p(\mathbb{R}^n)}\leq\int_0^T\|f(\cdot,t)\|_{L^p(\mathbb{R}^n)}dt$$ where $T\leq{\infty}$?

I can prove a similar inequality in the ase $T<\infty$, by using Holder's inequality and Fubini's theorem to get $$\left\|{\int_0^T}f(x,t)dt\right\|_{L^p(\mathbb{R}^n)}\leq{T^{1/p'}}\left(\int_0^T\|f(\cdot,t)\|_{L^p}^pdt\right)^{1/p}$$

but can't see how to reach the final inequality from here, and can't see that it would hold true for the case $T=\infty$.

Minkowski's integral inequality. – Mathemajician Sep 09 '20 at 08:32 — Mathemajician, Sep 09 '20 at 08:32
Ref. https://en.wikipedia.org/wiki/Minkowski_inequality – Kavi Rama Murthy Sep 09 '20 at 08:36 — Kavi Rama Murthy, Sep 09 '20 at 08:36

score 2 · Accepted Answer · answered Sep 09 '20 at 10:26

An elegant way to prove the inequality $\|\int f d\mu\|\le \int \|f\| d\mu$ for functions with values in a Banach space $(X,\|\cdot\|)$ is the norm formula $\|x\|=\sup\{ |\phi(x)|: \phi\in X^*, \|\phi\|^*\le 1\}$ (which is a consequence of Hahn-Banach). For $\phi\in X^*$ with $\|\phi\|^*\le 1$ you get $$ \left|\phi(\int fd\mu)\right|=\left|\int \phi(f)d\mu\right|\le \int \|f\|d\mu$$ where the last inequality uses just the monotonicity of the integral.

Swapping $L^p$ norm and integral

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