I've found this version from this Wikipedia page. I've re-written the proof to make my understanding clear. Could you confirm if my attempt is correct?
Let
-
$(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space and $(E, \| \cdot \|)$ a Banach space.
-
$\|\cdot \|_p$ the $L_p$-norm of $L_p(X, \mu,E)$.
-
$f_1, \ldots, f_n, g_1, \ldots, g_n: X \to E$ are $\mu$-measurable.
-
$p,q >0$ such that $1/p+1/q = 1$.
Then $$ \int_{X} \sum_{k=1}^n \| f_k (x)\| \cdot \|g_k (x) \| \mathrm d \mu (x)\leq \left ( \sum_{k=1}^n \| f_k \|^p_p \right )^{1/p} \left (\sum_{k=1}^n \| g_k \|_q^q \right )^{1/q}.$$
My proof: Let's first consider the case where $n=1$, i.e., $$\int_{X} {\| f (x)\|} \cdot {\|g (x)\|}{} \mathrm d \mu (x) \leq \|f\|_p \cdot \|g\|_q.$$ WLOG, we assume $0< \| f \|_p, \| g \|_q < \infty$. Then the above inequality is equivalent to $$\int_{X} \frac{\| f (x)\|}{\|f\|_p} \frac{\|g (x)\|}{\|g\|_q} \mathrm d \mu (x) \leq 1.$$
By Young's inequality, \begin{align}
\int_{X} \frac{\| f (x)\|}{\|f\|_p} \frac{\|g (x)\|}{\|g\|_q} \mathrm d \mu (x) &\le \int_{X} \frac{\| f (x)\|^p}{p\|f\|_p^p} \mathrm d \mu (x) + \int_X\frac{\|g (x)\|^q}{q\|g\|^q_q} \mathrm d \mu (x)\\
&=\frac{1}{p\|f\|_p^p} \int_{X} \| f (x)\|^p \mathrm d \mu (x) + \frac{1}{q\|g\|^q_q} \int_X \|g (x)\|^q \mathrm d \mu (x)\\
&= \frac{1}{p\|f\|_p^p} \|f\|_p^p + \frac{1}{q\|g\|_q^q} \|g\|_q^q = \frac{1}{p}+\frac{1}{q}=1.
\end{align}
This completes the proof for $n = 1$. Let $[n] \triangleq \{1, \ldots, n\}$ and $([n], \mathcal P([n]), \nu)$ be the finite counting measure space. Then the product measure space $(X \times [n], \mathcal A \otimes \mathcal P([n]), \mu \otimes \nu)$ is also $\sigma$-finite. To simplify notation, let's $\lambda \triangleq \mu \otimes \nu$. Certainly, we can apply the result from the case $n=1$ to this particular product measure space. We define $f,g: X \times [n] \to E$ by $f(x,k) \triangleq f_k(x)$ and $g(x,k) \triangleq g_k(x)$. Clearly, $f$ and $g$ are measurable. It follows that \begin{align}&\int_{X \times [n]} {\| f (x,k)\|} \cdot {\|g (x,k)\|}{} \mathrm d \lambda (x,k) \\
\leq &\left (\int_{X \times [n]}\|f(x,k)\|^p \mathrm d \lambda(x,k) \right )^{1/p} \cdot \left (\int_{X \times [n]}\|g(x,k)\|^q \mathrm d \lambda(x,k) \right )^{1/q}.\end{align}
Notice that all integrands are non-negative and measurable. So by Tonelli's theorem, we obtain \begin{align}&\int_{X} \int_{[n]} {\| f (x,k)\|} \cdot {\|g (x,k)\|}{} \mathrm d \nu(k) \mathrm d \mu (x) \\
\leq &\left (\int_{[n]} \int_{X} \|f(x,k)\|^p \mathrm d \mu (x)\mathrm d \nu(k) \right )^{1/p} \cdot \left (\int_{[n]} \int_{X} \|g(x,k)\|^q \mathrm d \mu (x) \mathrm d \nu(k) \right )^{1/q}.\end{align}
In above inequality, we integrate w.r.t. $\nu$ first and then $\mu$ on the LHS. On the RHS, we do the reverse, i.e., we integrate w.r.t. $\mu$ first and then $\nu$. This gives us \begin{align}&\int_{X} \sum_{k=1}^n {\| f_k(x)\|} \cdot {\|g (x,k)\|}{} \mathrm d \mu (x) \\
\leq &\left ( \sum_{k=1}^n \|f_k(x)\|^p \mathrm d \mu (x) \right )^{1/p} \cdot \left ( \sum_{k=1}^n \|g_k(x)\|^q \mathrm d \mu (x) \right )^{1/q}\end{align}
This completes the proof.
Best Answer
Here is a simpler proof:
Step 1: $\int_X \sum\limits_{k=1}^{n} \|f_k(x)\|\|g_k(x)\|d\mu(x) =\sum\limits_{k=1}^{n} \int_X \|f_k(x)\|\|g_k(x)\|d\mu(x)$
Step 2: $\int_X \|f_k(x)\|\|g_k(x)\|d\mu(x) \leq \|f_k\|_p\|g_k||_q$. [This is Holder's inequality].
Stpe 3: Finish the proof using the fact that $\sum\limits_{k=1}^{n} a_kb_k \leq (\sum\limits_{k=1}^{n}|a_k|^{p})^{1/p}) (\sum\limits_{k=1}^{n}|b_k|^{q})^{1/q})$