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I am looking to prove the inequality $$\|f \otimes g\|_{L^p} \leq \|f\|_{L^p} \|g\|_{L^p}$$ for $1 \leq p \leq \infty$. Here $f$ takes values in a Banach space $X_1$ and $g$ takes values in a Banach space $X_2$.

I suspect this is true from the Cauchy-Schwartz like identity that holds for operators $$\|S \otimes T\| \leq \|S\| \|T\|$$ that is outlined in the answer to this question Norm of a tensor product of operators, but I am not completely sure. Is this inequality true? Should it instead be Holder like as in $$\|f \otimes g\|_{L^p} \leq \|f\|_{L^q} \|g\|_{L^r}, \quad \frac1p = \frac1q + \frac1r?$$

CBBAM
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For all $S,T \in X_1 \times X_2$, $\|S \otimes T\| = \|S\|\|T\|$ (see the post you shared), so for all real number $t$, $\|f(t) \otimes g(t)\| = \|f(t)\|\|g(t)\|$ and by Hölder, if $p,q,r$ are such that $\frac{1}{p} + \frac{1}{q} = \frac{1}{r}$, \begin{align*} \|f \otimes g\|_{L^r} & = \|t \mapsto \|f(t) \otimes g(t)\|\|_{L^r}\\ & = \|t \mapsto \|f(t)\|\|g(t)\|\|_{L^r}\\ & \leqslant \|t \mapsto \|f(t)\|\|_{L^p}\|t \mapsto \|g(t)\|\|_{L^q}\\ & = \|f\|_{L^p}\|g\|_{L^q}. \end{align*} The first inequality, however, is not true in general. Take for example $p = 2$, $f_0 \in X_1$ of norm $1$, $g_0 \in X_2$ of norm $1$ and, $$ f : t \mapsto t^{-1/3}e^{-t^2}f_0, \qquad g : t \mapsto t^{-1/3}e^{-t^2}g_0. $$ Then, $\|f \otimes g\|_{L^2} = \left\|t \mapsto (t^{-1/3}e^{-t^2})^2\right\|_{L^2} = \sqrt{\|t^{-4/3}e^{-4t^2}\|_{L^1}} = +\infty$ while $\|f\|_{L^2}\|g\|_{L^2} = \|t \mapsto t^{-1/3}e^{-t^2}\|_{L^2}^2 = \|t \mapsto t^{-2/3}e^{-2t^2}\|_{L^2} < +\infty$.

Cactus
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