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Let $X$ be a measure space with measure $\mu$ and $\|f\|_p=\left(\int_X |f|^p\; d\mu\right)^{\frac{1}{p}}$ be the standard $p$-norm for complex-valued $f$. I want to prove the following result, given $0<r\leq s$:

For all $p\in [r,s]$, $\|f\|_p\leq \max(\|f\|_r,\|f\|_s)$.

Thanks to Hölder's inequality, I already proved that the function $p\mapsto \|f\|_p^p$ is convex on $[r,s]$, this giving $\|f\|_p^p\leq \max(\|f\|_r^r,\|f\|_s^s)$. But I didn't manage to remove the powers. Is there a simple way to do it?

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You may as well assume that $f \in L^r \cap L^s$ since otherwise the max is infinite. The inequality is trivial if $p=r$ or $p=s$ so you can assume $r < p < s$. Then $\dfrac 1s < \dfrac 1p < \dfrac 1r$ so there exists $0 < \theta < 1$ so that $\dfrac 1p = \dfrac \theta r + \dfrac{1-\theta} s$. Thus $1 = \dfrac{\theta p}{r} + \dfrac{(1-\theta)p}{s}$ so you may apply Hölder's inequality: $$\int_X |f|^p \, d\mu = \int_X |f|^{\theta p} |f|^{(1-\theta) p} \, d\mu \le \left( \int_X |f|^r \, d\mu \right)^{\theta p/r} \left( \int_X |f|^s \, d\mu\right)^{(1-\theta)p/s}.$$ That is, $$\|f\|_p \le \|f\|_r^\theta \|f\|_s^{1-\theta}.$$ The inequality is now straightforward.

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  • $\begingroup$ Scooped by Davide Giraudo! But here is the proof of what he suggested. $\endgroup$ – Umberto P. Oct 24 '13 at 13:52
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Take $\lambda\in (0,1)$ such that $p=\lambda r+(1-\lambda)s$. An application of Hölder's inequality gives $$\lVert f\rVert_p\leqslant \lVert f\rVert_r^{\theta}\lVert f\rVert_s^{1-\theta}$$ for some $\theta$ (we can compute it).

(we actually use a convexity argument)

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