An inequality $\| f \|_{L^p} \leq \| f \|_{L^\infty}^{1 - \frac{2}{p}} \| f \|_{L^2}^{\frac{2}{p}}$

Question

What is the name of this inequality $$\| f \|_{L^p(\Bbb R^n)} \leq \| f \|_{L^\infty(\Bbb R^n)}^{1 - \frac{2}{p}} \| f \|_{L^2(\Bbb R^n)}^{\frac{2}{p}}$$ for $p > 2$?And how can I prove this?

Answer 1

The following generalization is usually called the "log-convexity of $L^p$ norms" (see Lemma 2 and Lemma 1.11.5).

Since $p\frac{q-r}{q-p}+q\frac{r-p}{q-p}=r$ and $\frac{q-r}{q-p}+\frac{r-p}{q-p}=1$, Hölder's Inequality says that $$ \begin{align} \int_Mf^r\,\mathrm{d}\mu &=\int_Mf^{p\large\frac{q-r}{q-p}}f^{q\large\frac{r-p}{q-p}}\,\mathrm{d}\mu\\ &\le\left(\int_Mf^p\,\mathrm{d}\mu\right)^{\Large\frac{q-r}{q-p}} \left(\int_Mf^q\,\mathrm{d}\mu\right)^{\Large\frac{r-p}{q-p}}\tag{1} \end{align} $$ Raising to the $1/r$ power, $$ \left(\int_Mf^r\,\mathrm{d}\mu\right)^{\Large\frac1r} \le\left(\int_Mf^p\,\mathrm{d}\mu\right)^{\Large\frac1p\frac pr\frac{q-r}{q-p}} \left(\int_Mf^q\,\mathrm{d}\mu\right)^{\Large\frac1q\frac qr\frac{r-p}{q-p}}\tag{2} $$ Thus, $$ \|f\|_r\le\|f\|_p^{\Large\frac pr\frac{q-r}{q-p}}\|f\|_q^{\Large\frac qr\frac{r-p}{q-p}}\tag{3} $$ where $\frac pr\frac{q-r}{q-p}+\frac qr\frac{r-p}{q-p}=1$. Setting $\theta=\frac pr\frac{q-r}{q-p}$, $(3)$ can also be written as $$ \|f\|_r\le\|f\|_p^\theta\,\|f\|_q^{1-\theta}\tag{4} $$ where $$ \frac1r=\frac{\theta}{p}+\frac{1-\theta}{q}\tag{5} $$

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In the problem at hand, let $p=2$ and $q=\infty$. We get $\theta=2/r$ and $$ \|f\|_r\le\|f\|_2^{2/r}\,\|f\|_\infty^{1-2/r}\tag{6} $$

Answer 2

What is the name of this inequality?

A triviality?

To prove it, note that $|f|^p\leqslant\|f\|_\infty^{p-2}\,|f|^2=c^p\,|f|^2$ pointwise, with $c=\|f\|_\infty^{(p-2)/p}$, and integrate both sides of the inequality, this yields $$ \|f\|_p^p\leqslant c^p\|f\|_2^2, $$ which is equivalent to what you want.