Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
(Trivial case of) a basic interpolation inequality, also discussed here here, and here – user53153 Jan 25 '13 at 20:20
up vote 1 down vote accepted

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 $$ \frac1r=\frac{\theta}{p}+\frac{1-\theta}{q}\tag{5} $$

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.