Comparing $L^p$-norms of a function for two different values of $p$.

Question

Let $(S,\Sigma,\mu)$ be a measure space with $\mu(S)<\infty$. Let $p\in(1,\infty]$ and suppose $f\in L^p(\mu)$. A simple application of the Holder's inequality shows that for any $0<q<p$, $f$ is also in $L^q(\mu)$. Precisely, \begin{align} ||f||_{q}\leq C||f||_p & & (*) \end{align} where $0<C<\infty$ is a power of $\mu(S)$. On the other hand, if $p<q$ then we do not have such inequality. This is (roughly) because $f$ can have very extreme jump discontinuities.

However, my naive gut feeling tells me that if $f$ further satisfies some nice condition, then there is a hope for this to be true. Precisely, I wonder the answer of the following question:

Suppose that $S$ is also a compact topological space and $f$ is smooth (or at least, continuous) on $S$. Is there any chance for the inequality (*) to hold for $p<q$?

Preferably, the constant $C$ should independent of $f$. Also, if there is really such inequality, I do not expect it to hold for any $q>p$. Maybe $q$ has to satisfy some relation with $p$.

Any comment and answer are greatly appreciated.

Answer 1

Let $S=[0,1],\Sigma$ be the Borel sigma algebra and $\mu$ be Lebesgue measure. Take $p=1$ and $q=2$. Suppose there exists a consatnt $C$ such that $\|f\|_2 \leq C\|f\|_1$ for all smooth functions $f$. Apply this to $e^{nx}$ to see that $\frac {e^{2n}-1} {2n} \leq C^{2} (\frac {e^{n}-1} {n})^{2}$. It is easy to see that this inequality fails for large $n$.

Smoothness of function plays very little role here. The nature of the measure $\mu$ is more important.

Answer 2

In general this is not true. Here is an explicit counterexample, were we see that regularity is not that important. Consider some smooth function with support away from the boundary $\varphi :[-1;1] \rightarrow \mathbb{R}$ such that $\Vert \varphi\Vert_q=1$. For $n\in \mathbb{N}$ we define $\varphi_n(x) = \tilde{\varphi}(n x)$ (where $\tilde{\varphi}$ denotes the extension by zero). Then we can compute the $p$ and the $q$ norm. $$\Vert \varphi_n\Vert_q =\left( \int \vert \tilde{\varphi}(nx)\vert^q dx \right)^\frac{1}{q} =n^{-1/q} \Vert \tilde{\varphi}\Vert_q$$ Thus if $q>p$ then your inequality tells us $$ n^{-1/q} = \Vert \varphi_n\Vert_q \leq C\Vert \varphi_n \Vert_p =C n^{-1/p} \Vert \varphi \Vert_p $$ This would imply $$ n^{1/p-1/q} \leq C \Vert \varphi \Vert_p $$ which gives a contradiction as $1/p -1/q>0$ (as $q>p$) and we can pick $n$ as large as we want.

The inequality is true for all functions iff the measure is a finite linear combination of Dirac deltas. In fact we need $S$ to Hausdorff and $\Sigma$ to be the Borel sigma algebra for this (at least that's what I am going to use in the proof) Because we have for any measurable set $A\in \Sigma$ $$ m(A)^\frac{1}{q} = \Vert 1_A \Vert_q \leq C \Vert 1_A \Vert_p = C m(A)^\frac{1}{p} $$ Hence we have either $m(A)=0$ or $1/C \leq m(A)^{\frac{1}{p}-\frac{1}{q}}$. Thus, we get for every $A\in \Sigma$ $$ m(A)=0 \qquad \text{or} \qquad C^{-1/(\frac{1}{p}-\frac{1}{q})} \leq m(A). $$ Thus, there can only be finitely many points in the support of $\mu$ (here I use that I can separate finitely many points by disjoint nbhds). This implies that our measure space is isomorphic to some measure space of finite linear combinations of Dirac deltas.

Note that if you assume the inequality to be true for all continuous functions and if $S$ is locally compact and $\Sigma$ is the Borel sigma algebra, then $C_c(S)$ is dense in $L^r(S,\Sigma, \mu)$ for any $1\leq r <\infty$ (see https://planetmath.org/compactlysupportedcontinuousfunctionsaredenseinlp for a proof of this fact) and thus, the inequality is true for all $L^p$ functions.