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Here is the problem:

Let $f\in L^p(\mathbb R^n)\cap L^q(\mathbb R^n)$ and $s\in[p,q]$. Show that $f\in L^s(\mathbb R^n)$

I'm almost sure that this is a simple exercise on Hölder's inequality yet I can't find the right $a,b>1$ with $$\frac{1}{a}+\frac{1}{b}=1$$

to apply the inequality.

Whether you'll give a hint or a full solution (which I expect to be very short), please explain your thought process. I guess it's like with the $\varepsilon/\delta$-proofs where things happen to appear out of the blue but we know that it's because of some scratch-work before.

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3 Answers 3

up vote 2 down vote accepted

Write $s = tp+(1-t)q$. Consider $1/p'+1/q'=1$.

$$ \int |f|^{s} = \int |f|^{tp+(1-t)q} \le \left\{\int |f|^{tpp'} \right\}^{1/p'} \left\{\int |f|^{(1-t)qq'} \right\}^{1/q'} $$

Now you can choose $$ tpp'=p, (1-t)qq'=q\iff p'=\frac 1t, q'=\frac 1{1-t} $$ because $$ \frac 1{p'}+\frac 1{q'} = t+(1-t)=1 $$

Then $$ \int |f|^{s} \le ||f||_p^{p/p'} ||f||_q^{q/q'} <\infty$$

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Hint: It follows as a corollary of the following statement:

Let $\{f_i:i\in I\}$ be a family of functions with $f_i\in L^{p_i}(\Omega)$ and $\dfrac{1}{p}=\sum \dfrac{1}{p_i}\leq1$. Then $\prod f_i\in L^p(\Omega)$ and $$\|\prod f_i\|_{L^p(\Omega)}\leq\prod\|f_i\|_{L^{p_i}(\Omega)}$$

This is the interpolation inequality (H. Brezis "Functional Analysis, Sobolev Spaces and Partial Differential Equations").

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Here's an idea that might be helpful: that $L^p$ measures spikiness or broadness of a function in some sense. In particular, $L^{\infty}$ consists of functions with no spikes whatsoever, while $L^1$ functions can't be too broad; the higher $p$ is, the more control you have over spikes, and less control over broadness. Since $f \in L^p \cap L^q$, we've controlled both problems enough to force $f \in L^s$.

This motivates studying the places where $f$ is spiky and large, and small separately; to this end, let $$E = \{x : |f(x)| \ge 1\}$$ and consider $$f = f \chi_E + f \chi_{E^c}$$

Note that $E$ has finite measure, and use this to prove that each of these two functions lies in $L^s$. You might find it useful to compare $t^a$ and $t^b$ for $a < b$, depending on whether $t < 1$ or $t > 1$.

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