Problem of the Week #21 - October 22nd, 2012

[Chris L T521]

Here's this week's problem.

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Problem: Suppose that $(X,\mu)$ is a measure space. For $0<r<p<s<\infty$, assume that $f\in L_{\mu}^r(X)\cap L_{\mu}^s(X)$. Show that $f\in L_{\mu}^p(X)$ and that
\[\|f\|_{L_{\mu}^p(X)} \leq \|f\|_{L_{\mu}^r(X)}^{\theta}\|f\|_{L_{\mu}^s(X)}^{1-\theta}\qquad\text{for}\qquad \frac{1}{p}=\frac{\theta}{r}+\frac{1-\theta}{s}.\]

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Here's a hint for this week's question.

Spoiler

Use the generalized Hölder inequality for the second half of the problem.

Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Chris L T521]

No one answered this week's problem. Here's my solution.

Spoiler

Proof: Since $r<p<s$, we have the containment $L_{\mu}^r(X)\subset L_{\mu}^p(X)\subset L_{\mu}^s(X)$. Furthermore, since $f\in L_{\mu}^r(X)\cap L_{\mu}^s(X)$, we must also have that $f\in L_{\mu}^p(X)$.

Now, if we let $\theta=0$, we get the inequality $\|f\|_{L_{\mu}^p(X)}\leq \|f\|_{L_{\mu}^s(X)}$ and if $\theta=1$, we have $\|f\|_{L_{\mu}^p(X)}\leq \|f\|_{L_{\mu}^r(X)}$, which makes sense given how $f$ lies in each of these spaces. Now, suppose that $\theta\neq 0,1$. To prove the inequality, we need to use the general Hölder inequality
\[\|fg\|_{L^r}\leq \|f\|_{L^p}\|g\|_{L^q}\quad\text{for}\quad \frac{1}{r}=\frac{1}{p}+\frac{1}{q}.\]
In our situation, let $F$ and $G$ be functions such that $FG=f$, i.e. $F=f^{\theta}$ and $G=f^{1-\theta}$. Then by the general Hölder inequality with
\[\frac{1}{p}=\frac{\theta}{r}+\frac{1-\theta}{s}=\frac{1}{r/\theta}+\frac{1}{s/(1-\theta)}\]
we get
\[\begin{aligned}\|f\|_{L_{\mu}^p(X)} &= \|FG\|_{L_{\mu}^p(X)}\\ &\leq \|F\|_{L_{\mu}^{r/\theta}(X)} \|G\|_{L_{\mu}^{s/(1-\theta)}(X)}\\ &= \left(\int_E |F|^{r/\theta}\,d\mu\right)^{\theta/r} \left(\int_E |G|^{s/(1-\theta)}\,d\mu\right)^{(1-\theta)/s} \\ &= \left(\int_E |f|^{\theta(r/\theta)}\,d\mu\right)^{\theta/r} \left(\int_E |f|^{(1-\theta)[s/(1-\theta)]}\,d\mu\right)^{(1-\theta)/s}\\ &= \left[\left(\int_E |f|^r\,d\mu\right)^{1/r}\right]^{\theta} \left[\left(\int_E |f|^s\,d\mu\right)^{1/s}\right]^{1-\theta}\\ &=\|f\|_{L_{\mu}^r(X)}^{\theta}\|g\|_{L_{\mu}^s(X)}^{1-\theta}\end{aligned}\]

Therefore, $\|f\|_{L_{\mu}^p(X)}\leq \|f\|_{L_{\mu}^r(X)}^{\theta}\|f\|_{L_{\mu}^s(X)}^{1-\theta}$. Q.E.D.