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

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Chris L T521

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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.

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

Chris L T521

Well-known member
Staff member
No one answered this week's problem. Here's my solution.

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.

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