Spring 2010 - Problem 11

Solution.

1. We will first prove Young's convolution inequality: if $\frac{1}{p} + \frac{1}{q} = \frac{1}{r} + 1$, $f \in L^p\p{\R}$, $g \in L^q\p{\R}$, then $\norm{f * g}_{L^r} \leq \norm{f}_{L^p} \norm{g}_{L^q}$.

   Notice that because $1 \leq p, q \leq \infty$, we have $\frac{1}{p}, \frac{1}{q} \geq 1$, and so $\frac{1}{p}, \frac{1}{q} \geq \frac{1}{r}$ from the condition on $p, q, r$. Thus,

   $$\begin{aligned} 1 &= \frac{1}{p} + \frac{1}{q} - \frac{1}{r} \\ &= \p{\frac{1}{p} - \frac{1}{r}} + \p{\frac{1}{q} - \frac{1}{r}} + \frac{1}{r} \\ &= \frac{r - p}{pr} + \frac{r - q}{qr} + \frac{1}{r}. \end{aligned}$$

   Since all the terms are non-negative and sum to $1$, we see that $\frac{r-p}{pr}, \frac{r-q}{pqr} \in \br{0, 1}$, so if we set $\frac{1}{p'} = \frac{r-p}{pr}$ and $\frac{1}{q'} = \frac{r-q}{qr}$, then $\frac{1}{p'} + \frac{1}{q'} + \frac{1}{r} = 1$. We have

   $$\begin{gathered} 1 - \frac{p}{p'} = 1 - \frac{r - p}{r} = \frac{p}{r} \\ 1 - \frac{q}{q'} = 1 - \frac{r - q}{r} = \frac{q}{r}. \end{gathered}$$

   Thus, by Hölder's inequality,

   $$\begin{aligned} \abs{\p{f * \phi}\p{x}} &\leq \int_\R \abs{f\p{x - y}\phi\p{y}} \,\diff{y} \\ &= \int_\R \abs{f\p{x - y}}^{p/p'} \abs{f\p{x - y}}^{1-p/p'} \abs{\phi\p{y}}^{q/q'} \abs{\phi\p{y}}^{1-q/q'} \,\diff{y} \\ &= \int_\R \abs{f\p{x - y}}^{p/p'} \abs{\phi\p{y}}^{q/q'} \p{\abs{f\p{x - y}}^{p/r} \abs{\phi\p{y}}^{q/r}} \,\diff{y} \\ &\leq \p{\int_\R \abs{f\p{x - y}}^p \,\diff{y}}^{1/p'} \p{\int_\R \abs{\phi\p{y}}^q \,\diff{y}}^{1/q'} \p{\int_\R \abs{f\p{x - y}}^p \abs{\phi\p{y}}^q \,\diff{y}}^{1/r} \\ &= \norm{f}_{L^p}^{p/p'} \norm{\phi}_{L^q}^{q/q'} \p{\int_\R \abs{f\p{x - y}}^p \abs{\phi\p{y}}^q \,\diff{y}}^{1/r}. \end{aligned}$$

   Continuing,

   $$\begin{aligned} \norm{f * \phi}_{L^r}^r &\leq \int_\R \norm{f}_{L^p}^{pr/p'} \norm{\phi}_{L^q}^{qr/q'} \int_\R \abs{f\p{x - y}}^p \abs{\phi\p{y}}^q \,\diff{y} \,\diff{x} \\ &= \norm{f}_{L^p}^{r-p} \norm{\phi}_{L^q}^{r-q} \int_\R \abs{\phi\p{y}}^q \int_\R \abs{f\p{x - y}}^p \,\diff{x} \,\diff{y} && (\text{by Fubini-Tonelli}) \\ &= \norm{f}_{L^p}^{r-p} \norm{\phi}_{L^q}^{r-q} \norm{\phi}_{L^q}^q \norm{f}_{L^p}^p \\ &= \norm{f}_{L^p}^r \norm{\phi}_{L^q}^r \\ \implies \norm{f * \phi}_{L^r} &\leq \norm{f}_{L^p} \norm{\phi}_{L^q}. \end{aligned}$$

   To prove the inequality, observe that

   $$1 - \frac{1}{p} + \frac{1}{q} \geq 1 \implies \exists r \in \br{1, \infty} \text{ such that } 1 - \frac{1}{p} + \frac{1}{q} = \frac{1}{r} \implies \frac{1}{p} + \frac{1}{r} = \frac{1}{q} + 1.$$

   Thus, because $\phi$ has compact support, $\phi \in L^r\p{\R}$, so Young's convolution inequality gives

   $$\norm{f * \phi}_{L^q} \leq \norm{\phi}_{L^r}\norm{f}_{L^p}.$$

   Hence, $A = \norm{\phi}_{L^r}$ works, which completes the proof.

2. Let $\func{\phi}{\R}{\R}$ be defined as follows: $\phi\p{x} = 0$ for $x \notin \br{1, 2}$, $\phi\p{x} = 1$ for $x \in \br{0, 1}$, and linear on $\br{-1, 0}$ and $\br{1, 2}$.

   Since $p > q$, there exists $\alpha > 0$ such that $q < \frac{1}{\alpha} < p$, and set $f\p{x} = \frac{1}{x^\alpha}$ on $\pco{1, \infty}$ and $0$ otherwise. Observe that $q\alpha < 1 < p\alpha$, so $f \in L^p\p{\R}$, but $f \notin L^q\p{\R}$. Notice that for $x \geq 1$,

   $$\begin{aligned} \p{f * \phi}\p{x} &= \int_\R f\p{x - y}\phi\p{y} \,\diff{y} \\ &\geq \int_0^1 f\p{x - y} \,\diff{y} \\ &= \int_{x-1}^x \frac{1}{y^\alpha} \,\diff{y} \\ &\geq \frac{1}{x^\alpha}, \end{aligned}$$

   where the last inequality comes from the fact that $f$ is a decreasing function. Thus, $f * \phi \notin L^q\p{\R}$, so the inequality cannot hold.