---- > [!theorem] Theorem. ([[Young's convolution inequality]]) > Suppose $p,q,r \in [1, \infty]$ satisfy $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}+1$. [[Lp-norm|Then]] $\|f * g\|_{r} \leq \|f\|_{p} \|g\|_{q}$ for all functions $f \in L^{p}(\mathbb{R}^{n})$, $g \in L^{q}(\mathbb{R}^{n})$. Here, $f * g$ denotes the [[convolution]] of $f$ and $g$. > (Equivalently, for [[dual exponent|conjugates]] $p,q,r$ the assumption is $\frac{1}{r'}=\frac{1}{p'}+\frac{1}{q'}$) > [!proof]- Proof. ([[Young's convolution inequality]]) > > First assume $1<p,q,r < \infty$. Let $r'$ denote the [[dual exponent|conjugate index]] to $r$. Write $(f * g)(x)=\int _{\mathbb{R}^{n}} \underbrace{ [f(y)^{p/r}g(x-y)^{q/r}] }_{ =:A_{x}(y) }\underbrace{ [f(y)^{1-p/r}] }_{ =: B(y) }\underbrace{ [g(x-y)^{1-q/r}] }_{ C_{x}(y) } \, dy.$ > Put $\alpha=r$, $\beta=\frac{pr}{r-p}$, $\gamma=\frac{qr}{r-q}$. Noting that $\frac{1}{\beta}=\frac{1}{p}-\frac{1}{r}$ and $\frac{1}{\gamma}=\frac{1}{q}- \frac{1}{r}$, we see $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=1$ upon substitution of the assumption $\frac{1}{r}= \frac{1}{p}+\frac{1}{q}-1$. Thus we may apply [[Hölder's inequality|Hölder's inequality for 3 functions]] applied with exponents $\alpha, \beta, \gamma$ to obtain $|(f * g)(x)| \leq \|A_{x}\|_{\alpha} \|B\|_{\beta} \|C\|_{\gamma}.$ > > We compute $\begin{align} > \|A_{x}\|_{\alpha} &= \left( \int |f^{p} g(x - \cdot)^{q} |_{} \right)^{1/r} = \big( (|f|^{p} * |g|^{q})(x) \big)^{1/r}\\ > \|B\|_{\beta} &= \left( \int |f|^{(1-p/r)\beta} \right)^{1/\beta} = \left( \int |f|^{p} \right)^{(1/p)(1-p/r)}= \|f\|_{p}^{1-p/r} \\ > \|C\|_{\gamma}&= (\int |g(x - \cdot)|^{(1-q/r)\gamma} )^{1/\gamma} = \|g\|_{q}^{1-q/r}. > \end{align}$ > So $|(f * g)(x)| \leq \big( (|f|^{p} * |g|^{q})(x) \big)^{1/r} \|f\|_{p}^{1-p/r} \|g\|_{q}^{1-q/r}.$ > Exponentiating by $r$ on both sides and integrating wrt $x$, obtain $\int |f * g|^{r} \leq \|f\|_{p}^{r-p} \|g\|_{q}^{r-q} \left( \int_{} (|f|^{p} * |g|^{q})(x) \, dx \right) .$ > Using [[Fubini's Theorem]] and translation invariance of the [[integral|Lebesgue integral]] on $\mathbb{R}^n$, we get $\begin{align} > \int_{x \in \mathbb{R}^{n}} ( |f|^{p} * |g|^{q} )(x) \, dx &= \int _{x \in \mathbb{R}^{n}}\left( \int _{y \in \mathbb{R}^{n}} |f|^{p}(y) |g(x-y)|^{q}\, dy \right) \, dx \\ > &= \int _{y \in \mathbb{R}^{n}} |f(y)|^{p} \left( \int _{x \in \mathbb{R}^{n}} |g(x-y)|^{q} \, dx \, \right) \, dy \\ > &= \int _{y \in \mathbb{R}^{n}}|f(y)|^{p} \underbrace{ \left( \int _{x \in \mathbb{R}^{n}}| g(x)|^{q} \, dx \, \right) }_{ \|g\|_{q} ^{q}} \, dy \\ > &= \|f\|_{p}^{p} \|g\|_{q}^{q}. > \end{align}$ > So we have $\|f * g\|_{r}^{r} \leq \|f\|_{p}^{r-p} \|g\|_{q}^{r-q} \|f\|_{p}^{p} \|g\|_{q}^{q}=\|f\|^{r}_{p} \|g\|_{q}^{r}.$ The result follows. The missing cases were treated in lecture. > > ---- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```