----- > [!proposition] Proposition. ([[Young's inequality]]) > Suppose $1<p<\infty$. Then $ab \leq \frac{a^{p}}{p}+\frac{b^{p'}}{p'}$ for all $a \geq 0$ and $b \geq 0$. Here, $p'$ denotes the [[dual exponent]] of $p$. ^proposition > [!proof]- Proof. ([[Young's inequality]]) > Fix $b>0$ and define a function $f:(0, \infty) \to \mathbb{R}$ by $f(a)=\frac{a^{p}}{p}+\frac{b^{p'}}{p'}-ab.$ > We will show $f(a) \geq 0$ for all $a \in (0, \infty)$. [[derivative|Differentiating]] with respect to $a$ yields $f'(a)=a^{p-1}-b.$ > Observe $f'(a) <0$ when $a \in (0, b^{1/p-1})$ and $f'(a)>0$ when $a \in (b^{1/p-1}, \infty)$. Thus $f$ has a [[global extrema|global minimum]] at $b^{1/p-1}$, namely, $\begin{align} > f(b^{1/p-1})&=\frac{b^{p/p-1}}{p}+\frac{b^{\overbrace{ p' }^{ p/p-1 }}}{p'}-b^{\overbrace{ 1+ (1/p-1) }^{ p/p-1 }} \\ > &= \frac{pb^{p/p-1} + p'b^{p/p-1} - pp'b^{p/p-1}}{pp'} \\ > &= 0, > \end{align}$ > where the last inequality follows from $p+p'=pp'$. It follows that $f(a) \geq 0$ for all $a \in (0, \infty)$, and this implies the result. > ----- #### ---- #### 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 > ```