Exercise 6.10

Question

Exercise 10: Let $p$ and $q$ be positive real numbers such that $$\frac{1}{p}+\frac{1}{q}=1.$$ Prove the following statements.

(a) If $u\ge0$ and $v\ge0$, then $$uv\le\frac{u^p}{p}+\frac{v^q}{q}.$$ Equality holds if and only if $u^p=v^q$.

(b) If $f\in\mathscr R(\alpha)$, $g\in\mathscr R(\alpha)$, $f\ge0$, $g\ge0$, and $$\int_a^bf^p\,d\alpha=1=\int_a^bg^q\,d\alpha,$$ then $$\int_a^bfg\,d\alpha\le1.$$

(c) If $f$ and $g$ are complex functions in $\mathscr R(\alpha)$, then $$\bigg|\int_a^bfg\,d\alpha\bigg|\le\biggl(\int_a^b|f|^p\,d\alpha\biggr)^{1/p}
    \biggl(\int_a^b|g|^q\,d\alpha\biggr)^{1/q}.$$ This is {\it H\"{o}lder's inequality.} When $p=q=2$ it is usually called the Schwarz inequality.

(d) Show that H\"{o}lder's inequality is also true for the ``improper'' integrals defined in Exercises 7 and 8.

analambanomenos · Accepted Answer

Note that since $q$ and $p=q/(q-1)$ are positive, then $q-1$ is positive.

(a) Fix $u$ and let $f(v)=(u^p/p)+(v^q/q)-uv$. Then $f'(v)=v^{q-1}-u$ and $f^{\prime\prime}(v)=(q-1)v^{q-2}$ is non-negative for non-negative $v$, so the critical point $v=u^{1/(q-1)}$
    is a minimum. Hence
    \begin{align*}
        \frac{u^p}{p}+\frac{v^q}{q}-uv &\ge \frac{u^p}{p}+\frac{u^{q/(q-1)}}{q}-u^{1+1/(q-1)} \
        &=\biggl(\frac{1}{p}+\frac{1}{q}-1\biggr)u^p \
        &= 0
    \end{align*}
    so that $uv\le(u^p/p)+(v^q/q)$. Equality holds at the critical value $v=u^{1/(q-1)}$, or $v^q=u^{q/(q-1)}=u^p$.

(b) From part (a) we have
    \begin{align*}
        \int_a^bfg\,d\alpha &\le \int_a^b\biggl(\frac{f^p}{p}+\frac{g^q}{q}\biggr)\,d\alpha \
        &= \frac{1}{p}\int_a^b f^p\,d\alpha+\frac{1}{q}\int_a^bg^q\,d\alpha \
        &= \frac{1}{p}+\frac{1}{q} \
        &= 1.
    \end{align*}

(c) Define $$A=\biggl(\int_a^b|f|^p\,d\alpha\biggr)^{1/p}\quad\quad B=\biggl(\int_a^b|g|^q\,d\alpha\biggr)^{1/q}.$$ Then
    \begin{align*}
        \int_a^b\bigg|\frac{f}{A}\bigg|^p\,d\alpha &= \frac{1}{A^p}\int_a^b|f|^p\,d\alpha=1 \
        \int_a^b\bigg|\frac{g}{B}\bigg|^q\,d\alpha &= \frac{1}{B^q}\int_a^b|g|^q\,d\alpha=1.
    \end{align*}
    Applying part (b), we get $$\int_a^b\bigg|\frac{f}{A}\bigg|\cdot\bigg|\frac{g}{B}\bigg|\,d\alpha \le 1,$$ or $$\bigg|\int_a^bfg\,d\alpha\bigg|\le\int_a^b|fg|\,d\alpha\le AB=
    \biggl(\int_a^b|f|^p\,d\alpha\biggr)^{1/p}\biggl(\int_a^b|g|^q\,d\alpha\biggr)^{1/q}.$$

(d) Let $f\in\mathscr R$, $g\in\mathscr R$ on $[c,1]$ for all $c>0$ such that the improper integrals $\int_0^1|f|^p\,dx$ and $\int_0^1|g|^q\,dx$ exist. Then for all $c>0$ we have
    $$\bigg|\int_c^1fg\,dx\bigg|\le\biggl(\int_c^1|f|^p\,dx\biggr)^{1/p}\biggl(\int_c^1|g|^q\,dx\biggr)^{1/q}.$$ Since the right side increases monotonically as $cightarrow0$, we
    can take the limit of both sides to get the desired result.

Similarly, let $f\in\mathscr R$, $g\in\mathscr R$ on $[a,b]$ for all $b>a$ such that the improper integrals $\int_a^{\infty}|f|^p\,dx$ and $\int_a^{\infty}|g|^q\,dx$ exist. Then for
    all $b>a$ we have $$\bigg|\int_a^bfg\,dx\bigg|\le\biggl(\int_a^b|f|^p\,dx\biggr)^{1/p}\biggl(\int_a^b|g|^q\,dx\biggr)^{1/q}.$$ Since the right side increases monotonically as
    $bightarrow\infty$, we can take the limit of both sides to get the desired result.

Exercise 6.10

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Exercise 6.10

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