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Let $f,g,\alpha:[a,b]\rightarrow \mathbb{R}$ with $\alpha$ increasing and $f,g \in \mathscr{R}(\alpha)$, and $p,q>0$ with $\frac{1}{p}+\frac{1}{q}=1$. Prove that $$\left|\int_a^b f(x)g(x)d\alpha\right|\leq \left(\int_a^b \left|f(x)\right|^p d\alpha \right)^{1/p} \left(\int_a^b \left|g(x)\right|^q d\alpha \right)^{1/q}$$

I am using Young's inequality, which states that for $a,b>0$, $uv\leq \frac{1}{p}u^{p}+\frac{1}{q}v^{q}$. This gets me as far as showing that $$\left|\int_a^b f(x)g(x)d\alpha\right|\leq \int\left( \frac {1}{p}|f(x)|^p +\frac{1}{q}|g(x)|^q\right)d\alpha$$

But here I'm stuck. I'm vaguely thinking that I could use the fact that $\frac {1}{p}|f(x)|^p +\frac{1}{q}|g(x)|^q$ is a convex combination and so if I do some Jensen's inequality type thing, but I can't figure out a way to make it work out.

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Suppose $\displaystyle\int_a^b \left|f(x)\right|^p d\alpha\neq 0$ and $\displaystyle\int_a^b \left|g(x)\right|^q d\alpha\neq 0$. Otherwise, if $\displaystyle\int_a^b \left|f(x)\right|^p d\alpha=0$, then $f\equiv 0$ a.e. and the Holder's inequality is trivial in this case.

Now applying Young's inequality with $u=\displaystyle\frac{|f(x)|}{(\int_a^b \left|f(x)\right|^p d\alpha)^{\frac{1}{p}}}$ and $v=\displaystyle\frac{|g(x)|}{(\int_a^b \left|g(x)\right|^q d\alpha)^{\frac{1}{q}}}$, we have $$\frac{|f(x)|}{(\int_a^b \left|f(x)\right|^p d\alpha)^{\frac{1}{p}}}\cdot\frac{|g(x)|}{(\int_a^b \left|g(x)\right|^q d\alpha)^{\frac{1}{q}}}\leq\frac{1}{p}\frac{|f(x)|^p}{\int_a^b \left|f(x)\right|^p d\alpha}+\frac{1}{q}\frac{|g(x)|^q}{\int_a^b \left|g(x)\right|^q d\alpha}.$$ Integrating it from $a$ to $b$ with respect to $d\alpha$, we obtain $$\frac{\int_a^b|f(x)||g(x)|d\alpha}{(\int_a^b \left|f(x)\right|^p d\alpha)^{\frac{1}{p}}(\int_a^b \left|g(x)\right|^q d\alpha)^{\frac{1}{q}}}\leq\frac{1}{p}+\frac{1}{q}=1$$ which implies that $$ \tag{1}\int_a^b|f(x)||g(x)|d\alpha\leq\left(\int_a^b \left|f(x)\right|^p d\alpha \right)^{1/p} \left(\int_a^b \left|g(x)\right|^q d\alpha \right)^{1/q}.$$ Now the inequality which we want to prove follows from $(1)$ and the inequality $$\left|\int_a^b f(x)g(x)d\alpha\right|\leq\int_a^b|f(x)||g(x)|d\alpha.$$

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  • $\begingroup$ I can't see how we can integrate the first line of Young's inequality like that. Can you help me see that a little bit better? $\endgroup$ – crf Jan 25 '13 at 7:39
  • $\begingroup$ wow actually never mind, it's obvious. I just wasn't seeing straight. Thanks very much! $\endgroup$ – crf Jan 25 '13 at 8:26
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    $\begingroup$ P.S., how did anyone come up with that $\endgroup$ – crf Jan 25 '13 at 8:26
  • $\begingroup$ @crf, it's easiest to think about first proving the case for functions with norm 1, then using homogeneity to get the general case. $\endgroup$ – Scott Morrison Jul 31 '13 at 0:54
  • $\begingroup$ @paul what will be the inequality for the case $p=1$ and $q=\infty$? $\endgroup$ – Bhauryal Feb 13 '15 at 16:55
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I think following might be a way to come up with the proof of H$\ddot { o } $lder's inequality.

First, it's easy to show that $$f=-log(x)$$ is a convex function. (a function $f$ is convex if and only if $dom$ $f$ is convex and its Hessian is positive semidefinite: for all $x\in$$dom$ $f$).$${ \triangledown }^{ 2 }f(x)\ge 0$$

Then according to the definition of convex function: $$f(\theta a+(1-\theta )b)\le \theta f(a)+(1-\theta )f(b)$$ for all $a,b\in$$dom$ $f$, and $0\le \theta \le 1$

We will have: $$-log(\theta a+(1-\theta )b)\le -\theta log(a)-(1-\theta )log(b)$$ for $a,b\ge 0$

next take the exponential of both sides yields:$${ a }^{ \theta }{ b }^{ 1-\theta }\le \theta a+(1-\theta )b$$

applying this with:$$a=\frac { { \left| f(x) \right| }^{ p } }{ \int _{ a }^{ b }{ { \left| f(x) \right| }^{ p } } }, b=\frac { { \left| g(x) \right| }^{ q } }{ \int _{ a }^{ b }{ { \left| g(x) \right| }^{ q } } }, \theta =1/p$$

yields $$\frac { \left| f(x) \right| }{ { (\int _{ a }^{ b }{ { \left| f(x) \right| }^{ p }d\alpha } ) }^{ \frac { 1 }{ p } } } \cdot \frac { \left| g(x) \right| }{ { (\int _{ a }^{ b }{ { \left| g(x) \right| }^{ q }d\alpha } ) }^{ \frac { 1 }{ q } } } \le \frac { 1 }{ p } \frac { \left| f(x) \right| ^{ p } }{ \int _{ a }^{ b }{ { \left| f(x) \right| }^{ p }d\alpha } } +\frac { 1 }{ q } \frac { \left| g(x) \right| ^{ q } }{ \int _{ a }^{ b }{ { \left| g(x) \right| }^{ q }d\alpha } } $$

Finally, integrate it from a to b with respect to dα will obtain $H\ddot { o } lder$'s inequality.

Reference: 《Convex Optimization》Stephen Boyd and Lieven Vandenberghe $Chapter3,p78$

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In the vast majority of books dealing with Real Analysis, Hölder's inequality is proven by the use of Young's inequality for the case in which $p , q > 1$, and they usually have as an exercise the question whether this inequality is valid for $p =1$ which means that $q = \infty$. Well, if $f \in L_{1}$ and $g \in L_{\infty}$ then, \begin{equation} \Vert f \Vert_{1} = \int_{a}^{b} \vert f(x) \vert dx < \infty \text{ and } \Vert g \Vert_{\infty} = \sup_{x \in [a,b]} \lbrace g(x): x \in [a,b] \rbrace < \infty. \end{equation} So, using properties of the absolute value, we have \begin{eqnarray} \left\vert \int_{a}^{b} f(x)g(x) dx \right\vert &\leq& \int_{a}^{b} \vert f(x)g(x) \vert dx \\ &=& \int_{a}^{b} \vert f(x)\vert \vert g(x) \vert dx \\ &\leq& \int_{a}^{b} \vert f(x)\vert \sup_{x \in [a,b]} \lbrace \vert g(x) \vert : x \in [a,b] \rbrace dx \\ &=& \sup_{x \in [a,b]} \lbrace \vert g(x) \vert : x \in [a,b] \rbrace \int_{a}^{b} \vert f(x)\vert dx \\ &=& \Vert g \Vert_{\infty} \Vert f \Vert_{1}. \end{eqnarray} This is the answer to the mentionated case.

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