# Proving Hölder's Inequality

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.

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 $$\left|\int_a^b f(x)g(x)d\alpha\right|\leq\int_a^b|f(x)||g(x)|d\alpha.$$

• 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?
– crf
Jan 25, 2013 at 7:39
• wow actually never mind, it's obvious. I just wasn't seeing straight. Thanks very much!
– crf
Jan 25, 2013 at 8:26
• P.S., how did anyone come up with that
– crf
Jan 25, 2013 at 8:26
• @crf, it's easiest to think about first proving the case for functions with norm 1, then using homogeneity to get the general case. Jul 31, 2013 at 0:54
• @paul what will be the inequality for the case $p=1$ and $q=\infty$? Feb 13, 2015 at 16:55

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.

• Your $a^\theta b^{1-\theta} \le \theta a+(1-\theta)b$ is just weighted AM-GM
– D S
Feb 16 at 19:03

NOTE to the moderation: Please don't delete my answer for no reason I have spent a lot of time writing it and she goes further and use different ways than the other answers above.

-Let go further and here a little bit more general prove that hope will help futur reader and student like me.
Prove that with $$\frac{1}{p} + \frac{1}{q} = \frac{1}{r}$$ and $$p, q ,r \in [0; \infty ]$$ we have $$(\int |f(x)g(x)|^r dx)^\frac{1}{r} = \| fg\|_r \leq \| f \|_p \| g \|_q = (\int |f(x)|^p dx)^\frac{1}{p} (\int |g(x)|^q dx)^\frac{1}{q}$$ .
-For simplifying the scripture we write $$F = \| f \|_p , G = \| g \|_q$$.

I/Special cases

1- If $$F=0$$ or/and $$G=0 \Rightarrow 0 \leq \| fg\|_r \leq \| f \|_p \| g \|_q =0$$ because $$\| . \|_r \geq 0$$ as a measure. And thus $$\| fg\|_r = 0$$ too and so the inequality is respected.

2- If $$F= \infty$$ or/and $$G= \infty$$ in such a case we have that obviously $$\| fg\|_r \leq F \cdot G = \infty$$.

3- If for exemple $$q = \infty \Rightarrow \frac{1}{q}=\frac{1}{\infty}=0$$ by convention and so in such a case the inequality become $$p=r$$ and $$\| fg\|_r \leq \| f \|_r \| g \|_{\infty}$$ which is immediately verify because $$|f(x)g(x)|^p = |f(x)|^p |g(x)|^p \leq \|g(x) \|_{\infty} |f(x)|^p, \forall x$$.

II/In the following we will suppose that: $$0 and too that $$p,q < \infty$$

1- Now $$\frac{1}{p} + \frac{1}{q} = \frac{1}{r} \Rightarrow \frac{r}{p} + \frac{r}{q} = 1$$ we write $$a=\frac{r}{p} , b=\frac{r}{q}$$ and of course $$a,b > 0$$

2- Now let prove that: $$x^{a}y^{\left(1-a\right)} \leq ax+\left(1-a\right)y$$ for $$0 \leq a \leq 1$$ (remind that $$b=1-a$$).
We have that $$ax+(1-a)y = y(1+ a\frac{(x-y)}{y})$$, now let focus on $$(1+ a\frac{(x-y)}{y})$$ by the Bernoulli inequality we have that $$(1+ a\frac{(x-y)}{y}) \geq (1+\frac{(x-y)}{y})^a$$ we continue by multiplying both side by $$y$$ and we get $$ax+(1-a)y = y(1+ a\frac{(x-y)}{y}) \geq y(1+\frac{(x-y)}{y})^a= x^{a}y^{\left(1-a\right)}$$

3- Hence $$(\frac{|f(x)|^p}{F^p})^a(\frac{|g(x)|^q}{G^q})^b \leq a \frac{|f(x)|^p}{F^p} + b \frac{|g(x)|^q}{G^q}$$
Now let integrate both sides (remind that $$a=\frac{r}{p} , b=\frac{r}{q}$$ ): $$\int (\frac{|f(x)|^p}{F^p})^a(\frac{|g(x)|^q}{G^q})^b = \frac{1}{F^{pa} G^{bq}} \int |f(x)|^{ap} |g(x)|^{bq} = \frac{1}{(FG)^r} \int |f(x)|^{r} |g(x)|^{r} \leq a \int \frac{|f(x)|^p}{F^p} + b \int \frac{|g(x)|^q}{G^q} = a \frac{(\| f \|_p)^p}{F^p} + b \frac{(\| g \|_q)^q}{G^g}=1$$
So we can conclude $$(\| fg\|_r)^r \leq F^p G^q =(\| f \|_p)^p (\| g \|_q)^q \Rightarrow \| fg\|_r \leq \| f \|_p \| g \|_q$$.

III/When $$\mu (\left \{ f \neq 0 \right \}) \neq 0$$ we have that: $$\| fg\|_r = \| f \|_p \| g \|_q$$ iff $$\exists \lambda \leq 0 , |g|^q = \lambda |f|^p$$

$$(\Rightarrow)$$
From "II/3-" we focus on the measurable function $$a \frac{|f(x)|^p}{F^p} + b \frac{|g(x)|^q}{G^q} - (\frac{|f(x)|^p}{F^p})^a(\frac{|g(x)|^q}{G^q})^b$$ which is of integrable equal to $$0$$. Because its value are in $$[0; \infty]$$ it means that this function is mostly everywhere equal to $$0$$. We conclude that $$\frac{|f(x)|^p}{F^p} = \frac{|g(x)|^q}{G^q}$$ a.e. and after it is easy to conclude that $$\exists \lambda \leq 0 , |g|^q = \lambda |f|^p$$

$$(\Leftarrow)$$
Trivial.

• Your generalization is just $f(x) \mapsto f(x)^r, g(x) \mapsto g(x)^r, p \mapsto p/r, q \mapsto q/r$ in the original inequalitty
– D S
Feb 16 at 19:08

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, $$$$\Vert f \Vert_{1} = \int_{a}^{b} \vert f(x) \vert dx < \infty \text{ and } \Vert g \Vert_{\infty} = \sup_{x \in [a,b]} \lbrace g(x): x \in [a,b] \rbrace < \infty.$$$$ So, using properties of the absolute value, we have $$\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.