Proof of Hölder's Inequality

From a Pointwise Inequality to an Integral Bound

Hölder’s inequality was stated in the previous tutorial as a way to control an integral of a product using two different $L^p$ norms. Here we prove that estimate. The central idea is to first establish a scalar inequality for two nonnegative numbers, then normalize the functions so that their relevant integrals equal one. Integrating the scalar inequality then gives the desired bound.

Throughout the proof, $1<p<\infty$, and $q$ is the conjugate exponent, so $1/p+1/q=1$. The endpoint $p=1,\ q=\infty$ does not follow from the same scalar inequality and will be handled separately. We begin with the scalar estimate that drives the finite-exponent proof.

Theorem (Young’s Inequality): Let $p,q>1$ satisfy $1/p+1/q=1$. For all $a,b\geq0$, $$ ab\leq \frac{a^p}{p}+\frac{b^q}{q}. $$ Equality holds if and only if $a^p=b^q$.

Proof. Fix $b\geq0$, and consider the function $$ \phi(a)=ab-\frac{a^p}{p},\qquad a\geq0. $$ If $b=0$, then $\phi(a)=-a^p/p\leq0=b^q/q$, with equality only when $a=0$. Now suppose $b>0$. For $a>0$, $$ \phi'(a)=b-a^{p-1}. $$ This derivative is positive when $a<b^{1/(p-1)}$ and negative when $a>b^{1/(p-1)}$. Thus $\phi$ attains its maximum at $a=b^{1/(p-1)}$. Since $q=p/(p-1)$, the maximum is $$ \phi\bigl(b^{1/(p-1)}\bigr) =b^q-\frac{b^q}{p} =\frac{b^q}{q}. $$ Therefore $\phi(a)\leq b^q/q$, which rearranges to Young’s inequality. When $b>0$, equality occurs exactly at the maximizing value $a=b^{1/(p-1)}$, equivalently $a^p=b^q$. The case $b=0$ has equality exactly when $a=0$, which is also exactly the condition $a^p=b^q$. This proves both the inequality and its equality condition.

Worked Example: Checking Equality in Young’s Inequality

Take $p=3$, so its conjugate exponent is $q=3/2$, and choose $a=2$, $b=4$. The conjugacy relation holds because $1/3+2/3=1$. The powers are $$ a^p=2^3=8, \qquad b^q=4^{3/2}=8. $$ Young’s inequality therefore predicts equality. Direct substitution verifies it: $$ ab=2\cdot4=8, \qquad \frac{a^p}{p}+\frac{b^q}{q} =\frac{8}{3}+\frac{8}{3/2} =\frac{8}{3}+\frac{16}{3} =8. $$ The equality condition is a useful guide in the integral setting: equality requires the two normalized functions to satisfy the corresponding power relation almost everywhere.

Proof of Hölder’s Inequality

We now prove the finite-exponent form stated in the previous tutorial. The key step is normalization: divide each function by its own norm. The resulting functions have unit integrals of their $p$th and $q$th powers, so Young’s inequality can be integrated without leaving unwanted constants.

Theorem (Hölder’s Inequality, Finite-Exponent Proof): Let $1<p<\infty$, let $q$ be conjugate to $p$, and suppose $f\in L^p(X,\mu)$ and $g\in L^q(X,\mu)$. Then $fg\in L^1(X,\mu)$ and $$ \int_X|fg|\,d\mu\leq \|f\|_p\|g\|_q. $$

Proof. First suppose that both norms are positive. Define the nonnegative measurable functions $$ F=\frac{|f|}{\|f\|_p}, \qquad G=\frac{|g|}{\|g\|_q}. $$ By the definitions of the norms, $$ \int_X F^p\,d\mu=1, \qquad \int_X G^q\,d\mu=1. $$ Young’s inequality, applied pointwise to $F(x)$ and $G(x)$, gives $$ F(x)G(x)\leq \frac{F(x)^p}{p}+\frac{G(x)^q}{q}. $$ The right side is integrable and has integral $1/p+1/q=1$. By monotonicity of the nonnegative integral, $$ \int_X F G\,d\mu \leq \frac{1}{p}\int_XF^p\,d\mu+\frac{1}{q}\int_XG^q\,d\mu =\frac{1}{p}+\frac{1}{q}=1. $$ Since $|fg|=\|f\|_p\|g\|_q FG$, nonnegative homogeneity of the integral yields $$ \int_X|fg|\,d\mu \leq \|f\|_p\|g\|_q. $$ The right side is finite, so $fg$ is integrable.

It remains to consider zero norms. If $\|f\|_p=0$, then $\int_X|f|^p\,d\mu=0$. The Zero Integral Criterion implies that $f=0$ almost everywhere. Consequently, $|fg|=0$ almost everywhere and its integral is zero. The claimed inequality follows. If $\|g\|_q=0$, the same argument shows that $g=0$ almost everywhere and again $\int_X|fg|\,d\mu=0$. This proves the theorem in all cases.

Worked Example: A Product of Two Singular Functions

On $(0,1)$ with Lebesgue measure, let $f(x)=x^{-1/8}$ and $g(x)=x^{-1/3}$. Choose $p=4$ and $q=4/3$; these are conjugate because $1/4+3/4=1$. Their norms are finite: $$ \|f\|_4 =\left(\int_0^1 x^{-1/2}\,dx\right)^{1/4} =2^{1/4}, \qquad \|g\|_{4/3} =\left(\int_0^1 x^{-4/9}\,dx\right)^{3/4} =\left(\frac{9}{5}\right)^{3/4}. $$ Hölder’s inequality therefore guarantees that $fg$ is integrable and gives $$ \int_0^1|f(x)g(x)|\,dx \leq 2^{1/4}\left(\frac{9}{5}\right)^{3/4}. $$ Indeed, $f(x)g(x)=x^{-11/24}$, so a direct calculation also confirms integrability: $$ \int_0^1x^{-11/24}\,dx=\frac{1}{1-11/24}=\frac{24}{13}. $$ The theorem supplies an upper bound from the separate norms; it does not claim that this bound must equal the product integral.

The Endpoint and the Equality Condition

The endpoint pair $p=1,\ q=\infty$ uses the almost-everywhere bound built into the $L^\infty$ norm rather than Young’s inequality. Recall the Least Almost-Everywhere Bound theorem: if $g\in L^\infty$, then $|g|\leq\|g\|_\infty$ almost everywhere.

Theorem (Endpoint Hölder Inequality): If $f\in L^1(X,\mu)$ and $g\in L^\infty(X,\mu)$, then $fg\in L^1(X,\mu)$ and $$ \int_X|fg|\,d\mu\leq\|f\|_1\|g\|_\infty. $$

Proof. The Least Almost-Everywhere Bound theorem gives $|g|\leq\|g\|_\infty$ almost everywhere. Therefore $$ |fg|\leq \|g\|_\infty |f| \quad\text{almost everywhere}. $$ Monotonicity of the integral and nonnegative homogeneity now give $$ \int_X|fg|\,d\mu \leq\|g\|_\infty\int_X|f|\,d\mu =\|g\|_\infty\|f\|_1<\infty. $$ Thus $fg\in L^1$, and the estimate follows. Reversing the roles of the functions proves the other endpoint ordering.

Young’s equality condition also identifies exactly when the finite-exponent Hölder bound is attained. If either norm is zero, both sides of Hölder’s inequality are zero. Otherwise, use $F$ and $G$ as in the proof above. Pointwise, the difference $$ \frac{F^p}{p}+\frac{G^q}{q}-FG $$ is nonnegative by Young’s inequality. Its integral is $1-\int_XFG\,d\mu$. The difference has integral zero precisely when it is zero almost everywhere, by the Zero Integral Criterion. Young’s equality condition then gives the following characterization.

Theorem (Equality Condition in Hölder’s Inequality): Suppose $1<p<\infty$, $1/p+1/q=1$, and $f\in L^p,\ g\in L^q$, with both norms positive. Then $$ \int_X|fg|\,d\mu=\|f\|_p\|g\|_q $$ if and only if $$ \frac{|f|^p}{\|f\|_p^p} = \frac{|g|^q}{\|g\|_q^q} \quad\text{almost everywhere}. $$

Proof. The normalized functions $F,G$ satisfy $\int_XF^p\,d\mu=\int_XG^q\,d\mu=1$, and the normalized Hölder bound is $\int_XFG\,d\mu\leq1$. If equality holds, the nonnegative measurable function $$ H=\frac{F^p}{p}+\frac{G^q}{q}-FG $$ has integral $1-\int_XFG\,d\mu=0$. The Zero Integral Criterion implies $H=0$ almost everywhere. By the equality condition in Young’s inequality, $F^p=G^q$ almost everywhere. This is exactly the displayed condition in the theorem. Conversely, if $F^p=G^q$ almost everywhere, Young’s inequality is an equality almost everywhere. Integrating gives $$ \int_XFG\,d\mu =\frac{1}{p}\int_XF^p\,d\mu+\frac{1}{q}\int_XG^q\,d\mu =\frac{1}{p}+\frac{1}{q}=1. $$ Rescaling by the positive norms proves equality in Hölder’s inequality.

Worked Example: Equality and Strict Inequality on a Finite Space

Let $X=\{1,2\}$ with counting measure, and take $p=q=2$. First set $f=(1,2)$ and $g=(2,4)$. Then $$ \int_X|fg|\,d\mu=1\cdot2+2\cdot4=10, \qquad \|f\|_2=\sqrt{1^2+2^2}=\sqrt{5}, \qquad \|g\|_2=\sqrt{2^2+4^2}=\sqrt{20}. $$ Thus $\|f\|_2\|g\|_2=\sqrt{100}=10$, so equality holds. The equality condition also checks directly: $$ \frac{|f|^2}{\|f\|_2^2} =\left(\frac{1}{5},\frac{4}{5}\right) =\frac{|g|^2}{\|g\|_2^2}. $$ For contrast, take $u=(1,0)$ and $v=(0,1)$. Their norms are both $1$, but $$ \int_X|uv|\,d\mu=1\cdot0+0\cdot1=0<1=\|u\|_2\|v\|_2. $$ Their normalized squares are different at both points, so the equality condition correctly predicts strict inequality.

Why Normalization Matters

The proof separates the work into two parts. Young’s inequality is purely numerical, while normalization adapts it to functions with arbitrary finite norms. Without normalization, integrating Young’s inequality would produce terms involving $\int|f|^p$ and $\int|g|^q$ separately; dividing by the norms first makes each of those integrals equal to one and gives the sharp product of norms after rescaling.

A common pitfall is to apply Young’s inequality directly to $f$ and $g$ and expect the desired norm bound. Its terms involve the $p$th and $q$th powers, not the product of their norms. Another is to treat the endpoint as though it came from the same finite-exponent argument. At $p=1,\ q=\infty$, the essential bound on one factor is the key input. In every case, almost-everywhere inequalities suffice because changing a function on a null set does not change its integral.

Key takeaway: For finite conjugate exponents, normalize the functions and integrate Young’s inequality. Equality occurs precisely when the normalized $p$th and $q$th powers agree almost everywhere; the $L^1$-$L^\infty$ endpoint follows from an almost-everywhere bound.

Check Your Understanding

Use the scalar inequality, normalization argument, and equality condition to answer the following questions.

For $p=5/2$, what is the conjugate exponent $q$, and what relation characterizes equality in Young’s inequality?
Why does the normalized proof assume positive norms at first, and how are zero norms handled?
Which theorem supplies the almost-everywhere bound used in the endpoint proof?
For nonzero functions, what condition on their normalized powers is equivalent to equality in finite-exponent Hölder?
On a two-point space with counting measure, what feature of the pair $u=(1,0)$, $v=(0,1)$ makes their Hölder inequality strict?

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