Holder Inequality

Last updated Nov 1, 2022

Let $\infty \geq p,q \geq 1$ so that $\frac{1}{p} + \frac{1}{q} = 1$. Holders Inequality is broken into the following statements.

1. If $f,g \in L^{0}$, $||fg||{1} \leq ||f||{p}||g||_{q}$.
2. If $f \in L^{p}, g \in L^{q}$, $fg \in L^{1}$.
3. If $1 < p,q < \infty$ and $f \in L^{p}, g \in L^{q}$, then $||fg||{1} = ||f||{p} ||g||_{q}$ If and Only If either $f$ or $g$ is $0$ Almost Everywhere or there exist $\alpha, \beta > 0$ so $\alpha|f|^{p} = \beta |g|^{q}$ (Almost Everywhere).
4. If $f \in L^{\infty}, g \in L^{1}$, then $||fg||{1} = ||f||{\infty} ||g||_{1}$ If and Only If $g = 0$ Almost Everywhere or there exists $C \geq 0$ so that $C = |f|$ Almost Everywhere.

First note that if (1) holds, then (2) holds. This follows because if $f \in L^{p}$ and $g \in L^{q}$ so that $1 \leq p,q \leq \infty$ and $\frac{1}{p} + \frac{1}{q} = 1$, then $$||fg||{1} \leq ||f||{p} ||g||_{q} < \infty.$$ so $fg \in L^{1}$. $\checkmark$

Next, we establish (1). Getting some edge cases out of the way for $1 \leq p, q \leq \infty$:

1. If $||f||{p} = 0$, then $f=0$, so $||fg||{1} = 0$ and (1) holds, where we take $0 * \infty := 0$. Symmetry ensures (1) follows if $g = 0$.
2. If $||f||{p} = \infty$ and $||g||{q} \neq 0$, then because $||fg||{1} \leq \infty$ already, $||fg||{1} \leq ||f||{p} ||g||{q}$. Symmetry ensures (1) follows if we flip $f$ and $g$. $\checkmark$

It remains to show (1) for $f,g \in L^{0}$ so that $0 < ||f||{p}, ||g||{q} < \infty$. Note that if the result holds for all $f’,g’ \in L^{0}$ so $||f’||{p} = 1 = ||g’||{q}$, then we have for $f’ = \frac{f}{||f||{p}}$ and $g’ = \frac{g}{||g||{q}}$ $$||fg||{1} = ||f||{p} ||g||{q} ||f’g’||{1} \leq ||f||{p} ||g||{q} ||f’||{p}||g’||{q} = ||f||{p}||g||{q}.$$ Thus it suffices to prove $||fg||_{1} \leq 1$ for all $f \in S(L^{p}),g \in S(L^{q})$.

If $p= \infty$ and $q=1$, then $|f| \leq 1$ Almost Everywhere, which means $|fg| = |f||g| \leq |g|$ Almost Everywhere. Therefore $$||fg||{1} = \int\limits |fg| \leq \int\limits |g| = ||g||{1} = 1 = ||f||{\infty}||g||{1}$$ by Integration is Non-Decreasing. Symmetry ensures the result holds for $q=\infty$, $p = 1$. $\checkmark$

If $1 < p,q < \infty$, then by the Log Concavity Inequality, if $\frac{1}{p} + \frac{1}{q} = 1$, we have that $$|fg| = (|f|^{p})^{\frac{1}{p}} (|g|^{q})^{\frac{1}{q}} \leq \frac{1}{p} |f|^{p} + \frac{1}{q} |g|^{q}.$$ Because Integration is Non-Decreasing, $$||fg||{1} = \int\limits |fg| \leq \frac{1}{p} \int\limits |f|^{p} + \frac{1}{q} \int\limits |g|^{q} = \frac{1}{p} ||f||{p}^{p} + \frac{1}{q} ||g||{q}^{q} = 1 = ||f||{p}||g||_{q}.$$

This completes our proof of (1). $\checkmark$

Let $1 < p,q < \infty$ so that $\frac{1}{p} + \frac{1}{q} = 1$.

Let $f \in L^{p}, g \in L^{q}$ so that $||fg||{1} = ||f||{p}||g||_{q}$. We break this proof into the following cases:

# $f = 0$ or $g=0$ Almost Everywhere

Then $\Rightarrow$ direction trivially holds by assumption.

# $f,g \neq 0$ Almost Everywhere

For $f \in S(L^{p}), g \in S(L^{q})$ we would have $$\begin{align*} &||fg||{1} = \int\limits |fg| = \frac{1}{p} \int\limits |f|^{p} + \frac{1}{q} \int\limits |g|^{q} = 1 = ||f||{p} ||g||{q}\\ \Rightarrow&0 = \int\limits \frac{1}{p} |f|^{p} + \frac{1}{q} |g|^{q} - |fg|\\ \Rightarrow&\frac{1}{p} |f|^{p} + \frac{1}{q} |g|^{q} = |fg| \text{ a.e.} \end{align*}$$ because $|fg| \leq \frac{1}{p} |f|^{p} + \frac{1}{q} |g|^{q}$ and Integration of a Nonnegative Function is 0 iff the Function is 0 Almost Everywhere. By the Log Concavity Inequality, since $\frac{1}{p} \in (0,1)$, this means $|f|^{p} = |g|^{q}$ a.e.. For arbitrary $f \in L^{p}, g \in L^{q}$ because $||fg||{1} = ||f||{p} ||g||{q}$ means $||\frac{f}{||f||{p}} \frac{g}{||g||{q}}||{1} = || \frac{f}{||f||{p}} ||{p} || \frac{g}{||g||{q}} ||{q}$, we know that Almost Everywhere $$\begin{align*} &|\frac{f}{||f||{p}}|^{p} = |\frac{g}{||g||{q}}|^{q}\\ \Rightarrow &||g||{q}^{q} |f|^{p} = ||f||{p}^{p} |g|^{q}. \end{align*}$$ Letting $\alpha := ||g||^{q}{q}$ and $\beta := ||f||{p}^{p}$, and noting both $||g||{q} > 0$ and $||f||_{p} >0$, we have our result. $\checkmark$

# $f = 0$ or $g=0$ Almost Everywhere

Let $f \in L^{p}, g \in L^{q}$ so that $f=0$ Almost Everywhere. Then equality follows from noting $$||fg||{1} = 0 = ||f||{p} ||g||_{q}.$$ By symmetry on $p,q$, the proof also holds if $g=0$ Almost Everywhere.

# $\alpha |f|^{p} = \beta |g|^{q}$ for $\alpha, \beta > 0$ and $f,g \neq 0$ somewhere

The assumptions means neither $f=0$ or $g=0$ Almost Everywhere. Thus $||f||{p}, ||g||{q} > 0$.

Since $\alpha > 0$, we have that $| \frac{f}{||f||{p}} |^{p} = \frac{\beta}{\alpha ||f||{p}^{p} } |g|^{q}$. Let $C := (\frac{\beta}{\alpha ||f||{p}^{p} })^{\frac{1}{q}}$. Then we have $$1 = \int\limits | \frac{f}{||f||{p}} |^{p} = \int\limits |Cg|^{q}$$ and $Cg \in S(L^{q})$. Denote $f’ := \frac{f}{||f||{p}}$ and $g’ := Cg$. We have that $|f’|^{p} = |g’|^{q}$ Almost Everywhere, so by the Log Concavity Inequality: $$|f’g’| = \frac{1}{p} |f’|^{p} + \frac{1}{q} |g’|^{q}.$$ Then $$||f’g’||{1} = \frac{1}{p} + \frac{1}{q} = 1 = ||f’||{p}||g’||{q}.$$ Thus, $$\frac{C}{||f||{p}} ||fg||{1} = \frac{C}{||f||{p}} ||f||{p}||g||{q}$$ giving us $||fg||{1} = ||f||{p}||g||{q}$. $\checkmark$

We assume $f \in L^{\infty}, g \in L^{1}$ and $||fg||{1} = ||f||{\infty}||g||_{1}$

# $g =0$ Almost Everywhere

Then the result hold trivially by assumption.

# $g \neq 0$ somewhere

We will show $|f| = ||f||{\infty}$ Almost Everywhere. Let $f’ = \frac{f}{||f||{\infty}} \in S(L^{\infty})$. Then $$||f’g||{1} = \frac{||fg||{1}}{||f||{\infty}} = \frac{||f||{\infty} ||g||{1}}{||f||{\infty}} = ||g||{1}.$$ Since $|f’| \leq ||f’||{\infty} = 1$ Almost Everywhere, We have $|f’g| \leq |g|$ Almost Everywhere. Thus,

$$\begin{align*} &\int\limits |f’g| = ||f’g||{1} = ||g||{1} = \int\limits |g|\\ \Rightarrow&0 = \int\limits |g| - |f’g|\\ \Rightarrow&|g| = |f’||g| \text{ a.e.}\\ \Rightarrow&1 = |f’| \text{ a.e.} \end{align*}$$ because Integration of a Nonnegative Function is 0 iff the Function is 0 Almost Everywhere and $|g| \neq 0$ somewhere. Therefore $|f| = |f’| ||f||{\infty} = ||f||{\infty}$ Almost Everywhere. $\checkmark$

# $g =0$ Almost Everywhere

Then equality follows from noting $$||fg||{1} = 0 = ||f||{\infty} ||g||_{1}.$$

# $|f| = C$ for $C \geq 0$ Almost Everywhere

Then $||f||{\infty} = C$ so $$||fg||{1} = \int\limits|fg| = C\int\limits |g| = ||f||{\infty} ||g||{1}.$$ This completes the proof of (4) and the entire set of statement. $\checkmark$

$\blacksquare$

1. If $\infty > p,q > 0$ so that $\frac{1}{p} + \frac{1}{q} = 1$, then $q = \frac{p}{p-1}$.
2. If $\infty > p,q > 0$ so that $\frac{1}{p} + \frac{1}{q} =1$, then noting Remark (1) above, $q > 0 \Rightarrow 1-p > 0$, so $p > 1$. But our defining property of $p,q$ is symmetric, so $q > 1$ as well. Thus, we can relax our condition on $p,q$ to be $0 < p,q \leq \infty$.
3. Although we are using Lp Norms in the definition, we have not yet proven they satisfy the Triangle Inequality. We can however use their Non-Degeneracy and Homogeniety since that was proved in Lp Space without the use of Holders Inequality. However, in our statement for Holders Inequality.

• Measureable Function
• Lp Space
• Unit Sphere