Hölder's Inequality (General)

Recall from the Young's Inequality page that if $a, b \geq 0$ and $p, q > 1$ are such that $\displaystyle{\frac{1}{p} + \frac{1}{q} = 1}$ then:

(1)

\begin{align} \quad ab \leq \frac{a^p}{p} + \frac{b^q}{q} \end{align}

We now prove another very important inequality known as Hölder's Inequality.

Theorem 1 (Hölder's Inequality (General)): Let $1 \leq p < \infty$ and let $q$ be the conjugate index of $p$. If $(X, \mathfrak T, \mu)$ be a measure space and $f, g : X \to \mathbb{C}$ are measurable functions then $\displaystyle{\int_X |fg| \: d \mu \leq \| f \|_p \| g \|_q}$.

• Proof: There are two cases to consider.

• Case 1: Suppose that $p = 1$. Then $q = \infty$. Recall that:

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\begin{align} \quad \| g \|_{\infty} = \inf \left \{ M > 0 : |g(x)| \leq M \: \mu-\mathrm{a.e. \: on \:} X \right \} \end{align}

• Therefore, we have that $|g(x)| \leq \| g \|_{\infty}$ $\mu$-almost everywhere on $X$. Hence:

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\begin{align} \quad \int_X |fg| \: d \mu = \int_X |f||g| \: d \mu \leq \int_X |f| \| g \|_{\infty} \: d \mu = \left ( \int_X |f| \: d \mu \right ) \| g \|_{\infty} = \| f \|_1 \| g \|_{\infty} \end{align}

• Case 2: There are a few subcases to consider.

• Case 2.1: Suppose that $\| f \|_p = 0$. Then $\displaystyle{\left ( \int_X |f|^p \: d \mu \right )^{1/p} = 0}$ which means that $\displaystyle{\int_X |f|^p \: d \mu = 0}$. But this implies that $f(x) = 0$ $\mu$-almost everywhere on $X$. So $\displaystyle{\int_X |fg| \: d \mu =0}$ and trivially the inequality holds. An analogous argument can be made for the case when $\| g \|_q = 0$.

• Case 2.2: Suppose that $\| f \|_p = \infty$ or $\| g \|_q = 0$. Then trivially the inequality holds.

• Case 2.3: Suppose that $0 < \| f \|_p < \infty$ and $0 < \| g \|_q < \infty$. Let $x \in X$ and let:

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\begin{align} \quad a = \frac{|f(x)|}{\| f \|_p} \quad \mathrm{and} \quad b = \frac{|g(x)|}{\| g \|_q} \end{align}

• Note that $a, b > 0$. So by Young's inequality we have that:

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\begin{align} \quad ab = \frac{|f(x)|}{\| f \|_p} \cdot \frac{|g(x)|}{\| g \|_q} = \frac{|f(x)g(x)|}{\| f \|_p \| g \|_q} \leq \frac{1}{p} \frac{|f(x)|^p}{\| f \|_{p}^p} + \frac{1}{q} \frac{|g(x)|^q}{\| g \|_{q}^q} = \frac{a^p}{p} + \frac{b^q}{q} \end{align}

• We integrate both sides of the inequality above over $X$ to get that:

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\begin{align} \quad \int_X \frac{|fg|}{\| f\|_p \| g \|_q} \: d \mu \leq \int_X \left [ \frac{1}{p} \frac{|f|^p}{\| f \|_{p}^p} + \frac{1}{q} \frac{|g|^q}{\| g \|_{q}^q} \right ] \: d \mu &= \frac{1}{p} \int_X \frac{|f|^p}{\| f \|_{p}^p} \: d \mu + \frac{1}{q} \int_X \frac{|g|^q}{\| g \|_{q}^q} \: d \mu \\ &= \frac{1}{p \| f \|_p^p} \int_X |f|^p \: d \mu + \frac{1}{q \| g \|_q^q} \int_X |g|^q \: d \mu \\ &= \frac{1}{p \| f \|_p^p} \| f \|_p^p + \frac{1}{q \| g \|_q^q} \| g \|_q^q \\ &= \frac{1}{p} + \frac{1}{q} \\ &= 1 \end{align}

• Hence:

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\begin{align} \quad \int_X |fg| \: d \mu \leq \| f \|_p^p \| g \|_q^q \quad \blacksquare \end{align}