Hölder's Inequality for L1(Q) and Lp(E)

Hölder's Inequality for L1(Q) and Lp(E)

Recall from the Young's Inequality page that if $1 < p < \infty$ and $q$ is the conjugate index of $p$ and if $a, b \geq 0$ then:

(1)
\begin{align} \quad ab \leq \frac{a^p}{p} + \frac{b^q}{q} \end{align}

We will use Young's inequality to prove the significant Hölder's inequality for $L^1(E)$ and for $L^p(E)$ where $1 < p < \infty$.

Theorem 1 (Hölder's Inequality for $L^1(E)$ and $L^p(E)$): Let $E$ be a measurable set and let $1 \leq p < \infty$ with $q$ the conjugate index of $p$. If $f \in L^p(E)$ and $g \in L^q(E)$ then $fg \in L^1(E)$ and $\| fg \|_1 \leq \| f \|_p \| g \|_q$.
  • Proof: Observe that if $f = 0$ a.e. on $E$ or $g = 0$ a.e. on $E$ then Young's inequality holds trivially. So assume that $f \neq 0$ and $g \neq 0$ a.e. on $E$. We break the proof up into a few cases.
  • Case 1: Suppose that $p = 1$. Then $q = \infty$. Since $g \in L^q(E)$ we have that $|g(x)| \leq \| g \|_{\infty}$ a.e. on $E$. Then:
(2)
\begin{align} \quad \| fg \|_1 = \int_E |fg| = \int_E |f||g| \leq \int_E |f| \| g \|_{\infty} = \| g \|_{\infty}\int_E |f| = \| f \|_1 \| g \|_{\infty} \end{align}
  • Case 2: Suppose that $1 < p < \infty$. Let $f \in L^p(E)$], $g \in L^q(E)$ be such that $\| f \|_p = 1$ and $\| g \|_q = 1$. Since $|f(x)|, |g(x)| \geq 0$ for each $x \in E$, we have by Young's inequality that:
(3)
\begin{align} |f(x)g(x)| \leq \frac{|f(x)|^p}{p} + \frac{|g(x)|^q}{q} \end{align}
  • Taking the integral of both sides gives us that:
(4)
\begin{align} \quad \| fg \|_1 = \int_E |fg| \leq \int_E \left [ \frac{|f|^p}{p} + \frac{|g|^q}{q} \right ] = \frac{1}{p} \int_E |f|^p + \frac{1}{q} \int_E |g|^q = \frac{\| f \|_p}{p} + \frac{\| g \|_q}{q} = \frac{1}{p} + \frac{1}{q} = 1 = 1 \cdot 1 = \| f \|_p \| g \|_q \end{align}
  • So Young's Inequality holds if $\| f \|_p = 1$ and $\| g \|_q = 1$.
  • Let $f \in L^p(E)$ and let $g \in L^q(E)$. Define $F \in L^p(E)$ and $G \in L^q(E)$ by:
(5)
\begin{align} \quad F = \frac{f}{\| f \|_p} \quad , \quad G = \frac{g}{\| g \|_q} \end{align}
  • Then $\| F \|_p = 1$ and $\| G \|_q = 1$, and so:
(6)
\begin{align} \quad \| FG \|_1 \leq 1 \quad \Leftrightarrow \quad \biggr \| \frac{f}{\| f \|_p} \frac{g}{\| g \|_q} \biggr \|_1 \leq 1 \end{align}
  • Thus:
(7)
\begin{align} \quad \| fg \|_1 \leq \| f \|_p \| g \|_q \quad \blacksquare \end{align}
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