Hölder's Inequality for ℓ1 and ℓp

Hölder's Inequality for ℓ1 and ℓp

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 $\ell^1$ and for $\ell^p$ where $1 < p < \infty$.

 Theorem 1 (Hölder's Inequality for $\ell^1$ and $\ell^p$): Let $1 \leq p < \infty$ with $q$ the conjugate index of $p$. If $(a_i) \in \ell^p$ and $(b_i) \in \ell^q$ then $(a_ib_i) \in \ell^1$ and $\| a_ib_i \|_1 \leq \| a_i \|_p \| b_i \|_q$.
• Proof: Observe that if $(a_i) = (0)$ or $(b_i) = (0)$ then the inequality trivially holds. So assume that $(a_i) \neq (0)$ and $(b_i) \neq (0)$.
• Case 1: Suppose that $p = 1$. Then $q = \infty$. Since $(b_i) \in \ell^{\infty}$ we have that $(b_i)$ is bounded and $|b_k| \leq \| (b_i) \|_{\infty}$ for each $k \in \mathbb{N}$. Therefore:
(2)
\begin{align} \quad \| a_ib_i \|_1 = \sum_{k=1}^{\infty} |a_kb_k| = \sum_{k=1}^{\infty} |a_k||b_k| \leq \sum_{k=1}^{\infty} |a_k| \| (b_i) \|_{\infty} = \| (b_i) \|_{\infty} \sum_{k=1}^{\infty} |a_k| = \| (a_i) \|_1 \| (b_i) \|_{\infty} \end{align}
• Case 2: Suppose that $1 < p < \infty$. Suppose that $\| (a_i) \|_p = 1$ and $(b_i) \|_q = 1$. For all $i \in \mathbb{N}$ we have that $|a_i|, |b_i| \geq 0$. So by Young's inequality we have that:
(3)
\begin{align} \quad |a_ib_i| \leq \frac{|a_i|^p}{p} + \frac{|b_i|^q}{q} \end{align}
• Taking the sum from $i = 1$ to $\infty$ gives us:
(4)
\begin{align} \quad \| a_ib_i \|_1 = \sum_{i=1}^{\infty} |a_ib_i| \leq \sum_{i=1}^{\infty} \left [ \frac{|a_i|^p}{p} + \frac{|b_i|^q}{q} \right ] = \frac{1}{p} \sum_{i=1}^{\infty} |a_i|^p + \frac{1}{q} \sum_{i=1}^{\infty} |b_i|^q = \frac{1}{p} \| (a_i) \|_p + \frac{1}{q} \| (b_i) \|_q = \frac{1}{p} + \frac{1}{q} = 1 \leq \| (a_i) \|_p \| (b_i) \|_q \end{align}
• Now suppose that $(a_i) \in \ell^p$ and $(b_i) \in \ell^q$. Define $(A_i) \in \ell^p$ and $(B_i) \in \ell^q$ by:
(5)
\begin{align} \quad (A_i) = \left ( \frac{a_i}{\| (a_i) \|_p} \right ) \quad , \quad (B_i) = \left ( \frac{b_i}{\| (b_i) \|_q} \right ) \end{align}
• Then $\| (A_i) \|_p = 1$, $\| (B_i) \|_q = 1$, so from above:
(6)
\begin{align} \quad \| (A_iB_i) \|_1 \leq 1 \quad \Leftrightarrow \| (a_i)(b_i) \|_1 \leq \| (a_i) \|_p \| (b_i) \|_q \quad \blacksquare \end{align}