Minkowski's Inequality for ℓ1, ℓp, and ℓ∞

# Minkowski's Inequality for ℓ1, ℓp, and ℓ∞

Recall from the Hölder's Inequality for ℓ1 and ℓp page that if $1 \leq p < \infty$ and if $(a_i) \in \ell^p$, $(b_i) \in \ell^q$ then $(a_ib_i) \in L^1(E)$ and:

(1)
\begin{align} \quad \| (a_ib_i) \|_1 \leq \| (a_i) \|_p \| (b_i) \|_q \end{align}

We will now use this inequality to prove Minkowski's inequality, which will complete our proof in showing that $\ell^p$ is a normed linear space.

 Theorem 1 (Minkowski's Inequality for $\ell^1$, $\ell^p$, and $\ell^{\infty}$): Let $1 \leq p \leq \infty$. Then for all $(a_i), (b_i) \in \ell^p$ we have that $\| (a_i) + (b_i) \|_p \leq \| (a_i) \|_p + \| (b_i) \|_p$.
• So suppose that $1 < p < \infty$. Let $(a_i), (b_i) \in \ell^p$. Then for each $i \in \mathbb{N}$ we have that
(2)
\begin{align} \quad |a_i + b_i|^p = |a_i + b_i||a_i + b_i|^{p-1} \leq [|a_i| + |b_i|]|a_i + b_i|^{p-1} = |a_i||a_i + b_i|^{p-1} + |b_i||a_i + b_i|^{p-1} = |a_i||a_i + b_i|^{p/q} + |b_i||a_i + b_i|^{p/q} \end{align}
• Observe that $(a_i + b_i)^{p/q} \in L^q(E)$ since $\sum_{i=1}^{\infty} (|a_i + b_i|^{p/q})^q = \sum_{i=1}^{\infty} |a_i + b_i|^p < \infty$. Since $(a_i), (b_i) \in L^p(E)$, by Hölder's Inequality for L1(Q) and Lp(E) we have that:
(3)
\begin{align} \quad \| (a_i) + (b_i) \|_p^p = \sum_{i=1}^{\infty} |a_i + b_i|^p \leq \sum_{i=1}^{\infty} |a_i||a_i + b_i|^{p/q} + \sum_{i=1}^{\infty} |b_i||a_i + b_i|^{p/q} \leq \| (a_i) \|_p \| |(a_i) + (b_i)|^{p/q} \|_q + \| (b_i) \|_p \| |(a_i) + (b_i)|^{p/q} \|_q \end{align}
• Observe that:
(4)
\begin{align} \quad \| |(a_i) + (b_i)|^{p/q} \|_q = \left ( \sum_{i=1}^{\infty} (|a_i + b_i|^{p/q})^q \right )^{1/q} = \left ( \sum_{i=1}^{\infty} |a_i + b_i|^p \right )^{1/q} = \| (a_i) + (b_i) \|_p^{p/q} \end{align}
• Hence:
(5)
\begin{align} \quad \| (a_i) + (b_i) \|_p^p \leq [\| (a_i) \|_p + \| (b_i) \|_p] \| (a_i) + (b_i) \|_p^{p/q} \end{align}
• Dividing both sides by $\| (a_i) + (b_i) \|_p^{p/q}$ gives us that:
(6)
\begin{align} \quad \| (a_i) + (b_i) \|_p^{p - p/q} \leq \| (a_i) \|_p + \| (b_i) \|_p \end{align}
• But observe that $p - \frac{p}{q} = p - \frac{p(p-1)}{p} = p - (p - 1) = 1$. Hence:
(7)
\begin{align} \quad \| (a_i) + (b_i) \|_p \leq \| (a_i) \|_p + \| (b_i) \|_p \quad \blacksquare \end{align}