📂Banach Space

Proof that the p-norm becomes the maximum norm when p=∞

Theorem

Let’s define the sequence spaces $l^{p}$ and $1 < p_{0} < \infty$ such that $\left\{ x_{n} \right\}_{n \in \mathbb{N} } \in \mathcal{l}^{p_{0}}$. $$\lim_{p \to \infty} \left( \sum_{n \in \mathbb{N} } | x_{n} |^{p} \right)^{ {{1} \over {p}} } = \sup_{n \in \mathbb{N}} | x_{ n } |$$

Explanation

Despite being encountered early in analysis or linear algebra, the reason why the maximum norm is related to $\infty$ is often not well explained. Especially, professors tend to overlook this because it seems so obvious to them, but if it does not resonate, make sure to delve deeper.

Visualization ¹

No need for a thousand words. The picture above asymptotically converges to the shape shown when plotting the $p$-norm unit ball in the Euclidean space given $\mathbb{R}^{2}$.

Intuitive Mathematical Development

Approaching conceptually with mathematical formulas can be more helpful. When $\mathbf{x} \in \mathbb{R}^{n}$, $\left\| \mathbf{x} \right\|_{p}$ can be expressed as $$\left\| \mathbf{x} \right\|_{p} = \sqrt[p]{\left| x_{1} \right|^{p} + \cdots + \left| x_{n} \right|^{p}}$$ If the largest value among $\left| x_{1} \right|, \cdots , \left| x_{n} \right|$ is $\bar{x} = \max_{k = 1 , \cdots , n} \left| x_{k} \right|$, then the mathematical development might seem as follows. $$\begin{align*} \lim_{p \to \infty} \left\| x \right\|_{p} =& \lim_{p \to \infty} \sqrt[p]{\left| x_{1} \right|^{p} + \cdots + \left| x_{n} \right|^{p}} \\ =& \lim_{p \to \infty} \sqrt[p]{ \bar{x}^{p} \left( \left| {\frac{ x_{1} }{ \bar{x} }} \right|^{p} + \cdots + \left| {\frac{ x_{n} }{ \bar{x} }} \right|^{p} \right) } \\ =& \lim_{p \to \infty} \sqrt[p]{\bar{x}^{p}} \cdot \sqrt[p]{ \left( \left| {\frac{ x_{1} }{ \bar{x} }} \right|^{p} + \cdots + \left| {\frac{ x_{n} }{ \bar{x} }} \right|^{p} \right) } \\ \overset{?}{=} & \bar{x} \lim_{p \to \infty} \sqrt[p]{ 0 + \cdots + 1 + \cdots + 0 } \\ =& \bar{x} \lim_{p \to \infty} m^{{\frac{ 1 }{ p }}} \\ =& \max_{k = 1 , \cdots , n} \left| x_{k} \right| \cdot 1 \end{align*}$$ Although the rigorous proof does not proceed in this manner, having such intuition makes a significant difference in grasping the maximum norm.

Proof

Let’s say $\displaystyle M := \sup_{n \in \mathbb{N}} | x_{ n } |$. If $M=0$ then because of $\displaystyle 0 = \lim_{p \to \infty} \left( \sum_{n \in \mathbb{N} } | x_{n} |^{p} \right)^{ {{1} \over {p}} } = \sup_{n \in \mathbb{N}} | x_{ n } | = 0$, let’s assume $M > 0$. Defining a new sequence as $\displaystyle y_{n} : = {{x_{n}} \over {M}}$, since $\displaystyle \sup_{n \in \mathbb{N}} | y_{ n } | = 1$ and $\left\{ y_{n} \right\}_{ n \in \mathbb{N} } \in \mathcal{l}^{p_{0}}$, to show $\displaystyle \lim_{p \to \infty} \left( \sum_{n \in \mathbb{N} } | x_{n} |^{p} \right)^{ {{1} \over {p}} } = M$, it is sufficient to demonstrate $\displaystyle \lim_{p \to \infty} \left( \sum_{n \in \mathbb{N} } \left| {{x_{n} } \over {M}} \right|^{p} \right)^{ {{1} \over {p}} } = 1$.

• Part 1. $\displaystyle \liminf_{p \to \infty } \left( \sum_{n \in \mathbb{N} } | y_{n} |^{p} \right)^{ {{1} \over {p}} } \ge 1$

  Since $\displaystyle \sup_{n \in \mathbb{N}} | y_{ n } | = 1$, for any $\varepsilon > 0$, there exists $n_{0} \in \mathbb{N}$ satisfying $| y_{n_{0} } | > 1 - \varepsilon$. As $\displaystyle \liminf_{p \to \infty } \left( \sum_{n \in \mathbb{N} } | y_{n} |^{p} \right)^{ {{1} \over {p}} } \ge | y_{ n_{0} } | > 1 - \varepsilon$ holds for all $\varepsilon > 0$,

  $$\liminf_{p \to \infty } \left( \sum_{n \in \mathbb{N} } | y_{n} |^{p} \right)^{ {{1} \over {p}} } \ge 1$$

• Part 2. $\displaystyle \limsup_{p \to \infty} \left( \sum_{ n \in \mathbb{N} } | y_{n} |^{p} \right)^{ {{1} \over {p}} } \le 1$

  Since $\left\{ y_{n} \right\}_{ n \in \mathbb{N} } \in \mathcal{l}^{p_{0}}$, there exists $N \in \mathbb{N}$ satisfying $\displaystyle \sum_{ n > N } | y_{n} |^{p_{0}} < 1$. If we say $p > p_{0}$,

  $$\sum_{ n > N } | y_{n} |^{p} < \sum_{ n > N } | y_{n} |^{p_{0}} < 1$$

  Because $\displaystyle \left( \sum_{ n \in \mathbb{N} } | y_{n} |^{p} \right)^{ {{1} \over {p}} } \le \left( | y_{1} |^{p} + \cdots + | y_{N} |^{p} + 1 \right)^{ {{1} \over {p}} } \le \left( N + 1 \right)^{ {{1} \over {p}} }$ holds,

  $$\limsup_{p \to \infty} \left( \sum_{ n \in \mathbb{N} } | y_{n} |^{p} \right)^{ {{1} \over {p}} } \le \limsup_{p \to \infty} \left( N + 1 \right)^{ {{1} \over {p}} } = 1$$

• Part 3.
  Following Part 1. and Part 2., $$\limsup_{p \to \infty} \left( \sum_{ n \in \mathbb{N} } | y_{n} |^{p} \right)^{ {{1} \over {p}} } \le 1 \le \liminf_{p \to \infty } \left( \sum_{n \in \mathbb{N} } | y_{n} |^{p} \right)^{ {{1} \over {p}} }$$ as a result, we obtain the following. $$\lim_{p \to \infty} \left( \sum_{ n \in \mathbb{N} } | y_{n} |^{p} \right)^{ {{1} \over {p}} } =1$$

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