📂Analysis

Convergence of Norms of Function Sequences

Definitions

Suppose a sequence of functions $\left\{ f_{n} \right\}$ is given. If $\| f_{n} - f \|$ converges to $0$, then $f_{n}$ is said to converge in norm, denoted as follows.

$$f_{n} \to f \text{ in norm }$$ or $$\| f_{n} - f\| \to 0$$ or $$\lim \limits_{n \to 0} \| f_{n}-f\|=0$$

Explanation

To define the limit of a sequence, the concept of distance is necessary. Since distance in a function space is defined by the norm, the convergence of a sequence of functions is defined as above.

The norm of a function is defined by integration, so for $f_{n}$ to converge in norm to $f$ means that the average difference between $f_{n}$ and $f$ over a given interval converges to $0$.

$$f_{n}\rightarrow f\ \ \mathrm{in\ norm}\quad \iff \quad \int_{a}^b\left| f_{n}(x)-f(x) \right|^2 dx \rightarrow 0$$

It is important to note that norm convergence and pointwise convergence do not guarantee each other. On the other hand, uniform convergence guarantees norm convergence.

Example

Norm convergence does not imply pointwise convergence

Consider the interval $[0,1]$ where $f_{n}$ is as follows.

$$f_{n}(x) =\begin{cases} 1 & 0\le x \le \dfrac{1}{n} \\ 0 &\dfrac{1}{n}<x\le 1\end{cases}$$

Then, it can easily be shown that $\lim \limits_{n \to \infty} \| f_{n} \| = 0$ holds as follows.

$$\lim \limits_{n \to \infty} \|f_{n}-0 \|^2 = \lim \limits_{n \to \infty} \int_{0}^1 |f(x)-0|^2 dx = \lim \limits_{n \to \infty} \int_{0}^{\frac{1}{n}}dx = \lim \limits_{n \to \infty} \dfrac{1}{n} = 0$$

However, it is clear that for all $n$, since $f_{n}(0)=1$, $f_{n}(x)$ does not converge to $0$ at $x=0$.

Pointwise convergence does not imply norm convergence

Consider the interval $[0,1]$ where $g_{n}$ is as follows.

$$g_{n}(x) = \begin{cases} n & 0<x<\dfrac{1}{n} \\ 0 & \mathrm{elsewhere} \end{cases}$$

Then, since for all $n$, $g_{n}(0)=0$ holds, $\lim \limits_{n \to \infty} g_{n}(x) = 0$ follows. However, $\lim \limits_{n \to \infty} \| g_{n} \| =0$ does not hold.

$$\lim \limits_{n \to \infty} \| g_{n}-0 \|^2 = \lim \limits_{n \to \infty} \int_{0}^1 |g_{n}(x)-0|^2dx = \lim \limits_{n \to \infty} \int_{0}^{ \frac{1}{n} } n^2dx = \lim \limits_{n \to \infty} n = \infty \ne 0$$

Theorem

Suppose on the interval $[a,b]$ that $f_{n}(x)$ uniformly converges to $f(x)$. Then, $f_{n}$ norm converges to $f$.

Proof

According to the conditions for uniform convergence, for all $x \in [a,b]$, there exists $|f_{n}(x)-f(x)| \le M_{n}$ such that $\lim \limits_{n \to \infty} M_{n} = 0$ is satisfied. Therefore, the following holds.

$$\|f_{n}-f \|^2 = \int_{a}^b|f_{n}(x)-f(x)|^2dx \le \int_{a}^b {M_{n}}^2dx=(b-a){M_{n}}^2$$

Therefore, $f_{n}$ norm converges to $f$.

$$\lim \limits_{n \to \infty} \|f_{n}-f \|^2 \le \lim \limits_{n \to \infty} (b-a){M_{n}}^2 = 0$$

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