[Analysis] Uniform convergence

• Pointwise Convergence

Let $(N, d)$ be metric space. For sequence of functions

\[\begin{align} &f_n : A \longrightarrow N, \quad n =1, \cdots , n \\ &f\,\,\, : A \longrightarrow N \end{align}\]

The sequence of functions are said to converge pointwise to the function $f$, if for each $x \in A$, $f_n (x) \longrightarrow f(x)$,

i.e. for each $x\in A$ and $\epsilon >0$, $\exists L = L(x, \epsilon)$ such that $d(f_n (x), (x)) \leq \epsilon , \quad \forall n \geq L$

• Uniform Convergence

Let $(N, d)$ be metric space. For sequence of functions

\[\begin{align} &f_n : A \longrightarrow N, \quad n =1, \cdots , n \\ &f\,\,\, : A \longrightarrow N \end{align}\]

The sequence of functions are said to converge pointwise to the function $f$, if for each $x \in A$, $f_n (x) \longrightarrow f(x)$,

i.e. for each $x\in A$ and $\epsilon >0$, $\exists L = L( \epsilon)$ such that $d(f_n (x), (x)) \leq \epsilon , \quad \forall n \geq L$

which is same with

\[\underset{x \in A} {sup} \quad d(f_n (x) , (x)) < \epsilon, \quad \forall n \geq L\]

• Some relations

1. The derivatives of a pointwise convergent sequence of functions do not have to converge.

consider $X = \mathbb{R} $ and $f_n (x) = \frac{1}{n} sin(n^2 x)$ . Then,

\(\underset{n \rightarrow \infty}{lim} f_n (x) = 0\) intwise limit function is $f(x)=0$; the sequence of functions converges to 0. What about the derivatives of the sequence?

\[f_n '(x) = n cos (n^2 x)\]

and for most $x \in \mathbb{R}$, above derivative is unbounded, which means that it does not converge.

2. The integrals of a pointwise convergent sequence of functions do not have to converge.

Consider $X= [0,1],$ and $f_n (x) = \frac{2 n^2 x}{(1+n^2 x^2)^2}.$ Then,

\[lim_{n \rightarrow =\infty} f_n (x) = 0\]

However, the integrals are

\[\int^1_0 \frac{2 n^2 x dx}{(1+n^2 x^2)^2} \overset{u = 1+ n^2 x^2}{=} \int ^{1+n^2}_{1 }\frac{du}{u^2} = 1 - \frac{1}{1+n^2}\]

Therfore, even thought $lim_{n \rightarrow =\infty} f_n (x) = 0$ for all $x \in X$, the intergral is 1 as $n \rightarrow \infty$

3. The limit of a pointwise convergent sequence of continuous functions does not have to be contuniuous

$A = [ 0, 1]$ and $f_n(x) = x^n $. Then,

\(\underset{n \rightarrow \infty}{f_n (x)} = f(x) = \cases{0 \quad (0 \leq x <1)\\ 1 \quad (x=1)}\) It satisfies pointwise convegence, but limit is not continuous

4. The uniform convergence implies pointwise convergence, but not the other way around.

Same example with above one.

If $f_n(x)$ converges uniformly, then the limit function must be $f(x) =0$ for $x \in [0,1)$ and $f(1) = 1$. Uniform convergence implies that for any $\epsilon >0 $ there is $N_\epsilon \in \mathbb{N}$ such that $\vert x^n - f(x)\vert$ for all $n \geq N_\epsilon$ . Then, consier $\epsilon = \frac{1}{2}$. Then, there is $N $ such that for all $n \geq N$, $\vert x^n - f(x) \vert < \frac{1}{2}$. If we choose $n=N$ and $x = (\frac{3}{4})^N$, $f(x) = 0 $ and thus

\[\vert f_N (x) - f(x) \vert = \frac{3}{4}\]

contradicting our assumption.