Sequence of Functions

Types of Convergence

Pointwise convergence

$$\forall \epsilon \gt 0\; \forall x\in [a,b]\; \exists N \in \mathbb{Z^+}\; \forall n > N\;;\; \big|f_n(x)-f(x)\big| \lt \epsilon$$

Here $N$ depends on $\epsilon, x$.

Examples:

• $x^n$ on $[0,1]$

Uniformly convergence

$$\forall \epsilon \gt 0\; \exists N \in \mathbb{Z^+}\; \forall x\in [a,b]\; \forall n > N\;;\; \big|f_n(x)-f(x)\big| \lt \epsilon$$

Here $N$ depends on $\epsilon$ only. Implies pointwise convergence.

Examples:

• $\frac{x^2}{n}$ on $[0,1]$

Uniform convergence tests

Supremum test

Suppose $u_n(x)$ is sequence of bounded functions. $u_n(x)$ converges to $u(x)$ uniformly iff:

$$\lim_{n\to\infty} \sup_x\, \lvert u_n(x) - u(x) \rvert = 0$$

Proof Hint

Let $l_n=\lvert u_n(x)-u(x) \rvert$. $$

To prove $\implies$: $$

• Consider the epsilon-delta definition of uniform convergence
• $\frac{\epsilon}{2}$ is an upperbound of $l_n$
• $\sup_x l_n \le \frac{\epsilon}{2} \lt \epsilon$

To prove $\impliedby$: $$

• Consider the epsilon-delta definition of the above limit
• $l_n \lt \sup_x l_n \lt \epsilon$

Properties of uniform convergence

Continuity

If $u_n(x)$ is continuous and converging to $u(x)$, then $u(x)$ is also continuous.

Proof Hint

Consider the limit definitions of:

1. $u_n(x)$ converges to $u(x)$
2. $u_n(x)$ is continuous at $a$

Consider $\lvert u(x)-u(a)\rvert$. Introduce $u_n(x)$ and $u_n(a)$ in there. Split into 3 absolute values. Show that the sum is lesser than $3\epsilon$.

Integrability

Explained in Converging Functions | Riemann Integration.

Differentiability

Uniform convergence-differentiation pair doesn’t go as smooth like integration was.

Suppose $u_n(x)$ is a sequence of differentiable functions, and they uniformly converges to $u(x)$. Differentiability of $u(x)$ is not guaranteed. An example is:

$$u_n(x) = \sqrt{x^2 + \frac{1}{n}}$$

$u_n(x)$ is differentiable but $u(x)$ is only differentiable on $\mathbb{R}-\set{0}$.

Theorem

If (all conditions must be met):

1. $u_n(x)$ is differentiable on $[a,b]$
2. $u_n(x_0)$ converges (pointwise) for some $x_0 \in [a,b]$
3. $u_n'(x)$ converges to $f(x)$ uniformly on $[a,b]$

Then:

1. $u_n(x)$ converges to $u(x)$ uniformly on $[a,b]$
2. $u(x)$ is differentiable on $[a,b]$
3. $u'(x)=f(x)$ OR in other words $u_n'(x)$ converges to $u'(x)$ uniformly