# Pointwise convergence and properties of the limit (part 1)

We look here at the continuity of a sequence of functions that converges pointwise and give some counterexamples of what happens versus uniform convergence.

### Recalling the definition of pointwise convergence

We consider here real functions defined on a closed interval $$[a,b]$$. A sequence of functions $$(f_n)$$ defined on $$[a,b]$$ converges pointwise to the function $$f$$ if and only if for all $$x \in [a,b]$$ $$\displaystyle \lim\limits_{n \to +\infty} f_n(x) = f(x)$$. Pointwise convergence is weaker than uniform convergence.

### Pointwise convergence does not, in general, preserve continuity

Suppose that $$f_n \ : \ [0,1] \to \mathbb{R}$$ is defined by $$f_n(x)=x^n$$. For $$0 \le x <1$$ then $$\displaystyle \lim\limits_{n \to +\infty} x^n = 0$$, while if $$x = 1$$ then $$\displaystyle \lim\limits_{n \to +\infty} x^n = 1$$. Hence the sequence $$f_n$$ converges to the function equal to $$0$$ for $$0 \le x < 1$$ and to $$1$$ for $$x=1$$. Although each $$f_n$$ is a continuous function of $$[0,1]$$, their pointwise limit is not. $$f$$ is discontinuous at $$1$$. We notice that $$(f_n)$$ doesn't converge uniformly to $$f$$ as for all $$n \in \mathbb{N}$$, $$\displaystyle \sup\limits_{x \in [0,1]} \vert f_n(x) - f(x) \vert = 1$$. That's reassuring as uniform convergence of a sequence of continuous functions implies that the limit is continuous!

However, a sequence of continuous functions can converge pointwise to a continuous function although the convergence is not uniform. Consider the sequence $$(g_n)$$:
$g_n: \left|\begin{array}{lrl} [0,1] & \longrightarrow & \mathbb{R} \\ x & \longmapsto & 2nx \text{ if } x \in [0,\frac{1}{2n}] \\ x & \longmapsto & 2n(\frac{1}{n} – x) \text{ if } x \in [\frac{1}{2n},\frac{1}{n}] \\ x & \longmapsto & 0 \text{ if } x \ge \frac{1}{n} \end{array}\right.$ Each $$g_n$$ is continuous and the sequence $$(g_n)$$ converges pointwise to the function vanishing on $$[0,1]$$. We can see that as for all $$n \in \mathbb{N}$$ $$g_n(0) = 0$$ while for $$x \neq 0$$ $$g_n(x) = 0$$ for $$n \ge 1/x + 1$$. However $$\displaystyle \sup\limits_{x \in [0,1]} \vert g_n(x) \vert = 1$$ for all $$n \in \mathbb{N}$$, hence $$(g_n)$$ does not converge uniformly to its pointwise limit, i.e. the always vanishing function.

### A pointwise convergent sequence of functions need not be bounded

Consider the sequence $$(h_n)$$:
$h_n: \left|\begin{array}{lrl} [0,1] & \longrightarrow & \mathbb{R} \\ x & \longmapsto & 2 n^2 x \text{ if } x \in [0,\frac{1}{2n}] \\ x & \longmapsto & 2 n^2 (\frac{1}{n} – x) \text{ if } x \in [\frac{1}{2n},\frac{1}{n}] \\ x & \longmapsto & 0 \text{ if } x \ge \frac{1}{n} \end{array}\right.$ Using considerations similar to the sequence $$(g_n)$$, one can prove that $$(h_n)$$ is a sequence of continuous functions that converges pointwise to the always vanishing function. However the sequence is unbounded as $$g_n(\frac{1}{n})=n$$.

I’ll come back later on to other cases of pointwise convergence of sequences of continuous functions.