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.