# Does uniform convergence and pointwise continuity of a sequence of functions on $[0,1]$ imply continuity of the limit function?

If $$(f_n)$$ is a sequence of real functions on $$[0, 1]$$ converging uniformly to a function $$f$$ on $$[0, 1]$$, and if $$f_n$$ is continuous at $$x_n ∈ [0, 1]$$ for each $$n$$ with $$x_n\stackrel{n\to\infty}\longrightarrow x$$, must $$f$$ be continuous at $$x$$?

I believe that this statement is true but I am having trouble proving it. Could someone please help out?

• Do you really mean that $f_n$ is continuous at a single point $x_n$? – Christian Blatter Oct 19 '18 at 18:14
• @ChristianBlatter I believe it is safe to assume that $f_n$ is continuous at $x_n$ for each $n$. – Math1000 Oct 19 '18 at 18:17
• that's what i meant! sorry – MathematicianP Oct 19 '18 at 18:19

Consider the function $$f(x) = 1$$ for $$x>0$$ and $$f(x)=0$$ for $$x \le 0$$. Let $$f_n = f$$ for all $$f$$. Certainly $$f_n$$ converges uniformly to $$f$$. Now let $$x_n = 1/n$$. Then for every $$n$$ we have $$f_n$$ is continuous at $$x_n$$. Also $$x_n$$ converges to $$0$$, but $$f$$ is not continuous at $$0$$.

• That's so clear! Thank you! – MathematicianP Oct 21 '18 at 17:16

In the case where all the functions $$f_n$$ are continuous at a point $$x_0$$ (so $$x_0$$ is fixed and doesn’t depend on $$n$$) then this is true.

The sequence $$(f_n)$$ converges uniformly to the function $$f$$.

Hence we have :

$$\forall \epsilon > 0, \exists N, \forall n \geq N, \forall x \in [0,1], \mid f_n(x)-f(x) \mid < \epsilon$$

Moreover all the $$f_n$$ are continuous at $$x_0 \in [0,1]$$ which means :

$$\forall \epsilon > 0, \exists \delta, \forall x \in [x_0-\delta,x_0+\delta], \mid f_n(x_0)-f_n(x)\mid < \epsilon$$

Hence :

$$\forall \epsilon > 0, \exists \delta, \forall x \in [x_0-\delta,x_0+\delta], \mid f(x)-f(x_0) \mid = \mid f(x)-f_N(x)+ f_N(x)-f_N(x_0) + f_N(x_0) -f(x_0) \mid \leq \mid f(x)-f_N(x) \mid+ \mid f_N(x)-f_N(x_0) \mid + \mid f_N(x_0) -f(x_0) \mid \leq 3\epsilon$$

So $$f$$ is continuous at the point $$x_0$$.

• The statement is false as evidenced by @GEdgar's answer. – Math1000 Oct 19 '18 at 18:23
• Ok, I miss understood the question. I thought it was at a single point $x_0$ that doesn’t depend on $n$. – Thinking Oct 19 '18 at 18:24