# Error in uniform convergence question?

Suppose that $$(X,\ d)$$ and $$(Y,\ \rho)$$ are metric spaces, that $$f_n:X\to Y$$ is continuous for each $$n$$, and that $$(f_n)$$ convergence pointwise to $$f$$ on $$X$$. If there exists a sequence $$(x_n)$$ in $$X$$ such that $$x_n\to x$$ in $$X$$ but $$f_n(x_n)\not\to f(x)$$, show that $$(f_n)$$ does not converge uniformly to $$f$$ on $$X$$.

I've managed to "prove" that the question is self-contradictory, so please find my error.

$$\forall n\in\mathbb{N}$$, $$f_n$$ is continuous at $$x$$. Let $$\epsilon>0$$. $$\forall n\in\mathbb{N}$$ $$\exists\delta>0$$ s.t. $$\rho(f_n(y),\ f_n(x))<\epsilon/2$$ for all $$y\in X$$ s.t. $$d(y,\ x)<\delta$$. ------------ (1)

Since $$x_m\to x$$, $$\exists N_1\in\mathbb{N}$$ s.t. $$d(x_m,\ x)<\delta$$ $$\forall m\ge N_1$$. ------------- (2)

From (1) and (2),

$$\forall n$$, $$\forall m\ge N_1\implies d(x_m,\ x)<\delta\implies \rho(f_n(x_m),\ f_n(x))<\epsilon/2$$ ------------- (3)

Also, $$f_n(y)\to f(y)$$ for all $$y\in X$$ due to pointwise convergence. So, $$\exists N_2\in\mathbb{N}$$ s.t. $$\rho(f_n(y),\ f(y))<\epsilon/2$$ $$\forall n\ge N_2$$ and $$\forall y\in X$$. ----------- (4)

Let $$N_3=\max(N_1,\ N_2)$$. Suppose $$n\ge N_3$$. Then $$n\ge N_1$$ and $$n\ge N_2$$.

\begin{aligned} \implies\rho(f_n(x_n), f(x))&\le\rho(f_n(x_n),\ f_n(x))+\rho(f_n(x),\ f(x)) \\ &<\epsilon/2+\epsilon/2\text{ [Using (3) and (4)]} \\ &=\epsilon \end{aligned}

I have thus "proved" that $$f_n(x_n)\to f(x)$$, contradicting the question. Where have I gone wrong?

• In (1) you are assuming equicontinuity where $\delta$ does not depend on $n$ -- a stronger assumption than what you are given. Along with pointwise convergence this implies uniform convergence.
– RRL
Oct 21, 2018 at 20:31

If we are more precise, we may see where the error is. I will use $$|y-x|$$ to mean $$d(y,x)$$ since I think it makes it clearer than working with $$d$$ and $$\rho$$. It won't change any of the important details.

For each $$n$$, $$f_n$$ is continuous at $$x$$. Let $$\epsilon>0$$. Then, for each $$n\in\Bbb N$$, there is a $$\delta = \delta(n)$$, which depends on $$n$$, such that $$|f_n(y)-f_n(x)|<\epsilon/2$$ whenever $$|y-x|<\delta(n)$$.

Since $$x_m\to x$$, there is an $$N_1 = N_1(n)\in\Bbb N$$ (note that $$N_1$$ also depends on $$n$$ here!) such that $$|x_m - x| < \delta(n)$$ for each $$m\ge N_1$$.

Now you claim that for each $$k,m\ge N_1(n)$$, whenever $$|x_m-x|<\delta(n)$$, you have $$|f_k(x_m)-f_k(x)| < \epsilon/2$$.

This is the first serious point you erred. The reason this is an incorrect deduction is that for different $$n$$, you don't know that $$N_1 = N_1(n)$$ is large enough to guarantee that $$|f_k(x_m)-f_k(x)|<\epsilon/2$$. Only for the specific $$n$$ for which $$\delta = \delta(n)$$ is this true.

Edit: RRL summed it up nicely in the comment. You are assuming equicontinuity of the sequence $$\{f_n\}$$ when you are neglecting the $$n$$ that the $$\delta = \delta(n)$$ depends on.

This is an excellent example of how imprecise language/notation can lead to confusion.

In your first line you write "for all $$n$$, there exists $$\delta$$ such that..." This sort of suggests that the same $$\delta$$ can be used for all $$n$$, which is not correct and is what leads to your conclusion.

In the future it might be more useful to think of this as "for each $$n$$, there exists $$\delta$$...". At least to me, this suggests more strongly that the $$\delta$$ may be different for each $$n$$.

For a correct proof (by contradiction), show that $$f(x_n) \not\to f(x)$$ is impossible if convergence is uniform using

$$\rho(f_n(x_n), f(x)) \leqslant \rho(f_n(x_n), f(x_n)) + \rho(f(x_n), f(x))$$

Note that $$f$$ must be continuous when the sequence of continuous functions $$f_n$$ is uniformly convergent. Thus each term on the RHS is smaller than $$\epsilon/2$$ for sufficiently large $$n$$ -- the first by the uniform convergence $$f_n \to f$$ and the second by the convergence $$x_n \to x$$ and the continuity of $$f$$.

• I'm sorry but I don't see why given $\epsilon>0$, there must exist $N\in\mathbb{N}$ s.t. $\rho(f_n(x_n),\ f(x_n))<\epsilon/2$ whenever $n\ge N$. I know that for any $y\in X$, there must be an $N\in\mathbb{N}$ s.t. $\rho(f_n(y),\ f(y))<\epsilon/2$ whenever $n\ge N$ but I'm not sure how we can just replace $y$ by $x_n$ here. Oct 21, 2018 at 23:46
• @Thomas: We are assuming uniform convergence here and showing it leads to a contradiction for $f(x_n) \not\to f(x)$. So $\rho(f_n(y),\ f(y))<\epsilon/2$ when $n \geqslant N$ and for EVERY $y \in X$. This inequality must hold in particular for $x_n$. Note that $N$ does not depend on $y$.
– RRL
Oct 21, 2018 at 23:54