1
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – RRL
    Oct 21, 2018 at 20:31

3 Answers 3

2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $n$.

$\endgroup$
1
$\begingroup$

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 $f$.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Siddhartha
    Oct 21, 2018 at 23:46
  • 1
    $\begingroup$ @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 $y$. $\endgroup$
    – RRL
    Oct 21, 2018 at 23:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .