# Thread: Some problems concerning uniform convergence

1. ## Some problems concerning uniform convergence

Hi, I have some problems with a couple of exercises concerning uniform convergence, and am hoping for some helpful hints.

Problem 1: Let (X, d_X) be a metric space, f, f_n : X -> C (complex numbers) for n = 1,2,... and f_n converges uniformly to f on X. If each f_n is continuous at x in X, and if x_n -> x in X, prove that lim n->infinity f_n(x_n) = f(x).

Problem 2: Suppose (X, d_X) is a metric space, that f_n : X -> C is continuous for each n, and that {f_n} converges pointwise to f on X. If there exists a sequence {x_n} in X such that x_n -> x in X but f_n(x_n) does not converge to f(x), show that {f_n} does not converge uniformly to f on X.

I'm not sure how hard these really are, but they're so full of information that I can't figure out where to begin (plus, I'm in a hurry, exam's on Friday .

2. OK, I think I've solved the first one:

Fix $\epsilon >0$. Now choose $N_1$ such that $n \geq N_1$ implies $| f_n (x) - f(x) | \leq \frac{\epsilon}{2}$. (Possible since $f_n \to f$ uniformly.) Next, choose $\delta > 0$ such that $|p-x| < \delta$ implies $| f_n (p) - f_n (x) | < \frac{\epsilon}{2}$. (Possible since $f_n$ is continuous at $x$.) Now, choose $N_2$ such that $n \geq N_2$ implies $|x_n-x| < \delta$. (Possible since $x_n \to x$.) Let $N$ be the maximum of ${ N_1 , N_2 }$. Then $n \geq N$ implies $|x_n-x| < \delta$ which again implies $|f_n (x_n) - f_n (x) | < \frac{\epsilon}{2}$, but it also implies $| f_n (x) - f(x) | \leq \frac{\epsilon}{2}$. But then $| f_n (x_n) - f(x) | \leq | f_n (x_n) - f_n (x) | + | f_n (x) - f(x) | < \epsilon$, so $f_n (x_n) \to f(x)$.

After some consideration it seems to me that the second problem is a direct consequence of the first, because if we assume that ${f_n}$ does converge uniformly to $f$ we get a contradiction. Is this correct?

3. Originally Posted by claves
OK, I think I've solved the first one:

Fix $\epsilon >0$. Now choose $N_1$ such that $n \geq N_1$ implies $| f_n (x) - f(x) | \leq \frac{\epsilon}{2}$. (Possible since $f_n \to f$ uniformly.) Next, choose $\delta > 0$ such that $|p-x| < \delta$ implies $| f_n (p) - f_n (x) | < \frac{\epsilon}{2}$. (Possible since $f_n$ is continuous at $x$.) Now, choose $N_2$ such that $n \geq N_2$ implies $|x_n-x| < \delta$. (Possible since $x_n \to x$.) Let $N$ be the maximum of ${ N_1 , N_2 }$. Then $n \geq N$ implies $|x_n-x| < \delta$ which again implies $|f_n (x_n) - f_n (x) | < \frac{\epsilon}{2}$, but it also implies $| f_n (x) - f(x) | \leq \frac{\epsilon}{2}$. But then $| f_n (x_n) - f(x) | \leq | f_n (x_n) - f_n (x) | + | f_n (x) - f(x) | < \epsilon$, so $f_n (x_n) \to f(x)$.

After some consideration it seems to me that the second problem is a direct consequence of the first, because if we assume that ${f_n}$ does converge uniformly to $f$ we get a contradiction. Is this correct?
Just one thing: How do you claim that this $\delta$ works for all $f_n$ with $n>N$ (notice that it depends on your choice of $n$)?

4. Originally Posted by Jose27
Just one thing: How do you claim that this $\delta$ works for all $f_n$ with $n>N$ (notice that it depends on your choice of $n$)?
Thanks for pointing that out. I've tried to find a solution, but I am struggling with it, so any ideas would be greatly appreciated.

5. New try:

First observe that since each $f_n$ is continuous at $x$, $f$ is continuous at $x$.

Fix $\epsilon >0$. Now choose $N_1$ such that $n \geq N_1$ implies $| f_n (x_n) - f(x_n) | \leq \frac{\epsilon}{2}$. (Possible since $f_n \to f$ uniformly.) Next, choose $\delta > 0$ such that $|p-x| < \delta$ implies $| f (p) - f (x) | < \frac{\epsilon}{2}$. (Possible since $f$ is continuous at $x$.) Now, choose $N_2$ such that $n \geq N_2$ implies $|x_n-x| < \delta$. (Possible since $x_n \to x$.) Let $N$ be the maximum of $\{ N_1 , N_2 \}$. Then $n \geq N$ implies $|x_n-x| < \delta$ which again implies $|f (x_n) - f (x) | < \frac{\epsilon}{2}$, but it also implies $| f_n (x_n) - f(x_n) | \leq \frac{\epsilon}{2}$. But then $| f_n (x_n) - f(x) | \leq | f_n (x_n) - f (x_n) | + | f (x_n) - f(x) | < \epsilon$, so $f_n (x_n) \to f(x)$.

This is me, btw: $\Rightarrow$

6. Originally Posted by claves
New try:

First observe that since each $f_n$ is continuous at $x$, $f$ is continuous at $x$.

Fix $\epsilon >0$. Now choose $N_1$ such that $n \geq N_1$ implies $| f_n (x_n) - f(x_n) | \leq \frac{\epsilon}{2}$. (Possible since $f_n \to f$ uniformly.) Next, choose $\delta > 0$ such that $|p-x| < \delta$ implies $| f (p) - f (x) | < \frac{\epsilon}{2}$. (Possible since $f$ is continuous at $x$.) Now, choose $N_2$ such that $n \geq N_2$ implies $|x_n-x| < \delta$. (Possible since $x_n \to x$.) Let $N$ be the maximum of $\{ N_1 , N_2 \}$. Then $n \geq N$ implies $|x_n-x| < \delta$ which again implies $|f (x_n) - f (x) | < \frac{\epsilon}{2}$, but it also implies $| f_n (x_n) - f(x_n) | \leq \frac{\epsilon}{2}$. But then $| f_n (x_n) - f(x) | \leq | f_n (x_n) - f (x_n) | + | f (x_n) - f(x) | < \epsilon$, so $f_n (x_n) \to f(x)$.

This is me, btw: $\Rightarrow$

Now, if you're interested you could try an prove the converse, namely:

If for every sequence $x_n\rightarrow x$ we have $f_n(x_n)\rightarrow f(x)$ then $f_n\rightarrow f$ uniformly on compact subsets of $X$