real-analysisuniform-convergence
Sometimes I see in proofs they say "this function is pointwise convergent to limit a. Let's check if it's uniformly convergent to limit a." And then they do a uniform convergence check on a and prove that it cannot converge uniformly to limit a.
My question is, if we know a function converges pointwise to a limit, suppose it does converge uniformly, would it converge to the same limit?
For every irrational $x \in (0,1)$, $$\lim_{n \to \infty} f_n(x) = 0$$ is obvious. For a rational $x = \frac{p}{q}$, its again easy to see that $$\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{p}{q^n} = 0$$
Thus the pointwise limit is $0$ at all $x \in (0,1)$.
To show uniform convergence, note that $f_1(x) \leq 1$ for all $x$. Also $f_2(x) \leq \frac{1}{2}$ and so on, its easy to see (since for any rational $\frac{p}{q}$, $q \geq 2$) that $$f_n(x) \leq \frac{1}{2^n}$$
Now given $\epsilon > 0$, choose $N$ such that $$\frac{1}{2^N} < \epsilon$$ and you're done.
For your third question, JohnD's answer in the page you are quoting, contains a trick that's often used in Analysis courses. That is, try to find:
$$\sup_{x\in[0,1]}|f_n(x)-f(x)|$$
The supmemum occurs at $x$ such that:
$$\frac{df_n(x)}{dx}=0$$
Solving the above for $x$, you get:
$$x=\pm\frac{1}{n}$$
$\frac{1}{n}\in[0,1]$, so substitute back to the function to get:
$$\sup_{x\in[0,1]}|f_n(x)|=f_n\left(\frac{1}{n}\right)=\frac{1}{2}$$
And this is fixed and does not vanish, therefore convergence in $[0,1]$ is not uniform.
Addendum for comment:
I am adding a graphic, so you can see what's happening as a response to your second question.
This is the graph of $f_1(x)$, $f_2(x)$,..., $f_5(x)$, from right to left. Note that the supremum is given by $\left(\frac{1}{n},f_n\left(\frac{1}{n}\right)\right)=\left(\frac{1}{n},\frac{1}{2}\right)$ and is moved to the left on each iteration, but always stays at 1/2.
Best Answer
Yes, it does. Let $(f_n)$ be a series of functions converging uniformly to $f$. Then $\forall x$ we have $|f_n(x) - f(x)| \leq \sup_x{|f_n(x) - f(x)|}$. Thus $\lim_{n \to \infty}f_n(x) = f(x)$ (pointwise convergence). This limit is unique.