Unfortunately my analysis lecturer, as awesome as he is, lacks the structure in his lessons to provide worked out proofs for us to use as guidelines for proving other things. Hence, I am having a great deal of trouble proving this question (or even knowing what/how to prove).
Question: Show that a sequence $\left\{ f_{n}\right\} _{n=0}^{\infty}$ of functions which converges uniformly to the function $f$ on the interval $[a,b]$ converges pointwise to $f$ on that interval.
So far... I know that a sequence of functions converges uniformly if:
For $\forall \varepsilon>0$ $\exists N$ s.t $\forall x$ and $n \geq N$, we have $|f_{n}(x)-f(x)|< \varepsilon$ (Thank you wiki)
So, in order to show pointwise convergence, I need to show $\lim_{n\rightarrow\infty}f_{n}(x)=f(x)$ for every $x$.
I can clearly see that from the definition of uniform convergence, if my $n$ is tending to infinity, that is clearly an $n \geq N$, hence if I was to take:
$|f_{n}(x)-f(x)|< \varepsilon \implies -\varepsilon<f_{n}(x)-f(x)< \varepsilon$
But $\varepsilon > 0$, so:
$f_{n}(x)-f(x)< \varepsilon \implies f_{n}(x)<f(x)+ \varepsilon$
But if I consider that, as $n \rightarrow \infty$, $\varepsilon \rightarrow 0$, hence could I say:
$\lim_{n\rightarrow\infty}f_{n}(x)=f(x)$
Hence the sequence is pointwise convergent to $f$.
Problem: I feel like I'm missing steps, or logical connections, yet I'm at a loss to fill them. I feel like my basic idea is at least close to sound, but as I said, I may just be looking at the similarities and saying "It's clearly evident", but with more work. Would greatly appreciate a critique and aide, as we don't get a great deal of feedback or particular teaching on proofs - literally, our lecturer is great, but he tends to scrawl proofs rather vaguely, leaving out much of the "self evident" or "obvious" connections.
Thank you all!
I suggest that you should try to write everything in such a way as to clearly show what conditions bind each statement. For example:
Uniform convergence
$f_k \to f$ uniformly on $I$ iff:
For any real $ε > 0$:
For some natural $n$:
For any natural $k \ge n$:
For any $x \in I$:
$|f_k(x)-f(x)| < ε$
Pointwise convergence
$f_k \to f$ (pointwise) on $I$ iff:
For any $x \in I$:
For any real $ε > 0$:
For some natural $n$:
For any natural $k \ge n$:
$|f_k(x)-f(x)| < ε$
Observations
Observe that adjacent universal quantifications can be swapped (likewise for existential quantifications), so the only actual difference between the above two is the relative position of "For some natural $n$" and "For any $x \in I$". Then it is clear that the first implies the second because if some natural $n$ works for any $x \in I$, then I can use that same particular $n$ for any $x \in I$. Here is a completely formal proof of the fact that uniform convergence implies pointwise convergence:
Proof
For any functions $f$ and $( f_k : k \in \mathbb{N} )$ such that $f_k \to f$ uniformly on $I$:
For any $x \in I$:
[Here we pull in the definition of uniform convergence.]
For any real $ε > 0$:
For some natural $n$:
For any natural $k \ge n$:
For any $y \in I$:
$|f_k(y)-f(y)| < ε$
[Here we apply the above to the $x$ in the outer scope.]
Therefore $|f_k(x)-f(x)| < ε$
[Now we have exactly the definition for pointwise convergence.]
Therefore $f_k \to f$ (pointwise) on $I$
Exercise
Do the same for all the other notions like "uniform continuity", "continuity", "differentiability", "Lipschitz continuity" and so on. It would really help you know what each of them really means.