The uniform limit of Continuous Functions is continuous.
Proof
Let $\varepsilon > 0$. There exists $N$ in natural numbers such that $n>N$ implies
$|f_n(x)-f(x)| < \frac{\varepsilon}{3}$ for all $x$ in $S$.
In particular $|f_{N+1}(x)-f(x)| < \frac{\varepsilon}{3}$ for all $x$ in $S$.
Since $f_{N+1}$ is continuous at $x_0$ there is a $\delta>0$ such that,
$x$ in $S$ and $|x-x_0| < \delta$ imply $|f_{N+1}(x)-f_{N+1}(x_0)|< \frac{\varepsilon}{3}$
Now we conclude
$x$ in $S$ and $|x-x_0| < \delta$ imply $|f(x)-f(x_0)|< 3 \cdot \frac{\varepsilon}{3} = \varepsilon$.
Can someone explain what is happening here? I don't follow it all.
Best Answer
We want to show that $f$ is continuous at a point $x_0$, say. The condition for continuity says that
We want to prove this from stuff we know about the $f_n$. We know two things: firstly, that they are continuous, and secondly, that they converge uniformly to $f$.
Since $f_n$ converges uniformly to $f$, we can find an $N$ that is independent of $y$ so that $$\lvert f_n(y) - f(y) \rvert < \epsilon/3 $$ for any $n>N$, and any $y$ in the set. (In particular, this is true of both $y=x$ and $y=x_0$.)
Suppose we have such an $n$, $N+1$ will do, and now use that $f_{N+1}$ is continuous. Hence we can find a $\delta$ so that $$ \lvert f_{N+1}(x)-f_{N+1}(x_0) \rvert < \epsilon/3 $$ whenever $\lvert x - x_0 \rvert < \delta$.
We now use the triangle inequality: $$ \lvert f(x)-f(x_0) \rvert \leqslant \lvert f(x)-f_{N+1}(x) \rvert + \lvert f_{N+1}(x)-f_{N+1}(x_0) \rvert + \lvert f_{N+1}(x_0)-f(x_0) \rvert $$ Supposing now that $\lvert x - x_0 \rvert < \delta$, we apply the uniform convergence to the two end terms and the continuity of $f_{N+1}$ to the middle term, and find $$ \lvert f(x)-f_{N+1}(x) \rvert + \lvert f_{N+1}(x)-f_{N+1}(x_0) \rvert + \lvert f_{N+1}(x_0)-f(x_0) \rvert \leqslant \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon, $$ which is precisely what we wanted.