# Proof of uniform limit of Continuous Functions

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.

## 1 Answer

We want to show that $f$ is continuous at a point $x_0$, say. The condition for continuity says that

Given any $\epsilon>0$, we can find a $\delta>0$ so that the following statement is true:

If $\lvert x - x_0 \rvert < \delta$ then $\lvert f(x)-f(x_0) \rvert < \epsilon$.

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.