# 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.

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.