Question about proof: Uniform cauchy $\Rightarrow$ Uniform convergence

Question

I have one quick question regarding the proof of a theorem contained in these notes.

Theorem 5.13. A sequence $(f_n)$ of functions $f_n : A → R$ converges uniformly on $A$ if and only if it is uniformly Cauchy on $A$.

Question:

In the triangle inequality part

$|f_n(x) - f(x)| \leq |f_n(x) - f_m(x)| + |f_m(x) - f(x)|$

Why do we know that $f_m(x) \to f(x)$ as $m \to \infty$ so that $|f_m(x) - f(x)| < \frac{\epsilon}{2}$

I mean, isn't $|f_m(x) - f(x)|$ basically the same as $|f_n(x) - f(x)|$ on the left hand side? Just one is indexed by $m$ instead $n$? Why do we know that $|f_m(x) - f(x)| < \epsilon$ but we do not know immediately whether $|f_n(x) - f(x)| < \epsilon$

Answer 1

The subtlety is that $m$ depends on $x$. In the triangle inequality $$|f_n(x) - f(x)| \leq |f_n(x) - f_m(x)| + |f_m(x) - f(x)| \leq \frac\epsilon2 + |f_m(x) - f(x)| $$ $x$ is arbitrary. Initially $m$ is known to exceed $N$, but later $m$ is chosen so that, in addition, the second term on the RHS is less than $\frac\epsilon2$. (This is possible since, for each fixed $x$, the real sequence $\{f_n(x)\}$ is Cauchy and therefore converges to $f(x)$.) Note that this $m$ varies with $x$, but once the second term is replaced by $\frac\epsilon2$, the RHS no longer varies with $x$. After this the bound on $|f_n(x)-f(x)|$ is $\epsilon$, and is valid for every $x$, which was the goal.

Answer 2

Another way to look at it is as follows. This circumvents any consideration of the dependence of $m$ on $x$ which, clearly, is confusing.

It can be shown that for $n,m > N$ we have, for all $x \in A,$

$$|f_n(x) - f(x)| < \epsilon + |f_m(x) - f(x)|.$$

Hence,

$$|f_n(x) - f(x)| -\epsilon < |f_m(x) - f(x)|.$$

With $x$ fixed, we take the limit of both sides as $m \to \infty$. Since $f_m(x)$ converges pointwise to $f(x)$ and the LHS does not depend on $m$ it follows that for all $x \in A$ and $n > N$,

$$ |f_n(x) - f(x)| -\epsilon = \lim_{m \to \infty}\left(|f_n(x) - f(x)| -\epsilon\right) \leqslant \lim_{m \to \infty}|f_m(x) - f(x)| = 0 \\ \implies |f_n(x) - f(x)| \leqslant\epsilon .$$

Here we have used the following lemma which you can easily prove by contradiction:

$$a_m > a \, ,\forall m \in \mathbb{N} \implies \lim_{m \to \infty}a_m \geqslant a. $$