Exercise 6.3.7

Take any $𝜖>0$; we want to show that there is some $N$ so that for $n,m>N$, $|f_n(x)-f_m(x)|<𝜖$ (to use the Cauchy Criterion for Uniform Convergence).

Let $h_{n,m}(x)=f_n(x)-f_m(x)$, and note that $h_{n,m}^{\prime }$ converges uniformly to 0 as $n,m$ go to infinity (as a consequence of $(f_n^{\prime })$ converging uniformly). More formally, for any ${𝜖}_1>0$ we have some $N_1$ for which if $n,m>N_1$ then $\left|h_{n,m}^{\prime }(x)\right|<{𝜖}_1\forall <!--\; FUNCTION\; APPLICATION\; -->x\in [a,b]$.

By the Mean Value Theorem,

for some $x_1\in [a,b]$, and therefore if $n,m>N_1$ we have

This addresses the first term in the question hint.

Second, since $(f_n(x_0))$ converges as $n\to \infty$ and is thus a Cauchy sequence, we have that for any ${𝜖}_2$ there is an $N_2$ so that when $n,m>N_2$, $\left|f_n(x_0)-f_m(x_0)\right|<{𝜖}_2$. This addresses the second term.

Setting ${𝜖}_1=\frac{𝜖}{2(b-a)}$, ${𝜖}_2=\frac{𝜖}{2}$, and $N=\max <!--\; FUNCTION\; APPLICATION\; -->\{N_1,N_2\}$ completes the proof.