Exercise 3.5.3 (Weierstrass M-test)

Let us show that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^N{f}^{(i)}$ is a Cauchy sequence in $C(X\to ℝ)$.

Since ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{\parallel f^{(n)}\parallel }_{\infty }$ is convergent, we know that $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{\parallel f^{(n)}\parallel }_{\infty }=0$.

That is, given ${𝜖}^{\prime }>0$ there exists $N^{\prime }>0$ such that ${\parallel f^{(n)}\parallel }_{\infty }\mid <{𝜖}^{\prime }$ for all $n>N^{\prime }$.

So,

for all $n>N^{\prime }$.

Given $𝜖>0$ take ${𝜖}^{\prime }=\frac{𝜖}{n-m}$ and $N=N^{\prime }$. Then

Note that we assumed $n>m$, but the same argument gives us the case $n<m$, and $n=m$ is obvious.

Thus, ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^N{f}^{(i)}$ is a Cauchy sequence in $C(X\to ℝ)$ (I did not state this earlier, but a finite sum of continuous and bounded functions is also continuous and bounded, so each term is in $C(X\to ℝ)$ ).

And now using Theorem 3.4.5. we have that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{f}^{(n)}$ converges uniformly to some function $f$ on $X$, and that function $f$ is also continuous.