Show that $\sum_{k=m}^\infty f_k$ converges uniformly to a differentiable function $f:[a,b]\rightarrow R$

Question

Let $(f_k)_{k=m}^\infty$ be a sequence of differentiable functions $f_k:[a,b]\rightarrow R$ whose derivatives are continuous. Suppose there exists a sequence $(M_k)_{k=m}^\infty$ in $R$ with $|f_k'|\le M_k$ for all $x\in X, k\geq m,$ and such that $\sum_{k=m}^\infty M_k$ converges. Assume also that there is some $x_0\in [a,b]$ such that $\sum_{k=m}^\infty f_k(x_0)$ converges.

Show that $\sum_{k=m}^\infty f_k$ converges uniformly to a differentiable function $f:[a,b]\rightarrow R$ and that $f'(x)=\sum_{k=m}^\infty f_k'(x)$ for all $x\in [a,b]$.

$Remark:$ So you are showing that under these assumptions,

$\frac{d}{dx}\sum_{k=m}^\infty f_k= \sum_{k=m}^\infty \frac{d}{dx} f_k$

$Hint:$ Combine the Weirestrass M-test with Theorem $3.7.1$: Let $[a, b]$ be an interval, and for every integer $n ≥ 1$, let $f_n : [a, b] → R$ be a differentiable function whose derivative $f_n' : [a, b] → R$ is continuous. Suppose that the derivatives $f_n'$ converge uniformly to a function $g : [a, b] → R$. Suppose also that there exists a point $x_0$ such that the limit lim$_{n→∞} f_n (x_0)$ exists. Then the functions $f_n$ converge uniformly to a differentiable function $f$, and the derivative of $f$ equals $g$.

Answer 1

\begin{align} f(x)-f(x_0) & = \sum_{n=0}^{\infty}(f_k(x)-f_k(x_0))\\ & = \sum_{n=0}^{\infty}\int_{x_0}^{x}f_k'(t)dt. \end{align} The last sum converges absolutely because the integral is bounded in absolute value by $M_k|x-x_0|$. And the sum $\sum_k f_k(x_0)$ converges. Therefore $\sum_{n}f_k(x)$ converges for all $x$, and the convergence is uniform by the Weierstrass M-test. So $f(x)$ is a continuous function.

Answer 2

Here is a lemma you need to show that $f=\sum_k f_k$ is differentiable: suppose each $g_n$ is continuously differentiable, $\{g_n\}$ converges uniformly to $g$ and $\{g_n'\}$ converges uniformly to $h$. Then g is necessarily differentiable and $g'=h$. Once you prove this you can take $g_n$ to be the n-th partial sum of the series $\sum_k f_k$ to conclude that f is differentiable. To prove the lemma we have $g_n (x)-g_n (x_0 )=\int_{x_0} ^{x} g_n'(t)dt$. Letting $n \to \infty$ we get $g(x)-g(x_0)=\int_{x_0} ^{x} h(x)dx$. Being a uniform limit of continuous functions h is continuous. The indefinite integral of a continuous function is differentiable and we get $g'(x)=h(x)$ for all x.