Derivation in the sense of distributions

Question

Let $f_n(x)$ be a continuous sequence functions. Assume that the series $\sum_{n\in N} f_n(x)$ converges simply. Put $f(x)=\sum^\infty_{n=0} f_n(x)$. My question if the function $f$ is continuous. Is that $$\big(\sum^\infty_{n=0} f_n(x)\big)'=\sum^\infty_{n=0} f'_n(x)$$ in the sense of distributions.

Answer 1

For $n=0, 1, 2, \ldots$ let $f_n = F_n - F_{n-1}$ where $$ F_n(x) = \begin{cases} n^2 x & \text{when $0 \leq x \leq 1/n$}, \\ n(2-nx) & \text{when $1/n \leq x \leq 2/n$}, \\ 0 & \text{otherwise}. \end{cases} $$ Then $\sum_{k=0}^{n} f_k = F_n$ converges to $0$ pointwise but to $\delta$ as distribution. Therefore, $$ 0 = (\underbrace{\sum_{k=0}^\infty f_k}_{\text{pointwise limit}})' \neq \underbrace{\sum_{k=0}^\infty f_k'}_{\text{distributional limit}} = \delta'. $$

Pointwise convergence to $0$

For $x<0$ the "otherwise" case applies so $F_n(x)=0$ so here convergence to $0$ is trivial.

For $x=0$ the first case applies so $F_n(0)=n^2\cdot 0=0$ and convergence to $0$ is trivial.

For $x>0$ there exists some $n$ such that $x>2/n.$ The "otherwise" case then applies and convergence to $0$ is trivial.

Thus, $F_n \to 0$ pointwise.

Convergence to $\delta$

Let $\phi\in C^\infty_c(\mathbb R).$ Then $$ \langle F_n, \phi \rangle \int F_n(x) \, \phi(x) \, dx = \int_{0}^{1/n} n^2 x \, \phi(x) \, dx + \int_{1/n}^{2/n} n(2-nx) \, \phi(x) \, dx. $$ Now change variable to $t=nx.$ This gives $$ \langle F_n, \phi \rangle = \int_{0}^{1} t \, \phi(t/n) \, dt + \int_{1}^{2} (2-t) \, \phi(t/n) \, dt \to \int_{0}^{1} t \, \phi(0) \, dt + \int_{1}^{2} (2-t) \, \phi(0) \, dt = \left( \int_{0}^{1} t \, dt + \int_{1}^{2} (2-t) \, dt \right) \, \phi(0) = \phi(0) = \langle \delta, \phi \rangle $$ as $n\to\infty.$

This implies that $F_n'\to \delta'$ as distributions.