About possible pathological solutions to the DE $y=y'+y''+y'''+\ldots$ Is it possible to construct (or indirectly show the existence of) a function $y(x)\in\mathcal{C}^{\infty}(\mathbb{R})$ such that series
$$ S(x) = \sum_{n=1}^{+\infty}\frac{d^n y}{dx^n} $$
is pointwise convergent for any $x\in\mathbb{R}$, but $S(x)$ is not a differentiable function?

A more-or-less equivalent and strictly related question is the following one: is it possible for the operator
$$ T:\varphi(x) \mapsto e^x \int_{0}^{x} e^{-t}\varphi(t)\,dt $$
to produce $T(\varphi)\in \mathcal{C}^\infty$ without $\varphi$ being differentiable? Probably no, because...
Assuming that the integral appearing above is the Riemann integral, $T$ acts on the space of almost-everywhere continuous functions, so we are free to assume that $\varphi$ is a.e. $\mathcal{C}^0$, as well as
$$\Phi(x)=\frac{d}{dx}\int_{0}^{x}e^{-t}\varphi(t)\,dt. $$
By the remark in the comments, $\int_{0}^{x}\left(\Phi(t)-e^{-t}\varphi(t)\right)\,dt$ equals zero almost everywhere, so the fundamental theorem of Calculus "almost applies". $T(\varphi)$ is smooth by assumption, so it is $T(\varphi)e^{-x}=\int_{0}^{x}e^{-t}\varphi(t)\,dt$ and its derivative $\Phi(x)$. On the other hand $\Phi(x)=e^{-x}\varphi(x)$ almost everywhere, so $\varphi(x)$ is $\mathcal{C}^\infty$ almost everywhere.
 A: If you know that $S(x)$ is continuous, the answer is no. Assume that there is a point $x_{0}\in\mathbb{R}$ such that $S(x_{0})$ is not differentiable. Let $a,b\in\mathbb{R}$ such that $x_{0}\in\left[a,b\right]$. Since $$S(x)=\sum_{k\geq0}y^{(k)}\left(x\right)$$ for every $x\in\left[a,b\right]$ then $y^{(k)}\left(x\right)\rightarrow0$ as $k\rightarrow+\infty$. This means that the formal Taylor series $$\sum_{k\geq0}\frac{y^{(k)}\left(x\right)}{k!}(t-x)^{k}$$ has radius of convergence $\rho(x)=+\infty$, for all $x\in\left[a,b\right]$. In particular there exists some $\delta>0$ such that $\rho(x)\geq\delta$ for all $x\in\left[a,b\right]$ and this is enough to conclude that $y(x)$ is analytic for $x\in\left[a,b\right]$ (see, for example, here). Now, since$$\sup_{x\in\left[a,b\right]}\left|S(x)-\sum_{k=0}^{n}y^{(k)}\left(x\right)\right|=\left|S\left(x^{*}\right)-\sum_{k=0}^{n}y^{(k)}\left(x^{*}\right)\right|\rightarrow0$$ as $k\rightarrow+\infty$ then we can conclude that the convergence is uniform in $\left[a,b\right]$ and so $S(x)$ is analytic for $x\in\left[a,b\right]$; this is a contradiction. Furthermore, we observe that $$\int_{a}^{x}\sum_{k\geq0}y^{(k)}\left(t\right)dt=\int_{a}^{x}y\left(t\right)dt+\int_{a}^{x}\sum_{k\geq1}y^{(k)}\left(t\right)dt=Y(x)+\sum_{k\geq1}y^{(k-1)}\left(x\right)-\sum_{k\geq1}y^{(k-1)}\left(a\right)$$ $$=Y(x)+S(x)-S(a)$$ where $Y(x)=\int_{a}^{x}y\left(t\right)dt$, since the convergence is uniform and so we can exchange the integral with the series. But we also have $$\int_{a}^{x}\sum_{k\geq0}y^{(k)}\left(t\right)dt=\int_{a}^{x}S\left(t\right)dt$$ hence we have the ordinary differential equation of the first order $$S^{\prime}(x)-S(x)=-y(x)$$ and so we have the the family of solutions $$S(x)=e^{-x}\left(c-\int_{a}^{x}y(t)e^{-t}dt\right)$$ for $a\leq x\leq b$.
