$\text { Let } u_{n}(x)=\frac{1}{n^{2}} e^{-n^{2} x^{2}}, x \in \mathbb{R}$
(i) Calculate Var $u_{n}$
(ii) Let $u(x):=\sum_{n=1}^{\infty} u_{n}(x), x \in \mathbb{R} .$ Prove that $u \in B P V(\mathbb{R})$
(iii) Prove that $\sum_{n=1}^{\infty} u_{n}^{\prime}(x)$ does not converge uniformly in $[-1,1]$
(iv) Find a formula for $u^{\prime}$.
Well,I have solved (i) (ii).
For (iv),Since $\sum_{n=1}^{\infty} u_{n}^{\prime}(x)$ does not converge uniformly in $[-1,1]$,So we can't use term by term differential theorem .OTOH, we observe that $u(x)$ is monotonous in $[-\infty,0],[0,+\infty]$,So we can use Fubini term by term differential theorem(see https://en.wikipedia.org/wiki/Fubini%27s_theorem_on_differentiation) in $[-n,1-n],[n-1,n]$,Then we obtain that $u^{\prime}=\sum_{n=1}^{\infty} u_{n}^{\prime}(x) $ a.e. in $\mathbb{R}$.But we cannot find a formula of $u^{\prime} $for the whole real line in this way.
Can someone help me solve (iii) (iv)?Thanks in advance.
(iii) Note that for $x \in [0,1]$,
$$\left|\sum_{k=n+1}^\infty u_n'(x) \right| \geqslant \sum_{k=n+1}^{2n} 2xe^{-k^2x^2}\geqslant n \cdot 2xe^{-4n^2x^2},$$
and, since $\frac{1}{n} \in [0,1]$,
$$\sup_{x \in [0,1]}\left|\sum_{k=n+1}^\infty u_n'(x) \right| \geqslant n \cdot 2 \cdot \frac{1}{n}e^{-2n^2(1/n)^2}=2e^{-2}$$
Since the RHS does not converge to $0$ as $n \to \infty$ the Cauchy criterion is violated , and the series fails to converge uniformly on $[0,1]$ and, therefore, on $[-1,1]$.