$L^2$ boundness of derivatives of a uniformly convergent sequence?

Question

Let $f_n\colon (a,b)\rightarrow \mathbb{R}$ be a sequence of functions in $C^\infty_c(a,b)$, that converges uniformily to a function $f$. Is the sequence of the derivatives bounded in $L^2(a,b)$? In other words, is it true that $$ \sup_{n\in\mathbb{N}} \int\limits_a^b \vert f_n'(x) \vert^2 dx <+\infty \quad? $$

Answer 1

No. Consider $g_n$ a $C^\infty$ function such that:

• $g_n(x)=1$ for all $x\in \left[\frac1n, 1\right]$;
• $g_n(x)=0$ for all $x\in \left(-\infty,\frac1{n+1}\right]\cup [2,\infty)$;
• $g_n(x)$ is increasing on $\left[\frac1{n+1},\frac1{n}\right]$ and decreasing on $[1,2]$

Also, let's assume that $g_n(x)=g_1(x)$ for all $n\in\Bbb N$ and for all $x\in[1,\infty)$.

Then consider $u_n(x)=g_n(x)x^{-1/2}-g_n(4-x)(4-x)^{-1/2}$.

The sequence $f_n(x)=\int_{-\infty}^x u_n(t)\,dt$ converges uniformly on $(0,4)$ (in fact, on $\Bbb R$) to the continuous function $f(x)=\int_0^x u(t)\,dt$, where $$u(x)=\begin{cases}0&\text{if }x<0\lor x>4\\ x^{-1/2}&\text{if }0\le x\le 1\\ u_1(x)x^{-1/2}&\text{if }1<x\le 2\\ -u_1(4-x)(4-x)^{-1/2}&\text{if }2\le x<3\\ -(4-x)^{-1/2}&\text{if }3\le x\le 4\end{cases}$$ Notice, in fact, that for all $x\in (0,4)$, $$\lvert f_n(x)-f(x)\rvert\le \left\lvert\int_0^{1/n} t^{-1/2}\,dt\right\rvert+\left\lvert\int_{4-\frac1n}^4 (4-t)^{-1/2}\,dt\right\rvert=4n^{-1/2}$$ However, since $\lvert u_n\rvert$ is an increasing sequence of functions, by Beppo Levi $$\int_0^4 \lvert f'_n(x)\rvert^2\,dx=\int_0^4 \lvert u_n(x)\rvert^2\,dx\ge \int_0^1 \lvert u_n(x)\rvert^2\,dx\stackrel{\text{Beppo Levi}}{\longrightarrow} \int_0^1 \frac1x\,dx=\infty$$