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Let $u:(0,\infty) \to \mathbb{R}$ be such that $u' \in L^2(0,\infty)$, but $u \notin L^2(0,\infty)$. However $u \in L^2(0,T)$ for all finite $T$.

Is it possible to find a sequence of functions $u_n \in L^2(0,\infty)$ with $u_n' \to u'$ in $L^2(0,\infty)$?

I'm thinking about multiplying $u$ with a smooth cut-off function, so $u_n = u\psi_n$ but I don't see any properties of the cut-offs $\psi_n$. Can someone give me some help or reference? Thanks.

aaa
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1 Answers1

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You're on the right track, but using the cut-off function on $u$ itself will lead you into trouble with the derivative. Instead, use it on $u'$ and take the integral.

Let $\varphi_n\in C^\infty_c(0,\infty)$ be such that $\varphi=0$ on $(0,\frac1n)\cup(2n,\infty)$, $\varphi=1$ on $(\frac2n,n)$ and $0\le\varphi\le1$ otherwise. For $x\in(0,\infty)$ define $$u_n(x):=\int_{2n}^xu'\varphi_n\ \mathrm{d}t=\int_0^xu'\varphi_n\ \mathrm{d}t-\int_0^{2n}u'\varphi_n\ \mathrm{d}t$$ Observe that $|u_n|\le2\left(\int_0^{2n}|u'|\right)\mathbf1_{(0,2n)}$, and so $$\|u_n\|_2\le4n\int_0^{2n}|u'|\le4n\|u'\|_2(2n)^{1/2}<\infty$$ where the last step comes from Holder's inequality. Hence $u_n\in L^2(0,\infty)$. Moreover, if $\phi\in C^\infty_c(0,\infty)$ then letting $c_n=\int_0^{2n}u'\varphi_n$ we have \begin{align*} \int u_n\phi'&=\int_0^\infty\int_0^xu'(t)\varphi_n(t)\phi'(x)\ \mathrm{d}t\mathrm{d}x-c_n\int_0^\infty\phi'(x)\ \mathrm{d}x\\ &=\int_0^\infty\int_t^\infty u'(t)\varphi_n(t)\phi'(x)\ \mathrm{d}x\mathrm{d}t\\ &=-\int_0^\infty u'(t)\varphi_n(t)\phi(t)\ \mathrm{d}t \end{align*} and so $u_n$ is weakly differentiable with $u_n'=u'\varphi_n$. The dominated convergence theorem implies $u'\varphi_n\to u'$ in $L^2$, so we are done.

Jason
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  • Thanks for this answer. If I had a uniform bound on the cutoff $psi_n$ and on its derivative, then my suggestion of the cutoff on $u$ itself should work, do you think? – aaa Feb 10 '15 at 12:17
  • I don't think so. We would need that $\varphi_n'u\to0$ which is not something we can really achieve. – Jason Feb 10 '15 at 13:07
  • For example (see http://math.stackexchange.com/questions/197431/construction-of-cut-off-function), we would have $\varphi_n' \to 0$ pointwise, and the uniform bound gives the convergence to zero. – aaa Feb 10 '15 at 13:28
  • Okay, fair enough. It didn't occur to me to control how quickly we dropped from $1$ to $0$. A sequence based on that function would also work. The advantage of mine is that we not require $u$ to be $L^2(0,T)$, only locally integrable. – Jason Feb 10 '15 at 22:45
  • Yes I like your approach. But I did not follow your last sentence: you are saying that $u \in L^2(0,T)$ for each $T$ is not the same as $u$ is locally integrable? – aaa Feb 11 '15 at 11:09
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    Not quite - locally integrable allows the function to blow up at zero as fast as you like. So $u(x)=\frac1x$ is locally integrable, but $u\notin L^2(0,T)$ for any $T>0$. – Jason Feb 11 '15 at 11:15
  • Although, come to think of it, your approach probably only requires $u$ to be locally integrable as well, provided you cut off the bump function at $frac1n$ as well. So maybe there isn't much difference. Note that $u$ is continuous so both sequences use continuous functions, which is nice. – Jason Feb 11 '15 at 11:19
  • Ah ok, I guess the point $0$ is excluded in the locally integrable version since the compact set $K$ is a subset of $(0,\infty)$ not $[0,\infty).$ Re: your second comment, it is interesting that cutting it off at $1/n$ as you suggest give sthe same result (I think)! – aaa Feb 11 '15 at 11:34