Convergence in L^2

Question

Define a function $f_\varepsilon: \mathbb{R} \to \mathbb{R}$ as $$\begin{equation} f_\varepsilon(x) = \begin{cases} -1 & \text{if}\; \,x<-\varepsilon \\ \sin \left(\frac{\pi x}{2 \varepsilon}\right) & \text{if}\;\, |x| \le \varepsilon\\ 1 & \text{if}\;\, x>\varepsilon \end{cases} \end{equation} $$ Then $f_\varepsilon(x)$ converges pointwise to $\mbox{sgn}(x)$ as $\varepsilon \to 0$. And it can be easily proven that $ \int_{\mathbb{R}}|f_\varepsilon(x)- \mbox{sgn}(x)|^2\,dx= C\varepsilon $ .

My question is given $u \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ and $\varphi \in C^\infty_c(\mathbb{R})$ can we prove $$\int_{\mathbb{R}}|f_\varepsilon(u(x))- \mbox{sgn}(u(x))|^2\varphi(x)\,dx \le C\varepsilon^r$$ for some r>0. I actually tried to mimic the proof but the main issue over here is we cannot apply the change of variable formula. And I'm always ending up in an upper bound like $\varepsilon^r+c$, which is not desired.

If the fact you can't apply change of variable formula brothers you, I'd suggest try to prove it first for u smooth and then just approach your general function by smooth functions on the support of $\phi$. — Keen, Jan 09 '19 at 13:42
@Keen $u$ is required to be one-one to apply change of variable, which need not be true here. — Savannah, Jan 09 '19 at 17:59

score 2 · Accepted Answer · answered Jan 09 '19 at 23:21

2

Take, for every $0 < t < 1$, $u(t)=e^{-1/t}$ and $u=0$ everywhere else. Take $\varphi$ to be a smooth approximation of $1_{[0,1]}$, that is not lower than this function. On $[0,1]$, $\{0 < u \leq r\}$ has measure $|\ln{r}|^{-1}$.

Thus your integral is not lower than $\int_{0<|u| \leq \epsilon/2}{|1-sin(\pi/4)|^2} \geq \frac{c}{|\ln(\epsilon/2)|}$ where $c >0$ is a numeric constant. So the answer to your question is negative.

answered Jan 09 '19 at 23:21

Aphelli

34,439

Why is the measure $|\log r|^{-1}$? – Savannah Jan 10 '19 at 09:36
$0 <u(t) \leq r$ iff $0 < t < 1$ and $e^{-1/t} \leq r$ and you can figure out the rest. – Aphelli Jan 10 '19 at 09:43
But $e^{-1/t} \leq r$ gives $ t \ge -1/\log r$ if $r >1$ and $t< -1/ \log r$ if $r<1$. – Savannah Jan 10 '19 at 09:55
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But note that $r < 1$ always because $0 \leq u \leq e^{-1}$. – Aphelli Jan 10 '19 at 10:02
And why does the last inequality contradicts? – Savannah Jan 10 '19 at 10:17
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When $r>0$ goes to $0$, then for any $a<0$, $|\ln(r)| << r^a$, so this integral is $>>\epsilon^{-a}$ for any $a <0$. – Aphelli Jan 10 '19 at 10:24

Convergence in L^2

1 Answers1