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Let $w \in C_0^\infty(\mathbb{R},\mathbb{R}^+)$ be a function with $\int_\mathbb{R} w(x) = 1$.

What can we say about the first derivative, or what can we say about $\int_\mathbb{R} |\partial_xw|$ ?

I am especially interested in the case where $w$ can be written as $w(x) = \frac{1}{\epsilon^2} v(\frac{x}{\epsilon})$ with $v \in C_0^\infty(\mathbb{R},\mathbb{R}^+), v(x)=v(-x)$ and $\int_\mathbb{R} v(x) = 1$. Is there an estimation like $|w|_{1,1}\leq c\epsilon$ ?

Thanks!

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Are you sure about the $\epsilon^2$ ? I doubt you get a normalized $w$. –  vanna Sep 20 '12 at 16:27
    
You're right, it is just $\epsilon$. –  AlexisZorbas Sep 21 '12 at 9:22

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It looks doubtful that the estimate you suggest will hold.

I'm assuming that as @vanna suggests, the $\epsilon^2$ is not valid: it looks like it should be $w(x)=\frac{1}{\epsilon}v(\frac{x}{\epsilon})$. In this case (let $x=\epsilon y$ in the $v$ integral and then change the name of $y$ to $x$), $$\int_\mathbb{R}w(x)dx=\int_\mathbb{R}v(x)dx=1.$$ With $w(x)=\frac{1}{\epsilon^2}v(\frac{x}{\epsilon})$, we'd have $$\int_\mathbb{R}w(x)dx=\frac{1}{\epsilon}\int_\mathbb{R}v(x)dx.$$

Writing $v'=r$, we then have $w'(x)=\frac{1}{\epsilon^2}r(\frac{x}{\epsilon})$, leading to $$\int_\mathbb{R}|w'(x)|dx=\frac{1}{\epsilon}\int_\mathbb{R}|r(x)|dx.$$ Since this $L^1$ norm of $r$ is independent of $\epsilon$, $|w|_{1,1}$ looks like it can be arbitrarily large.

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So the estimation is simply $|w|_{1,1} \leq \frac{c}{\epsilon}$ ? –  AlexisZorbas Sep 21 '12 at 9:27
    
Yes - it might be more accurate to say that $|w|_{1,1}\leq\frac{c}{\epsilon}$ is a valid estimate for $w\in C^\infty_0$ that can be written as $w=\frac{1}{\epsilon}v(\frac{x}{\epsilon})$ for $v\in C^\infty_0$. –  user12477 Sep 21 '12 at 13:08

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