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Is this proposition correct? Will you please give a contour example if it is wrong?

If $J \in L^2(\mathbb{R}) \cap C^1 (\mathbb{R})$, $f \in C^{\infty}(\mathbb{R})$, $|f'|\leq K$, where $K > 0$ is a constant, then

(1) $J' \in L^2(\mathbb{R})$;

(2) $f \cdot J' \in L^2 (\mathbb{R})$.

I was considering as follows when I asked this question:

(1) If $J \in L^2(\mathbb{R})$, then $\exists N>0$ sufficiently large and $\exists \epsilon>0$ sufficiently small, s.t. $J=o(x^{-(1/2+ϵ)})$ for $|x|> N$.

Because $J \in C^1(\mathbb{R})$, $J'$ is continuous for $|x| \leq N$.

As $J=o(x^{-(1/2+ϵ)})$, $J' = o(x^{-(3/2+ϵ)})$ for $|x| > N$. $J' \in L^2(\mathbb{R})$.

(2) Because $f \in C^{\infty}(\mathbb{R})$ and $|f'|\leq K$, $|f| \leq K|x|+1 $, for all $|x|> N$.

$J' = o(x^{-(3/2+ϵ)})$, $|f| \leq K|x|+1 $, so $f \cdot J' = o(x^{-(1/2+ϵ)})$, for $|x| > N$. Because $f \cdot J'$ is continuous for $|x| \leq N $, $f \cdot J' \in L^2(\mathbb{R})$.

I tried this example $J = \frac{\sin (x)}{ x }$, $f = x $.

But I found my thoughts were wrong which are marked in bold.

By the comment from @PhoemueX, a contour example for this proposition was constructed as $J = \frac{\sin (x) \sin (e^x) }{ x }$.

William Wu
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    What have you tried? What are your thoughts? For (a) try to build a fast decaying function with large derivative, e.g. by multiplying a "nice" function by $\sin(e^x)$. – PhoemueX Jul 15 '14 at 08:03
  • What means $fJ'$: $f\cdot J'$ or $f\circ J'$? – daw Jul 15 '14 at 08:17
  • @daw By $f J'$, I mean the multiplication of the two functions $f$ and $J'$. Thanks. – William Wu Jul 15 '14 at 08:38
  • @PhoemueX Thank you very much. My thoughts are as follows:

    $J\in L^2 \cap C^1$, then $\exists N > 0$ sufficiently large and $\exists \epsilon>0$ sufficiently small, s.t. $J = o(x^{-(1/2 + \epsilon)})$, and $J'= o(x^{-(3/2 + \epsilon)})$ for all $|x| \geq N$.

    $f \in C^{\infty}$ and $|f'| \leq K$, then $|f|≤K|x|+c$, where $c$ is a real constant, for all $|x| \geq N$.

     – William Wu Jul 15 '14 at 09:38
  • I've been thinking about this question and since I don't have a background in Sobolev spaces, I would like to know: What is the motivation/intuition behind this question ? – Srinivas K Jul 16 '14 at 07:08
  • @SrinivasK Thank you very much. I would like to consider the transport equation by Galerkin method, while the equation is $\frac{\partial}{\partial t} J + f \cdot \nabla J = 0 $ with initial condition $J(0)=J_0$. I am considering what space the solution $J$ belongs to (in order to use variation in Galerkin method). – William Wu Jul 16 '14 at 07:22
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    My idea is that if we can prove $J'$ is not bounded,then it cannot be L2.(I'm not sure if there's a counter example for this. Correct me if this is wrong.)We can take $J$ to be a bounded,L2 and C1 function with an unbounded derivative using the ideas in http://math.stackexchange.com/questions/257584/bounded-functions-have-bounded-derivatives. – Srinivas K Jul 16 '14 at 07:39
  • @SrinivasK Thank you very much. PhoemueX tells a very good method which is multiplying a $L^2$ function by $\sin(e^x)$. A contour example: $J(x) = \frac{\sin (x) sin (e^x)}{x}$. – William Wu Jul 16 '14 at 07:55
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    @pomelo: I think you're looking for a counterexample. – Eric Stucky Jul 24 '14 at 04:01

1 Answers1

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This counterexample is in terms of continuous but non-differentiable functions because it's easy to describe. At the points of non-differentiability one can smooth out the function to the degree required, e.g. by convolution with mollifiers etc.

Take $f\equiv 1$. $J$ is a function whose graph is an infinite sequence of hats (by "hat" I mean like the graph of $1-|x|$) of height 1 on the $x$-axis, consecutive hats being separated by unit length and $J(x)=0$ in between hats. The base width of the $n^{\rm th}$ hat is $1/n$. Thus $J$ is $L^2$ and $|J'(x)| = n$ a.e. on the hats (and $0$ in-between). So, $J'$ is not $L^2$.

InTransit
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