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How to prove that $x_1=signt \notin W{}_2^1[-1;1] $ ?

I know that signt=$\frac{d}{dt}\mid t\mid$ and $ \mid t\mid '=\frac{t}{\mid t\mid}$, derivative of module is not defined in t=0. Am I thinking right?

Thanks for any help

T. Elmira
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  • By $W_2^1$ do you mean $W^{1,2}$? –  Dec 24 '18 at 22:04
  • @ZacharySelk yes – T. Elmira Dec 24 '18 at 22:09
  • What do you mean by $x_1$? Is that the name of your function depending on $t$? However, you need to prove that either your function does not live in $L^2[-1,1]$ or its weak derivative does not. Hint: Its the second case. – YoungMath Dec 24 '18 at 22:28
  • @YoungMath $x_1$ is the name of my function depending on t. Can you please explain how to prove that signt does not live in $L^2[-1;1]? – T. Elmira Dec 24 '18 at 22:33
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    Well, $\text{sign}$ is indeed in $L^2[-1,1]$. But its weak derivative is not. Hint two: The weak derivative of $\text{sign}$ only exists in a distributional manner. It is (twice) the famous Dirac-distribution. – YoungMath Dec 24 '18 at 22:40

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Problem. A weak derivative $v(t)$ of $sign(t)$ will satisfy, for all smooth functions $\phi$ such that $\phi(-1) = \phi(1) = 0$, $$\int_{-1}^1sign(t)\phi'(t)\,dt = -\int_{-1}^1v(t)\phi(t)\,dt$$ so we want to see that it can't exist.

Reason. You can evaluate $$\int_{-1}^1 sign(t)\phi'(t)dt = \int_0^1\phi'(t)dt - \int_{-1}^0\phi'(t)dt = -2\phi(0)$$ As mentioned in the comments above this is a $\delta$ distribution, $v(t) = 2\delta_0$ which is not a function.

Finishing the proof. If you haven't done so already, the most straightforward way to prove that $\delta$ distributions are not representable is by applying the dominated convergence theorem to a sequence $\varphi_n$ of bump functions $0 \leq \varphi_n(t) \leq 1$ whose support shrinks to a point. Someone wrote this down over here: Why Dirac's Delta is not an ordinary function?

Ben
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  • That is not entirely true. As far as I know, $\phi$ has to be a bump function, i.e. $\phi$ has compact support in the interior of $[-1,1]$. Thus, $\phi(-1)=\phi(1)=0$. – YoungMath Dec 25 '18 at 11:25
  • @youngmath Sorry, I don’t understand your comment. – Ben Dec 25 '18 at 12:21
  • Long story short: $\phi$ has to vanish on the boundary. See for instance aforementioned link to Wikipedia about weak derivatives. – YoungMath Dec 25 '18 at 12:32
  • @YoungMath Oh I see what you are saying - thanks! I'll fix that. – Ben Dec 25 '18 at 12:35
  • @YoungMath, if you shift the test function $\varphi$ by a constant to $\varphi + C$ it doesn't affect the left hand side but the right hand side changes by $C \cdot \int v,dt$, so yes what I wrote the first time was wrong. – Ben Dec 25 '18 at 12:39