Weak derivative of $u(x)=|x|$ not belong in $W^{1,p}(-1,1)$
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Consider the functión $u in W^{1,p}(-1,1) $, defined by $u(x)=|x|$, we know its weak derivative is
$$g(x)=left { begin{matrix} 1 & text{if }xin(0,1) \ -1 & text{if } x in (-1,0) end{matrix} right..$$
By intregation by parts is straightforward verify this. But I want to prove that $g notin W^{1,p}(-1,1)$. If we suposse that $g$ has a weak derivative then exist a $h in L^{p}(-1,1)$, which satisfies
$$varphi(1)-varphi(0)+varphi(-1)-varphi(0)=int_{-1}^{1} h(t)varphi(t)dt$$for any $varphi in C_{c}^{1}(-1,1)$. I tried to get a contradiction evaluating by concrete test functions like $varphi(t)=t$ or $varphi(t)=1$ but I don't get any interesting. Which kind functions would help me? or there is another aproach?
sobolev-spaces weak-derivatives variational-analysis
asked Jan 10 at 3:16
Pablo HerreraPablo Herrera
394113
$endgroup$
add a comment |
$begingroup$
Consider the functión $u in W^{1,p}(-1,1) $, defined by $u(x)=|x|$, we know its weak derivative is
$$g(x)=left { begin{matrix} 1 & text{if }xin(0,1) \ -1 & text{if } x in (-1,0) end{matrix} right..$$
By intregation by parts is straightforward verify this. But I want to prove that $g notin W^{1,p}(-1,1)$. If we suposse that $g$ has a weak derivative then exist a $h in L^{p}(-1,1)$, which satisfies
$$varphi(1)-varphi(0)+varphi(-1)-varphi(0)=int_{-1}^{1} h(t)varphi(t)dt$$for any $varphi in C_{c}^{1}(-1,1)$. I tried to get a contradiction evaluating by concrete test functions like $varphi(t)=t$ or $varphi(t)=1$ but I don't get any interesting. Which kind functions would help me? or there is another aproach?
sobolev-spaces weak-derivatives variational-analysis
asked Jan 10 at 3:16
Pablo HerreraPablo Herrera
394113
$endgroup$
add a comment |
$begingroup$
Consider the functión $u in W^{1,p}(-1,1) $, defined by $u(x)=|x|$, we know its weak derivative is
$$g(x)=left { begin{matrix} 1 & text{if }xin(0,1) \ -1 & text{if } x in (-1,0) end{matrix} right..$$
By intregation by parts is straightforward verify this. But I want to prove that $g notin W^{1,p}(-1,1)$. If we suposse that $g$ has a weak derivative then exist a $h in L^{p}(-1,1)$, which satisfies
$$varphi(1)-varphi(0)+varphi(-1)-varphi(0)=int_{-1}^{1} h(t)varphi(t)dt$$for any $varphi in C_{c}^{1}(-1,1)$. I tried to get a contradiction evaluating by concrete test functions like $varphi(t)=t$ or $varphi(t)=1$ but I don't get any interesting. Which kind functions would help me? or there is another aproach?
sobolev-spaces weak-derivatives variational-analysis
asked Jan 10 at 3:16
Pablo HerreraPablo Herrera
394113
$endgroup$
Consider the functión $u in W^{1,p}(-1,1) $, defined by $u(x)=|x|$, we know its weak derivative is
$$g(x)=left { begin{matrix} 1 & text{if }xin(0,1) \ -1 & text{if } x in (-1,0) end{matrix} right..$$
By intregation by parts is straightforward verify this. But I want to prove that $g notin W^{1,p}(-1,1)$. If we suposse that $g$ has a weak derivative then exist a $h in L^{p}(-1,1)$, which satisfies
$$varphi(1)-varphi(0)+varphi(-1)-varphi(0)=int_{-1}^{1} h(t)varphi(t)dt$$for any $varphi in C_{c}^{1}(-1,1)$. I tried to get a contradiction evaluating by concrete test functions like $varphi(t)=t$ or $varphi(t)=1$ but I don't get any interesting. Which kind functions would help me? or there is another aproach?
sobolev-spaces weak-derivatives variational-analysis
sobolev-spaces weak-derivatives variational-analysis
asked Jan 10 at 3:16
Pablo HerreraPablo Herrera
394113
asked Jan 10 at 3:16
Pablo HerreraPablo Herrera
394113
asked Jan 10 at 3:16
Pablo HerreraPablo Herrera
394113
asked Jan 10 at 3:16
Pablo HerreraPablo Herrera
394113
asked Jan 10 at 3:16
Pablo HerreraPablo Herrera
394113
394113
add a comment |
add a comment |
2 Answers
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For any $varphi in C_c^1((-1,1))$ satisfying $varphi(0) = 0$, we have $int_{-1}^1 h(t) varphi(t),dt = 0$. However, the set of all such $varphi$ is dense in $L^q((-1,1))$. Hence we must have $h=0$ which is absurd.
(Or, find a sequence of such $varphi_n$ which converges a.e. and boundedly to $operatorname{sgn} h$, and use dominated convergence to conclude $int_{-1}^1 |h(t)|,dt = 0$.)
answered Jan 10 at 3:26
Nate EldredgeNate Eldredge
64.5k682174
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Any function in $W^{1,p}(-1,1)$ is absolutely continuous (to be precise, any $fin W^{1,p}(-1,1)$ has an absolutely continuous representative).
Your function $g$ is clearly not even continuous, hence it cannot belong to $W^{1,p}(-1,1)$.
answered Jan 10 at 5:34
BigbearZzzBigbearZzz
9,01021652
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
For any $varphi in C_c^1((-1,1))$ satisfying $varphi(0) = 0$, we have $int_{-1}^1 h(t) varphi(t),dt = 0$. However, the set of all such $varphi$ is dense in $L^q((-1,1))$. Hence we must have $h=0$ which is absurd.
(Or, find a sequence of such $varphi_n$ which converges a.e. and boundedly to $operatorname{sgn} h$, and use dominated convergence to conclude $int_{-1}^1 |h(t)|,dt = 0$.)
answered Jan 10 at 3:26
Nate EldredgeNate Eldredge
64.5k682174
$endgroup$
add a comment |
$begingroup$
For any $varphi in C_c^1((-1,1))$ satisfying $varphi(0) = 0$, we have $int_{-1}^1 h(t) varphi(t),dt = 0$. However, the set of all such $varphi$ is dense in $L^q((-1,1))$. Hence we must have $h=0$ which is absurd.
(Or, find a sequence of such $varphi_n$ which converges a.e. and boundedly to $operatorname{sgn} h$, and use dominated convergence to conclude $int_{-1}^1 |h(t)|,dt = 0$.)
answered Jan 10 at 3:26
Nate EldredgeNate Eldredge
64.5k682174
$endgroup$
add a comment |
$begingroup$
For any $varphi in C_c^1((-1,1))$ satisfying $varphi(0) = 0$, we have $int_{-1}^1 h(t) varphi(t),dt = 0$. However, the set of all such $varphi$ is dense in $L^q((-1,1))$. Hence we must have $h=0$ which is absurd.
(Or, find a sequence of such $varphi_n$ which converges a.e. and boundedly to $operatorname{sgn} h$, and use dominated convergence to conclude $int_{-1}^1 |h(t)|,dt = 0$.)
answered Jan 10 at 3:26
Nate EldredgeNate Eldredge
64.5k682174
$endgroup$
For any $varphi in C_c^1((-1,1))$ satisfying $varphi(0) = 0$, we have $int_{-1}^1 h(t) varphi(t),dt = 0$. However, the set of all such $varphi$ is dense in $L^q((-1,1))$. Hence we must have $h=0$ which is absurd.
(Or, find a sequence of such $varphi_n$ which converges a.e. and boundedly to $operatorname{sgn} h$, and use dominated convergence to conclude $int_{-1}^1 |h(t)|,dt = 0$.)
answered Jan 10 at 3:26
Nate EldredgeNate Eldredge
64.5k682174
answered Jan 10 at 3:26
Nate EldredgeNate Eldredge
64.5k682174
answered Jan 10 at 3:26
Nate EldredgeNate Eldredge
64.5k682174
answered Jan 10 at 3:26
Nate EldredgeNate Eldredge
64.5k682174
64.5k682174
add a comment |
add a comment |
$begingroup$
Any function in $W^{1,p}(-1,1)$ is absolutely continuous (to be precise, any $fin W^{1,p}(-1,1)$ has an absolutely continuous representative).
Your function $g$ is clearly not even continuous, hence it cannot belong to $W^{1,p}(-1,1)$.
answered Jan 10 at 5:34
BigbearZzzBigbearZzz
9,01021652
$endgroup$
add a comment |
$begingroup$
Any function in $W^{1,p}(-1,1)$ is absolutely continuous (to be precise, any $fin W^{1,p}(-1,1)$ has an absolutely continuous representative).
Your function $g$ is clearly not even continuous, hence it cannot belong to $W^{1,p}(-1,1)$.
answered Jan 10 at 5:34
BigbearZzzBigbearZzz
9,01021652
$endgroup$
add a comment |
$begingroup$
Any function in $W^{1,p}(-1,1)$ is absolutely continuous (to be precise, any $fin W^{1,p}(-1,1)$ has an absolutely continuous representative).
Your function $g$ is clearly not even continuous, hence it cannot belong to $W^{1,p}(-1,1)$.
answered Jan 10 at 5:34
BigbearZzzBigbearZzz
9,01021652
$endgroup$
Any function in $W^{1,p}(-1,1)$ is absolutely continuous (to be precise, any $fin W^{1,p}(-1,1)$ has an absolutely continuous representative).
Your function $g$ is clearly not even continuous, hence it cannot belong to $W^{1,p}(-1,1)$.
answered Jan 10 at 5:34
BigbearZzzBigbearZzz
9,01021652
answered Jan 10 at 5:34
BigbearZzzBigbearZzz
9,01021652
answered Jan 10 at 5:34
BigbearZzzBigbearZzz
9,01021652
answered Jan 10 at 5:34
BigbearZzzBigbearZzz
9,01021652
9,01021652
add a comment |
add a comment |
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