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Tom Boardman
Tom Boardman
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If it's in $H^1$ it's a.e. differentiable, with weak differential a.e. equal its differential, which for an indicator function is a.e. zero. So if you integrate any candidate for your weak derivative multiplied by a compactly supported test function you should get zero. Now if you use the right test function and the definition of a weak derivative you ought to be able to contrive a contadiction of the form "$0=1$".

If it's in $H^1$ it's a.e. differentiable, with weak differential a.e. equal its differential which for an indicator function is a.e. zero. So if you integrate any candidate for your weak derivative multiplied by a compactly supported test function you should get zero. Now if you use the right test function and the definition of a weak derivative you ought to be able to contrive a contadiction of the form "$0=1$".

If it's in $H^1$ it's a.e. differentiable, with weak differential a.e. equal its differential, which for an indicator function is a.e. zero. So if you integrate any candidate for your weak derivative multiplied by a compactly supported test function you should get zero. Now if you use the right test function and the definition of a weak derivative you ought to be able to contrive a contadiction of the form "$0=1$".

Source Link
Tom Boardman
Tom Boardman
  • 3.2k
  • 5
  • 27
  • 23

If it's in $H^1$ it's a.e. differentiable, with weak differential a.e. equal its differential which for an indicator function is a.e. zero. So if you integrate any candidate for your weak derivative multiplied by a compactly supported test function you should get zero. Now if you use the right test function and the definition of a weak derivative you ought to be able to contrive a contadiction of the form "$0=1$".