index of 1st-order diff. operator, Ind(D) = dim(Ker(D)) -...

Michael_J12

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Mar 10, 2008
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Hi,

What is the index of the first order differential operator? D which linearly maps from C[sup:2c7li053]1[/sup:2c7li053]([0,1],R) to C([0,1],R). with index I refer to

Ind(D) = dim(Ker(D)) - dim(Y/Im(D))

I suppose my observations that it cant be bijective is correct? so the index cant be zero?
would appreciate if someone could help me and show how you get to the answer!

Michael J
 
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