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Sep 11, 2018 at 2:09 comment added Ning Bao Is there perhaps some interesting subclasses of small detailed cases where this is true? Again, what (nontrivial) restrictions on $K$ could one put on equations of this form to get a unique solution?
Sep 11, 2018 at 0:56 comment added Christian Remling If you write $Tf=\int Kf\, dx$, then you want $1+T$ to have trivial kernel, so $-1$ must not be an eigenvalue of $T$, and whether or not this holds will depend on small details, so there's not a whole lot you can say in this generality.
Sep 10, 2018 at 22:58 history edited Konstantinos Kanakoglou
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Sep 10, 2018 at 22:49 history edited Ning Bao CC BY-SA 4.0
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Sep 10, 2018 at 22:46 history edited LSpice CC BY-SA 4.0
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Sep 10, 2018 at 22:35 history asked Ning Bao CC BY-SA 4.0