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Sep
24 |
awarded | Autobiographer |
Apr
15 |
awarded | Scholar |
Apr
15 |
accepted | Nonintegrable inverse powers as distributions |
Apr
15 |
comment |
Nonintegrable inverse powers as distributions
Oh, gotcha. So Using Shanlin's answer from below, every such $T$ is just a particular extension plus a linear combination of $\delta$ and its derivatives, since the support of the difference is just the origin. But any derivative of the delta other than zeroth order clearly doesn't work, so we end up with just a multiple of $\delta$ in the linear combination. Thanks! |
Apr
15 |
awarded | Student |
Apr
15 |
comment |
Nonintegrable inverse powers as distributions
They don't say anything that wasn't mentioned above, save for a hint about using the following theorem: If $T$ is zero on the nullspaces of the distributions $S_1 \dots S_m$ then $T$ is a linear combination of the $S_i$. I am familiar with the characterization of point supported distributions as linear combinations of the delta's derivatives, but those that agree with $T_f$ are not point supported, i can't see how to apply that. |
Apr
15 |
asked | Nonintegrable inverse powers as distributions |