Miguel
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 Jan 2 comment What is a distribution in $H^{-1}(\Omega)$? Ok, I just saw your other question, where you explicitly state that your definition of $H^{-1}$ is not as the dual of $H^1_0$. But if $H^{-1}(\Omega)$ is defined as the dual of $H^1_0(\Omega)$, then the statement "$L$ is in $H^{-1}(\Omega)$" should simply mean that $L$ is not only continuous on $C^{\infty}_0$ but on all of $H^1_0$. But I guess this was obvious, sorry Jan 2 answered difference between the dual space of $H^1(\Omega)$ and the dual of $H^1_0(\Omega)$ Dec 25 comment Solving multiple matrix equations What's the rank of your matrices? It could be that your system is overdetermined / incompatible. Dec 24 comment Fractal signal analysis You are welcome :) Fractal geometry and analysis on fractals are whole fields of mathematics! Don't take me wrong, but you cannot expect to receive any useful answer to such a broad question (or I can't give it, especially since I'm no expert). I suggest you have a look at those books and then you post a more precise question. Dec 23 revised Hausdorff Measure under linear maps Grammar and spelling Dec 23 comment Find on $C[0,1]$ closed and bounded set $A$ that there are no such $f,g \in A$ that imply $\operatorname{diam}(A)=d(f,g)$ Indeed. I wasn't precise enough, sorry. Looks like I can't edit the comment any longer, though Dec 23 suggested approved edit on Hausdorff Measure under linear maps Dec 23 answered Fractal signal analysis Dec 23 comment Show the following is bounded in $n$ (and actually converges to $0$) I'd try to look at the behaviour of numerator and denominator to see if L'Hôpital's theorem applies (looks like it). Dec 23 revised Is $f^p$ Lipschitz for every $p \geq 1$ whenever $f$ is? Formatting of title Dec 23 suggested approved edit on Is $f^p$ Lipschitz for every $p \geq 1$ whenever $f$ is? Dec 23 suggested rejected edit on Find on $C[0,1]$ closed and bounded set $A$ that there are no such $f,g \in A$ that imply $\operatorname{diam}(A)=d(f,g)$ Dec 23 comment Find on $C[0,1]$ closed and bounded set $A$ that there are no such $f,g \in A$ that imply $\operatorname{diam}(A)=d(f,g)$ Just for the record, one could also use $f_n(x)=max(1-n x,0)$, again bounded and closed, with $diam=1$ but no two elements realising this distance. Dec 23 comment Find on $C[0,1]$ closed and bounded set $A$ that there are no such $f,g \in A$ that imply $\operatorname{diam}(A)=d(f,g)$ Faster by one minute! ;) One remark: I'd change the conclusion to say something along the lines of: The set $A$ is bounded by 1 and it is closed since any (not eventually constant) subsequence in it has as only possible limit the function which is constantly one (the pointwise limit of $(f_n)$). However since $f_n(0)=0$ always, there is no convergence in the $sup$ metric to $f=1$. Dec 10 comment Trying to prove $f_{yx}(0,0)=f_{xy}(0,0)$ Hint/spoiler: look up Schwarz' theorem for the interchangeability of partial derivatives. Dec 8 answered Polar radius of a general ellipsoid Dec 4 comment If all derivative are uniformly bounded, there is a subsequence that converges uniformly to an infinitely differentiable function You need to somehow use the fact that your derivatives are bounded. You can try with the mean value theorem... Nov 23 revised existence of compact set in Lusin theorem spelling, grammar, latex Nov 23 suggested approved edit on existence of compact set in Lusin theorem Nov 22 awarded Yearling