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 Yearling
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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
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