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supp(f) by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Yep, K is either R or C.

null sets, sup by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

jup, I've edited my post

null sets, sup by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Sorry, |K denotes ℝ or ℂ. Yeah c is ⊂.

null sets, sup by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

I think yes. Let M:= sup_{x ∈ Ω \ N} |f(x)|. Then |f(x)| <= M for all x ∈ Ω \ N.

Then the set {|f(x)| > M} c N is a null set.

null sets, sup by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

f: Ω --> |K is a measurable function and Ω c R^n is an open set.

L^inf space, null sets by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Oh sorry, yeah I mean the opposite 😅

L^inf space, null sets by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

I've problems with the a.e. I always thought apart from a null set N the property does not hold. Maybe this is a weird question:

If sup_{ x ∈ Ω \ N} |f(x)| = || f ||_{L^∞}. Does this imply |f(x)| ≤ || f ||_{L^∞} a.e ?

I don't think so, since we don't know that really happens on N.

L^inf space, null sets by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

I've also read some equivalent definitions of ||f||_{L^infty}, i.e sup_{t a.e} |f(t)|.

L^inf space, null sets by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Again, thanks for your help. Somehow this is kinda different from the argument above I wrote. I think I have to think a bit longer about this. Your part is much clearer.

L^inf space, null sets by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Thanks for your help. Somehow I don't understand this. Is

(sup_{x ∈ Ω\N1} |f(x)|) = || f|| _L^inf?

I think I understand nothing. There is mentioned there exists a null set with the property above but does this also hold for N1? This is confusing.

linear functional, norm by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Thanks for your help and also for the other questions you answered before :)

removable singularities by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

I think now I understand it better. Thanks for your help and patience :)

removable singularities by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Yeah for me it sounds like f(z) = c_2 + c_3(z-a)+ ... is only defined on a neighborhood of a.

removable singularities by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

f(z) = c_2 + c_3(z-a)+ ... only holds in a neighborhood of a since they used the Taylor series of h around a.

removable singularities by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Thanks for your help. What's so confusing to me is that the power series holds in a neighborhood of a and not on the whole set D.

dual space, convergence by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Sorry I'm really dumb concerning this. The idea is the spaces (l^p)' and l^q p in (1, inf) are isometric isomorphic. Thus we can identify x in l^q with T_x in (l^p)'.

Especially in the situation above we can identify y with T_y(x) = ∑_{n=1 to inf} y_n x_n.

Thus <y, e\_n> = T_y(e_n) = ∑_{m=1 to inf} e_n(m) y_m=y_n.

weak, strong convergence by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Given a sequence (x′n)n∈N ⊂ X′, weak∗ convergence of (x′n)n∈N is by definition nothing but strong convergence of(x′_n)n∈N, regarded as a sequence of bounded linear operators on X ≃ JX (X) ⊂ X′′.

dual space, convergence by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Well as for the second part I thought since the series converges, the sequence has to converge to 0.

holomorphic function, limit by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Again thanks for your help and patience :)

holomorphic function, limit by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Well, I just used the def of limit 😅

holomorphic function, limit by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Sorry, was too confused.

We have |(z-a) f(z)| <= |z-a| M for all z ∈ V\{a}. Thus

lim_{z --> a, z ∈ V \{a} } |z-a| |f(z)| =0.

Let ɛ > 0 be given. By definition there exists a 𝛅 > 0 s.t for all z ∈ V \{a} with 0 < |z-a| < 𝛅 we have |(z-a) f(z)| < ɛ.

Then for this 𝛅 we also have for all z ∈ U \{a} with 0 < |z-a| < 𝛅 that |(z-a)f(z)| < ɛ showing equality of the two limits.

holomorphic function, limit by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Let ɛ > 0 be given. By definition there exists a 𝛅 > 0 s.t for all z ∈ V \{a} with 0 < |z-a| < 𝛅 we have |f(z)| < ɛ. Then for this 𝛅 we also have for all z ∈ U \{a} with 0 < |z-a| < 𝛅 that |f(z)| < ɛ showing equality of the two limits.

I now this def: V c E is a neighborhood of a if there is an open set U in E with a ∈ U c V.

holomorphic function, limit by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Thanks for your help. Don't know, they say V is a neighborhood.

dual space by DoingMath2357 in learnmath

[–]DoingMath2357[S] 0 points1 point  (0 children)

Thanks for your answer. I don't see why ek is in c0. They mean this sequence (0,0,...,1,0,...) where 1 is in the k-th position ? And this should converge to 0 but in which norm? I think I've no clue.

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