Mathematics

12:00 AM

i noticed that invariance of domain required that the map be a homeomorphism too. i was a little surprised that there wasn't already a google-able question like this that i could find on stackexchange

masfenix

Hi guys looking to get some sort of hint here, practicing for my FA comp. Let $H$ be a Hilbert space and let $M$ be a closed linear subspace of $H$. Let $f \in M^*$ be a linear functional defined. $f : M \rightarrow \mathbb{C}$. Show that there exists $g \in H^*$ such that $g(x) = f(x)$ for all $x \in M$ and $||f|| = ||g||$.

I am pretty sure the answer here is a one liner: "Apply the Hanh-Banach theorem, and we are done". But is there a way to do this without using Hanh Banach?

evinda

@PedroTamaroff Do you maybe have an idea what property should be satisfied so that a point $\in \mathbb{P}^2(\mathbb{C})$ ?

Daniel Fischer

@masfenix Sure, we don't need Hahn-Banach if $H$ is a Hilbert space. Orthogonal projection.

Or Riesz representation theorem, since we're looking at maps to $\mathbb{C}$.

Mary Star

@DanielFischer How can we conclude that $$\mathbb{F}_{p^n}\cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{\gcd (m,n)}}$$ ??

masfenix

12:16 AM

@DanielFischer well, so if its as easy as I think it is. You can define a $g: H \rightarrow \mathbb{C}$ by $g(x) = <x, y>$ for all $x \in H$. This gives us $||g|| = ||y||$

Daniel Fischer

@masfenix And what is $y$ there?

masfenix

$y$ is given to us. Let $f \in M^*$. This implies, that for all $x \in M$ there exists a unique $y \in M$ such that $f_y(x) = <x, y>$

Daniel Fischer

@MaryStar Have you understood why we have $$\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} \subset \mathbb{F}_{p^{n-m}}$$ if $n > m$?

masfenix

but my concern now is that, we've only $y$ that belongs to $M$.

Daniel Fischer

@masfenix Well, $M\subset H$.

masfenix

12:19 AM

Yes, but $H*$ is isomorphic to $H$. I mean i dont know how to voice my concerns, but if we only have $y \in M$, dosnt that break the isomorphism?

cxseven

@TedShifrin the "bounded distance away" is what I was looking for

masfenix

@DanielFischer

@DanielFischer infact even the proof of Reisz Representation theorem only finds such a $y$ in a closed subspace. We know $T: H \rightarrow H^*$ is a isomorphism. But ... ahh im getting confused here.

I dont know how to voice my concern.

Daniel Fischer

@masfenix I don't understand. By Riesz, since $M$ itself is a Hilbert space, we know that there is a unique $y\in M$ with $f(x) = \langle x, y\rangle$ for all $x\in M$. So define $g(x) = \langle x, y\rangle$ for all $x\in H$.

masfenix

@DanielFischer but what if there arnt enough "$y \in M$" to construct such a $g$.

Daniel Fischer

@masfenix You only need one.

masfenix

12:27 AM

@DanielFischer OH I see!. So basically (not related to his particular question), we have a space of linear functions $H^*$. Each linear functional is associated with a particular unique $y \in H$. But the keypoint is all the functionals could be associated with the same $y$.. you only need one. Its just that, if you pick a particular $f \in H^*$ then this $f$ is only associated with ONE $y \in H$.

I hope I make sense.

This clears up a LOT of confusion :)

Daniel Fischer

@masfenix No, each $f\in M^\ast$ is associated to a different $y\in M$. The Riesz representation theorem gives you an antilinear bijection between $M$ and $M^\ast$.

But you have only one $f\in M^\ast$ to start with. That gives you one $y\in M$ with $f = \langle\,\cdot\,, y\rangle_M$.

And now, since $M\subset H$, you simply define $g = \langle\,\cdot\,, y\rangle_H$.

"Its just that, if you pick a particular $f\in H^\ast$ then this $f$ is only associated with ONE $y\in H$." <- That part is right.

masfenix

I see. so this may be a stupid question, but does $M$ have an infinite number of elemnets?

Transcript for

$Mathematics$ Mathematics