@DavidGao$:$ This is clear to me as $\mathcal G_x$ and $\mathcal G^x$ are discrete in the subspace topology if the underlying groupoid $\mathcal G$ is étale. The idea is to consider open bisections on which the range and the source map restrict to homeomorphisms. But here we also have the supremum in front and someway or the other I need to have a uniform control over the number of terms appear over $\mathcal G^{(0)}.$

The proof that $\lambda(f)$ is bounded is already contained in Lemma 5.6.12 of Brown-Ozawa.

Sorry, I realized what I was suggesting didn’t work, mostly because the Cauchy-Schwarz estimate you used is a bit too weak. Just read the proof of Lemma 5.6.12. I would just be repeating that proof if I write an answer anyway.

@DavidGao$:$ This Lemma does not provide a proof of $\|\lambda (f)\| = \sup\limits_{x \in \mathcal G^{(0)}} \|\lambda_x (f)\|.$ Is it easy?

(Note that your desired inequality simply does not work. If $\mathcal{G}$ is an infinite discrete group and $f = 1_{\{e\}}$, then the sum you wrote is already infinite.)

@DavidGao$:$ I can see why $$\|\lambda (f)\| \leq \sup\limits_{x \in \mathcal G^{(0)}} \|\lambda_x (f)\|.$$

@DavidGao$:$ Good point.

That’s nearly just definition. The part after the supremum in your formula for $\|\lambda(f)\xi\|^2$ is just $\|\lambda_x(f)\xi_x\|^2$.

And in any case, $\leq$ is already enough to prove boundedness.

@DavidGao$:$ I think by $\xi_x$ you mean $\xi \rvert_{\mathcal G_x}.$ But for the other part we need to find an extension of $\xi \in \ell^2(\mathcal G_x)$ to a function in $C_c (\mathcal G).$

As for your question on whether one can find the exact norm $\|\lambda(f)\|$, in general, no. You can’t do this even in the case of a general discrete group.

@DavidGao$:$ Could you please help me proving the equality? I have been trying to prove it but somehow I can't.

Any real/complex-valued function defined on a closed subspace can be extended. This is standard point set topology. Though, again, to prove boundedness, $\leq$ is already enough.

en.wikipedia.org/wiki/Tietze_extension_theorem

@DavidGao$:$ Yeah I see now. But will the extension, thus obtained, be compactly supported?

Just multiply a compactly supported function to restrict the support. This also allows you to ensure $\xi_y$ has norm at most $1$ in $\ell^2(\mathcal{G}_y)$ for each $y$.

@DavidGao$:$ Sorry I don't get your point. What do you mean by multiplying a compactly supported function with the extension? That compactly supported function should be identically $1$ on $\mathcal G_x.$ Right? For otherwise the extension won't restrict to $\xi$ on $\mathcal G_x.$ Could you please elaborate it a little more? Thanks in advance. Sorry for my stupid question.

I should have been clearer about this. Since you only need a vector that approximately achieve the norm of $\lambda_x(f)$, you can assume $\xi$ has finite support from the start. Then by local compactness, you can arrange a compactly supported continuous function restricting to $1$ on the support of $\xi$ (which is finite and does not need to be the entirety of $\mathcal{G}_x$).

Because you also need to control the $\ell^2$ norm of what happens on each fiber, what should be done is something like restricting to a relatively compact neighborhood $U$ of the (finite) support of $\xi$ s.t. $U$ is a disjoint union of finitely many open sets, on each of which the source map is a homeomorphism. Then use Tietze extension to get a function restricting to $1$ on the support of $\xi$ and $0$ outside of $U$, as well as being bounded by the inverse of the $\ell^2$ norm of the previous extended function on each fiber.

@David Gao$:$ I don't get your point as to why we may assume that $\text {supp}\ (\xi)$ is finite. Given $\xi_x \in \ell^2 (\mathcal G_x)$ are you someway trying to paste them in a certain way so that the resultant function is a compactly supporeted continuous function and it restricts to $\xi_x$ on $\mathcal G_x$ for each $x \in \mathcal G^{(0)}.$ If you wish, you can add it as answer. I don't think this platform is suitable for extended discussions. Sooner or later some moderator will appear and move all the comments in chat without any intimation.

I’m not saying that happens for each $x$, I’m just saying that, given a finitely supported $\xi_x \in \ell^2(\mathcal{G}_x)$ with $2$-norm at most $1$, there is a $\xi$, compactly supported, so that its $2$-norm on each fiber is at most $1$ and it restricts to $\xi_x$ on $\mathcal{G}_x$. I don’t understand what your confusion is, this is all very standard. (This hardly justifies an answer, and it’s also quite different from your original question.)

Let me clarify what my doubt exactly is. I would like to show that $\sup\limits_{x \in \mathcal G^{(0)}} \|\lambda_x(f)\| \leq \|\lambda f\|.$ Right?

@DavidGao: Are you here?

@Anacardium That is the case, yes.

For this it is enough to show that $\|\lambda_x f\| \leq \|\lambda (f)\|$ for any fixed but arbitrary $x \in \mathcal G^{(0)}.$ This is all I am trying.

Yes? That is also what I’ve been proving.

“I’m not saying that happens for each $x$, I’m just saying that, given a finitely supported $\xi_x \in \ell^2(\mathcal{G}_x)$ with $2$-norm at most $1$, there is a $\xi$, compactly supported, so that its $2$-norm on each fiber is at most $1$ and it restricts to $\xi_x$ on $\mathcal{G}_x$.”

So my idea is to start with $\xi_x \in \ell^2 (\mathcal G_X)$ and then extend it continuously to a function on $\mathcal G$ such that the extended function is compactly supported.

I only have doubts in finite supports. If we start with an arbitrary $\xi_x \in \ell^2 (G_x)$ with 2 norm being equal to $1$ then how can I assure it is finitely supported?

Just to clarify: do you not understand why my message in quotation marks is true? Or do you not understand how the message in quotation marks implies the result?

@Anacardium You cannot. But you just need to prove ||\lambda(f)|| >= ||\lambda_x(f)||, so you might as well choose a finite supported \xi_x in l^2(G_x) s.t. ||\lambda_x(f)\xi_x|| \geq (1-\epsilon)||\lambda_x(f)|| and extend that finitely supported \xi_x

Oh I see. This is the first thing that I missed.

Easy indeed. Now let me check your argument again.

Now you are covering supp (\xi) by finitely many open bisections.

On each of the open bisection s is a homeomorphism.

Yes, that is the case

But on the union s might not be a homeomorphism. Right?

That’s correct. Unless supp(\xi_x) is just a singleton, s cannot possibly be a homeomorphism on the union

I think you can just omit that part. You can find such a continuous function by Urysohn lemma and then multiply it by the Tietze extension of \xi_x.

So the only part to invesigate is whether any finite set can be included inside a relatively compact open bisection?

Which is what you claimed.

Any single point has that property.

for obvious reasons.

More or less, yes. You need to control the l^2 norm on each fiber to ensure that you get something in C_c(G) with norm 1, but that can be easily arranged since Tietze extension allows you to control the sup norm

That's not my question. My question is how to incorporate the finite support of \xi_x inside an open bisection

Should I need open bisection at all here?

? I don’t understand how that is an issue? You know every point has a relatively compact neighborhood on which s is a homeomorphism, just take the union.

On the union s might not be a homeomorphism. You just said.

I think we might be using different terminologies, what do you mean by “bisection” here?

I don’t need s to be homeomorphism on the union? Why is that needed in the first place?

Bisection means all those subsets E of G such that r and s restrict to homeomorphism.

Then we are using the same terminology. I’m confused. Why exactly do you want a single bisection to cover the support? That obviously is not going to happen, nor is it needed.

That's my question. What do I need s to be a homeomorphism on the union?

Why?

@DavidGao Not at all. I know that can't happe.

What I said that you don't even need bisections to cover the support. What is the use of that.

I think you have a fundamental misunderstanding as to how the proof is supposed to go. Let me just run through the proof. Let {g_1, …, g_n} be the support of \xi_x. Then you can choose U_i, relatively compact open bisection containing g_i. Furthermore, by shrinking U_i if necessary, we may assume they are disjoint. Then by Tietze extension, you can choose f_i, continuous, which equals \xi_x(g_i) at g_i, is bounded by |\xi_x(g_i)|, and is 0 outside U_i. Then \xi = f_1 + … + f_n is the extension

you want. Now \xi is compactly supported. The important part of why I need each U_i to be a bisection and that they are disjoint is to ensure that \xi has norm 1.

My idea is different than that of yours.

Once you take $\xi_x$ with l^2-norm 1 on $G_x$ Tietze guarantees an extension \xi which is continuous on G and bounded by 1. Now finitely many open set covering the support of \xi_x and let U be the union of those open sets. Then by Urysohn lemma we can define a compactly supported continuous function g on G which takes the value 1 on the support of \xi_x and takes the value 0 outside U. Now I take $\xi_x g.$

”Bounded by 1” does not ensure the 2-norm on each fiber is bounded by 1. There are multiple points on each fiber

Okay. Here the norm is not the sup norm. Rather it is induced by the C_0(G^(0))-valued inner product.

Yes? But that is exactly the supremum of 2-norms on each fiber

Your process uses Titeze + Urysohn simultaneously, I guess. Tietze does not specify how the extension should look like except the norm boundedness condition.

You can take extend the function that is \xi_x(g_i) at g_i and 0 outside U_i, {g_i} union the complement of U_i is closed, so just Tietze extension works.

I mean, this is really just standard point-set topology. I’m obviously not writing down all details as to how to use Tietze extension

Right. You can think of this way as well. Of course details are required. I am not the one who will believe in everything blindly without any reason. It might be obvious to you. But not to me. That's why I am veryfying all the details. There is no harm in doing so.

You can take U_i's to be disjoint as the underlying space is regular.

But the Tietze extension needs the space to be normal.'

So does Urysohn’s lemma. The standard way around this is that compact Hausdorff spaces are normal, so you can just work in the one-point compactification.