Show that one of these functionals is unbounded

[Fredrik]

Suppose that $\mathcal H$ is a Hilbert space, and that $A:\mathcal H\rightarrow\mathcal H$ is linear and unbounded. Is it safe to conclude that $y\mapsto\langle x,Ay\rangle$ is unbounded for at least one $x\in\mathcal H$? How do you prove this?
(My inner product is linear in the second variable).For each $x\in\mathcal H$, let $\phi_x$ be the linear functional $y\mapsto\langle x,Ay\rangle$. Suppose that $\phi_x$ is bounded for all $x\in\mathcal H$. (This is what I'd like to disprove, so I'm hoping to obtain a contradiction). Then for each $x\in\mathcal H$, there exists a unique $x'\in\mathcal H$ such that $\phi_x=\langle x',\cdot\rangle$. This means that for all $x\in\mathcal H$, we have $\langle x,Ay\rangle=\phi_x(y)=\langle x',y\rangle$. Note that x' depends on x. We also have

$$|\langle x,Ay\rangle|=|\langle x',y\rangle|\leq \|x'\|\,\|y\|=\|\phi_x\|\,\|y\|$$

for all $x,y\in\mathcal H$. This is where I'm stuck. Can you really get a contradiction from this?

 

[micromass]

Is this an exercise of Conway? If so, which one?

Maybe you could show that $$\{\phi_x~\vert~\|x\|=1\}$$ violates the uniform boundedness principle...

 

[Fredrik]

No, it's not an exercise. It's just something I started thinking about while reading about the definition of the adjoint, and yes, mainly in Conway. If $\phi_x$ is bounded, then there's a unique x' such that $\phi_x=\langle x',\cdot\rangle$, and we can define the adjoint of A as the map $x\mapsto x'$. The domain of A* is the set of all x such that $\phi_x$ is invertible. So what I'm wondering is if there exists an unbounded linear operator defined on the entire Hilbert space with an adjoint that's defined on the entire Hilbert space too. I expect the answer to be no, but I'm not sure.

The uniform boundedness principle you say...that sounds interesting, mainly because it's a theorem I just a had a quick look at, and then moved on to study other things. I'll have a closer look at it tomorrow. Now I need to go to bed. Thanks for the tip.

 

[micromass]

No, it's not an exercise. It's just something I started thinking about while reading about the definition of the adjoint, and yes, mainly in Conway. If $\phi_x$ is bounded, then there's a unique x' such that $\phi_x=\langle x',\cdot\rangle$, and we can define the adjoint of A as the map $x\mapsto x'$. The domain of A* is the set of all x such that $\phi_x$ is invertible. So what I'm wondering is if there exists an unbounded linear operator defined on the entire Hilbert space with an adjoint that's defined on the entire Hilbert space too. I expect the answer to be no, but I'm not sure.

The uniform boundedness principle you say...that sounds interesting, mainly because it's a theorem I just a had a quick look at, and then moved on to study other things. I'll have a closer look at it tomorrow. Now I need to go to bed. Thanks for the tip.

Well, I seem to remember vaguely that if an operator has an adjoint which is defined on the entire Hilbert space, then the operator must be bounded. But don't take my word for it...

 

[Fredrik]

I looked up the uniform boundedness principle before I went to bed. If I understand it correctly, what it says about my $\phi_x$ is that $\{\|\phi_x\|\,|x\in\mathcal H\}$ is bounded from above if and only if for every $y\in\mathcal H$, $\{|\phi_x(y)|\,|x\in\mathcal H\}$ is bounded from above. Alternatively, let S be the unit sphere in H, and try to use this instead: $\{\|\phi_x\|\,|x\in S\}$ is bounded from above if and only if for every $y\in\mathcal H$, $\{|\phi_x(y)|\,|x\in S\}$ is bounded from above.

I had actually shown that the statements "on the left" in these two statements are false, before you suggested I should look at the uniform boundedness principle, and I had obtained results that are at least as strong as the negations of the statements "on the right", without using the theorem, but I didn't see how to use them. Of course, after typing most of the next paragraph of this post, to explain why those results are useless, I did find a way to use one of them.


Suppose that I show that $\{\|\phi_x\|\,|x\in S\}$ isn't bounded from above, and conclude that it's not true that for every $y\in\mathcal H$, $\{|\phi_x(y)|\,|x\in S\}$ is bounded from above. Then there's a $y_0\in\mathcal H$ such that for each $K>0$, there's an $x_K\in S$ such that $|\langle x_K,Ay_0\rangle|=|\phi_{x_K}(y_0)|>K$. [strike]This doesn't appear to tell us anything useful.[/strike] If we choose $K=\|Ay_0\|$, we get $|\langle x_K,Ay_0\rangle|>\|Ay_0\|$, which contradicts the CBS inequality.


Just to get my thoughts in order, here's the first part of the proof. Suppose that $\{\|\phi_x\|\,|x\in S\}$ is bounded from above. Then there's a real number M such that for all $x\in S$, $\|\phi_x\|\leq M$. The inequality implies that for all $y\in S$, $|\phi_x(y)|\leq\|\phi_x\|\,\|y\|\leq M$. So for all $x,y\in S$, $|\langle x,Ay\rangle|\leq M$. Choose $x=Ay/\|Ay\|$. Then for all $y\in S$, $\|Ay\|\leq M$, contradicting the assumption that $A$ is unbounded.

So now we know that $\{\|\phi_x\|\,|x\in S\}$ isn't bounded from above. This means that for each K>0, there's an $x_K\in S$ such that $\|\phi_{x_K}\|>K$. But $\|\phi_{x_K}\|$ is the least upper bound for $\{|\phi_{x_K}(y)|\,|y\in S\}$, so there's a $y_K\in S$ such that $|\langle x_K,Ay_K\rangle|=|\phi_{x_K}(y_K)|>K$. Hm, on second thought, this doesn't appear to be as strong as what I get from the uniform boundedness principle, since $x_K$ is determined by K. So maybe I do need the uniform boundedness principle.

 

[micromass]

So, you've shown that $$\{\|\phi_x\|~\vert~x\in S\}$$ isn't bounded. So, it suffices to show that for every y, $$\{|\phi_x(y)~\vert~x\in S\}$$ is bounded. But this follows from CBS by $$|\phi_x(y)|\leq |<x,Ay>|\leq \|x\|\|A_y\|=\|Ay\|$$ which is an upper bound for the family...

 

[Fredrik]

Yes, that simplifies what I said in the third paragraph of #5 a bit. Thanks again for the help. This solves the problem I posted in #1. I will continue to think about whether it's possible to do it without the uniform boundedness principle.

 

[micromass]

Let me know if you find anything, since I find the problem quite intriguing. And you're correct, using the UFP is quite unsatisfying...

 

[Fredrik]

The more I think about it, the more I think it's impossible. I was hoping that maybe some property of the inner product or the specific family of functionals we're dealing with would simplify the proof of the implication

For all $y\in\mathcal H$ $\{|\phi_x(y)|\,|\,x\in S\}$ is bounded from above. $\Rightarrow$ $\{\|\phi_x\|\,|\,x\in S\}$ is bounded from above.​

(the non-trivial part of the uniformed boundedness theorem, for this specific family of functionals), so that we don't have to use any fancy theorems about Banach spaces. But the properties of the inner product and the $\phi_x$ functionals are precisely what I used to prove that the statement on the right is false, so I don't think that approach can work.

 

[Fredrik]

I cleaned up the proof for my notes, so I might as well copy-and-paste it into this thread.

Theorem: If $y\mapsto\langle x,Ay\rangle$ is bounded for all $x\in\mathcal H$, then A is bounded.

Proof:
Let S be the unit sphere in $\mathcal H$. For all $y\in\mathcal H$ and all $x\in S$,

$$|\phi_x(y)|=|\langle x,Ay\rangle|\leq\|x\|\,\|Ay\|=\|Ay\|.$$

So for each $y\in H$, $\|Ay\|$ is an upper bound for the set $\{|\phi_x(y)|\,|\,x\in S\}$. By the principle of uniform boundedness, this implies that $\{\|\phi_x\|\,|\,x\in S\}$ is bounded from above. So there exists an $M\in\mathbb R\,$ such that for all $x\in S$, $\|\phi_x\|\leq M$. This implies that for all $y\in\mathcal H$ and all $x\in S$,

$$|\langle x,Ay\rangle|=|\phi_x(y)|\leq\|\phi_x\|\,\|y\|\leq M\|y\|.$$

This implies that for all $y\in\mathcal H$,

$$\|Ay\|=\Big\langle\frac{Ay}{\|Ay\|},Ay\Big\rangle\leq M\|y\|,$$

and this means that A is bounded.