Proving that an operator is unbounded

[SemM]

Hi, after reading on boundedness and unboundedness, I like to prove that an operator, S, is unbounded. However I am not sure this is good enough?
\begin{equation}
||S\psi|| \leqslant c ||\psi||
\end{equation}
\begin{equation}
||S\psi|| = \bigg( \int_a^b (S \psi)^2 dx \bigg)^{1/2}
\end{equation}
\begin{equation}
\bigg( \int_a^b (S \psi)^2 dx \bigg)^{1/2} \leqslant c ||\psi||
\end{equation}
\begin{equation}
\int_a^b (S \psi)^2 dx \leqslant (c ||\psi||)^{2}
\end{equation}
Now comes something which I am not sure on, I cancel the integral by derivating on both sides.
\begin{equation}
(S \psi)^2 \leqslant c^2 \frac{d}{dx} ||\psi||^2 dx
\end{equation}
\begin{equation}
||\psi||^2 = \int_a^b |\psi|^2 dx
\end{equation}
Which finally gives:
\begin{equation}
(S \psi)^2 \leqslant c^2 \frac{d}{dx} \int_a^b |\psi|^2 dx
\end{equation}
which gives:
\begin{equation}
(S \psi)^2 \leqslant c^2 |\psi|^2
\end{equation}
Something is not right here, because it seems too simple integrals vanishing and derivatives canceling intergrals, and it doesn't really say whether S is unbounded, as long as $\psi$ is kept general. I can of course test this on some function, but that would be no proof!

Is this possible to be used as a basis for proving unboundedness? If it was, the final answer does not say anything, because the operators action on the square of $\psi$ still does not guarantee that it is smaller than c.

Thanks!

 

[fresh_42]

Hi, after reading on boundedness and unboundedness, I like to prove that an operator, S, is unbounded. However I am not sure this is good enough?
\begin{align*}
||S\psi|| &\leqslant c ||\psi||\\
||S\psi|| &= \bigg( \int_a^b (S \psi)^2 dx \bigg)^{1/2}\\
\bigg( \int_a^b (S \psi)^2 dx \bigg)^{1/2} &\leqslant c ||\psi||\\
\int_a^b (S \psi)^2 dx &\leqslant (c ||\psi||)^{1/2}
\end{align*}
Now comes something which I am not sure on, I cancel the integral by derivating on both sides.
\begin{align*}
(S \psi)^2 &\leqslant c^2 \frac{d}{dx} ||\psi||^2 dx\\
||\psi||^2 &= \int_a^b |\psi|^2 dx
\end{align*}
Which finally gives:
\begin{align*}
(S \psi)^2 &\leqslant c^2 \frac{d}{dx} \int_a^b |\psi|^2 dx
\end{align*}
which gives:
\begin{align*}
(S \psi)^2 &\leqslant c^2 |\psi|^2
\end{align*}
Something is not right here, because it seems too simple integrals vanishing and derivatives canceling intergrals, and it doesn't really say whether S is unbounded, as long as $\psi$ is kept general. I can of course test this on some function, but that would be no proof!

Is this possible to be used as a basis for proving unboundedness? If it was, the final answer does not say anything, because the operators action on the square of $\psi$ still does not guarantee that it is smaller than c.

Beside the typo in the fourth line which should be $2$ instead of $1/2$, you started with the definition of a bounded operator and ended with the same inequality. Nothing in between has happened. The first line $||S\psi|| \leqslant c ||\psi||$ could be read as the assumption of the contrary of what has to be proven, and a contradiction is supposed to be deduced. But this should be mentioned and as nothing actually happens, there can't be a contradiction either. You showed: if $\mathcal{A}$ then $\mathcal{A}$.

If we assume $S$ to be bounded (unbounded), what would happen if it were unbounded (bounded), resp. where has the actual definition of the operator $S$ been used?

 

[SemM]

Beside the typo in the fourth line which should be $2$ instead of $1/2$, you started with the definition of a bounded operator and ended with the same inequality. Nothing in between has happened. The first line $||S\psi|| \leqslant c ||\psi||$ could be read as the assumption of the contrary of what has to be proven, and a contradiction is supposed to be deduced. But this should be mentioned and as nothing actually happens, there can't be a contradiction either. You showed: if $\mathcal{A}$ then $\mathcal{A}$.

If we assume $S$ to be bounded (unbounded), what would happen if it were unbounded (bounded), resp. where has the actual definition of the operator $S$ been used?

The only thing that would change would be the $\leqslant$ sign according to either of the two. But how does one prove unboundedness for a wavefunction made general? You gave me an excellent example of proving linearity by using $F(g)+F(f)=F(g+f)$ which works for virtually any operator, is there such a form one can use instead of the reasoning I gave, which as you said, proves that A = A?

Say that I insert the derivative operator for S:

\begin{equation}
(\frac{d}{dx} \psi)^2 \leqslant c^2 |\psi|^2
\end{equation}

that would show that d/Dx is probably unbounded, but that is not really proved, its only assumed

 

[fresh_42]

The only thing that would change would be the $\leqslant$ sign according to either of the two. But how does one prove unboundedness for a wavefunction made general? You gave me an excellent example of proving linearity by using $F(g)+F(f)=F(g+f)$ which works for virtually any operator, is there such a form one can use instead of the reasoning I gave, which as you said, proves that A = A?

As there are bounded operators as well as unbounded operators in this world, the definition of a specific operator to be one or the other has to appear somewhere. For linear operators $S$ the following are equivalent:

• $S$ is bounded
• $S$ is continuous
  
• $S$ is continuous at $0$

So e.g. to show that $S$ is unbounded, it is sufficient to show that for a sequence $\psi_n \to 0$ we have $S\psi_n \nrightarrow 0$ or to find a certain $\psi_0$ on which $S$ isn't bounded, i.e. for all $c>0$ we have $||S\psi_0 || > c\cdot ||\psi_0 ||$. Usually it are the bounded operators which are considered, i.e. the continuous ones.

 

[Mark44]

$\int_a^b (S \psi)^2 dx \leqslant (c ||\psi||)^{2}$

Now comes something which I am not sure on, I cancel the integral by derivating on both sides.

$(S \psi)^2 \leqslant c^2 \frac{d}{dx} ||\psi||^2 dx$

Aside from the points that @fresh_42 made, the second step isn't valid. The definite integral on the left side of the first equation is a number, so its derivative would be zero. Also, the "dx" on the right side in the second equation shouldn't be there.

As shown, you differentiated (not derivated -- no such word in English) the left side, but you found the differential on the right side. These are different operations.

One part of the Fundamental Theorem of Calculus, which you should review, says this:
$\frac d {dx} \int_a^x f(t) dt = f(x)$
It does not say that $\frac d {dx} \int_a^b f(t) dt = f(x)$, which is essentially what you did in going from the first equation to the second.

 

[SemM]

So e.g. to show that $S$ is unbounded, it is sufficient to show that for a sequence $\psi_n \to 0$ we have $S\psi_n \nrightarrow 0$

For this case,

$\psi_n \to 0$ we have $S\psi_n \nrightarrow 0$

if S = d/dx

This is a great way to show that S is unbounded in an infinite space $\mathscr{H}$ , and that the same unbounded operator in $\mathscr{H}$ can be bounded in $\mathscr{D}(S) \subset \mathscr{H}$ .

But this is not the case for T if it is (d/dx + x), because then $T\psi_n \rightarrow x\psi$.For this case:

or to find a certain $\psi_0$ on which $S$ isn't bounded, i.e. for all $c>0$ we have $||S\psi_0 || > c\cdot ||\psi_0 ||$. Usually it are the bounded operators which are considered, i.e. the continuous ones.

I am not sure how this actually SHOWS that d/dx is unbounded, although it is correct. If the operation on $\psi$ gave $\phi$ and one could see that $c*\phi$ with some given value of x was actually smaller then integral of $\phi$ ($\psi$) of the one same given value of x then it would be evident. But this proof is not as evident as the former you gave.

 

[SemM]

Aside from the points that @fresh_42 made, the second step isn't valid. The definite integral on the left side of the first equation is a number, so its derivative would be zero. Also, the "dx" on the right side in the second equation shouldn't be there.

As shown, you differentiated (not derivated -- no such word in English) the left side, but you found the differential on the right side. These are different operations.

One part of the Fundamental Theorem of Calculus, which you should review, says this:
$\frac d {dx} \int_a^x f(t) dt = f(x)$
It does not say that $\frac d {dx} \int_a^b f(t) dt = f(x)$, which is essentially what you did in going from the first equation to the second.

Yes, this confirms my thought "Something is not right here, because it seems too simple integrals vanishing and derivatives canceling intergrals,". Thanks

 

[Mark44]

Say that I insert the derivative operator for S:
$(\frac{d}{dx} \psi)^2 \leqslant c^2 |\psi|^2$

that would show that d/Dx is probably unbounded, but that is not really proved, its only assumed

With suitable assumptions on the quantities above being nonnegative, the inequality above can be rewritten as $\frac {d\psi} {dx} \le c \psi$
, a first-order differential equation that can be solved for $\psi$.

The only way that you can show that $\frac d {dx}$ is unbounded, based on your initial assumption, is to arrive at a contradiction. This is a point that @fresh_42 made earlier. Since you were apparently doing a proof by contradiction, your proof should have started like this:
Suppose that $||S\psi|| \leqslant c ||\psi||$...

 

[SemM]

With suitable assumptions on the quantities above being nonnegative, the inequality above can be rewritten as $\frac {d\psi} {dx} \le c \psi$
, a first-order differential equation that can be solved for $\psi$.

The only way that you can show that $\frac d {dx}$ is unbounded, based on your initial assumption, is to arrive at a contradiction. This is a point that @fresh_42 made earlier. Since you were apparently doing a proof by contradiction, your proof should have started like this:
Suppose that $||S\psi|| \leqslant c ||\psi||$...

Yes, that is what I was going to do but saying that the "proof is not satisfied".

But your point on setting $\frac {d\psi} {dx} \le c \psi$ was interesting and solve it as ODE. I will try that, although the operator at hand looks more like the T operator given above.
THanks

 

[Mark44]

Yes, that is what I was going to do but saying that the "proof is not satisfied".

Of course the proof was not satisfied -- you didn't end up with a contradiction, but ended with more-or-less a restatement of your initial assumption.
Do you understand the basic technique of a proof by contradiction?

But your point on setting $\frac {d\psi} {dx} \le c \psi$ was interesting and solve it as ODE. I will try that, although the operator at hand looks more like the T operator given above.

Well, if the operator is different, the proof of whether the operator is bounded or not will be different, as well.
One point that fresh_42 made was that you hadn't used the definition of the operator in your proof (post #2).

 

[Mark44]

In post #1 you said

Something is not right here, because it seems too simple integrals vanishing and derivatives canceling intergrals

Right after your first inequality you should have had this:
$\sqrt{\int_a^b (S\psi)^2 dx} \le c\sqrt{\int_a^b \psi^2 dx}$
From which you could square both sides to get this:
$\int_a^b (S\psi)^2 dx \le c^2\int_a^b \psi^2 dx$
Replacing S by d/dx yields this:
$\int_a^b (\frac {d\psi}{dx})^2 dx \le c^2\int_a^b \psi^2 dx$
Finally, using the property that if $\int_a^b f(x) dx \le \int_a^b g(x) dx$ for nonnegative functions f and g, then $f(x) \le g(x)$, we get
$(\frac {d\psi}{dx})^2 \le c^2 \psi^2$
That's where the integrals went. It had nothing to do with derivatives "cancelling" integrals.

 

[fresh_42]

I am not sure how this actually SHOWS that d/dx is unbounded, although it is correct.

Look at $S = \dfrac{d}{dx}\, : \,L_2((-2,2)) \longrightarrow L_2((-2,2))$ with
$$
\psi_n(x) =
\begin{cases}
\sqrt{\dfrac{n}{3}}(\cos (n \pi x) +1) & \text{if } x \leq \dfrac{1}{n} \\
0 & \text{elsewhere }
\end{cases}
$$
and compute $||\psi_n||_2^2$ and $S\psi_n$.

 

[SemM]

In post #1 you said

Right after your first inequality you should have had this:
$\sqrt{\int_a^b (S\psi)^2 dx} \le c\sqrt{\int_a^b \psi^2 dx}$
From which you could square both sides to get this:
$\int_a^b (S\psi)^2 dx \le c^2\int_a^b \psi^2 dx$
Replacing S by d/dx yields this:
$\int_a^b (\frac {d\psi}{dx})^2 dx \le c^2\int_a^b \psi^2 dx$
Finally, using the property that if $\int_a^b f(x) dx \le \int_a^b g(x) dx$ for nonnegative functions f and g, then $f(x) \le g(x)$, we get
$(\frac {d\psi}{dx})^2 \le c^2 \psi^2$
That's where the integrals went. It had nothing to do with derivatives "cancelling" integrals.

Thanks Mark, This was critical!

Cheers

 

[SemM]

Look at $S = \dfrac{d}{dx}\, : \,L_2((-2,2)) \longrightarrow L_2((-2,2))$ with
$$
\psi_n(x) =
\begin{cases}
\sqrt{\dfrac{n}{3}}(\cos (n \pi x) +1) & \text{if } x \leq \dfrac{1}{n} \\
0 & \text{elsewhere }
\end{cases}
$$
and compute $||\psi_n||_2^2$ and $S\psi_n$.

Thanks for this example! Will use a slight simpler form of psi because the operator is very complex. I take the function selected MUST be infinitely differentiable.

This appears very similar to this approach. however the last part (the quotient) in this attachment is new to this discussions rationale. It was posted several weeks ago by me, in December I think, when two mentors (can't recall who they were) said it was not correct.

 

[fresh_42]

I take the function selected MUST be infinitely differentiable.

No, this is not required. The functions only have to be within the domain of the differential operator, of course, and square integrable for the norm.

 

[SemM]

No, this is not required. The functions only have to be within the domain of the differential operator, of course, and square integrable for the norm.

Weird, did a full numerical calculation on the $S\psi$ and $||\psi||^2$ and it gives the result that

$S\psi$= -6.85
$||\psi||^2$=6.26

so $-6.85< 6.25c^2$ gives a complex value of c. Wasn't aware that c could be complex.

 

[fresh_42]

I have something different, but it is hard to guess what you might have done. I have $||\psi_n||_2^2=1$ and $S\psi_n$ is dependent on $n$ which doesn't vanish in the norm.

 

[SemM]

I have something different, but it is hard to guess what you might have done. I have $||\psi_n||_2^2=1$ and $S\psi_n$ is dependent on $n$ which doesn't vanish in the norm.

This thread was to derive a general idea on proving boundedness /unboundedness.

The operator that I tested is different, it's this:

$S=i\hbar\frac{d^3}{dx^3}+x^2$

It gives a complex value for c for $S: L^2[0,2\pi] \rightarrow L^2[0,2\pi]$.

where

\begin{equation}
\psi = cos(n \pi x) \ \ \ \ \ \ \ 0 \leqslant x \leqslant 2\pi/n \\
\psi = 0 \ \ \ \ elsewhere
\end{equation}

 

[Mark44]

From post #12

and compute $||\psi_n||_2^2$ and $S\psi_n$.

Let me correct the above to $||S\psi_n||_2^2$.

The operator that I tested is different, it's this: $S=i\hbar\frac{d^3}{dx^3}+x^2$

Please show us your work (not just your results) for what @fresh_42 asked for.

 

[Mark44]

This thread was to derive a general idea on proving boundedness /unboundedness.

The operator that I tested is different, it's this:
$S=i\hbar\frac{d^3}{dx^3}+x^2$

It gives a complex value for c for $S: L^2[0,2\pi] \rightarrow L^2[0,2\pi]$.

That makes no sense. You're comparing two norms: $||S\psi||$ and $||\psi||$. Each of these would be real-valued. If S is bounded, there should be a real number c for which $||S\psi|| \leqslant c ||\psi||$.

 

[SemM]

From post #12
Let me correct the above to $||S\psi_n||_2^2$.
Please show us your work (not just your results) for what @fresh_42 asked for.

Sure, here is the note. Thanks to the final clarification by Orodruin, that worked out.

 

[Mark44]

Sure, here is the note. Thanks to the final clarification by Orodruin, that worked out.

Your work has mistakes.
1. $||H_s\psi|| = \int_0^{2\pi/n}\sqrt{[H_s(\cos(n\pi x)]^2}dx$
2.$c||\psi|| = \int_0^{2\pi/n}\cos(n\pi x)dx = \sin(n\pi x)|_0^{2\pi/n}$
In 1, that's not the $L^2$ norm. The integral should be $\sqrt{\int_0^{2\pi/n}[H_s(\cos(n\pi x)]^2dx}$. Where the square root is placed makes a difference. For example $\sqrt{\int_a^b x^2 dx} \ne \int_a^b \sqrt{x^2}dx$, which can be verified by carrying out each integration.
In 2, you have the same mistake as in #1, of the misplaced square root, and the integration is incorrect. $\int \cos(n\pi x)dx \ne \sin(n\pi x) + C$. You can verify this by differentiating $\sin(n\pi x)$ and seeing that you don't get $\cos(n\pi x)$. You have also lost the c that's on the left side.

My advice is to stop what you're doing, and go back and review first-year calculus.

 

[SemM]

My advice is to stop what you're doing, and go back and review first-year calculus.

Thanks for this fantastically stimulating comment!

Cheers

 

[SemM]

Your work has mistakes.
1. $||H_s\psi|| = \int_0^{2\pi/n}\sqrt{[H_s(\cos(n\pi x)]^2}dx$
2.$c||\psi|| = \int_0^{2\pi/n}\cos(n\pi x)dx = \sin(n\pi x)|_0^{2\pi/n}$
In 1, that's not the $L^2$ norm. The integral should be $\sqrt{\int_0^{2\pi/n}[H_s(\cos(n\pi x)]^2dx}$. Where the square root is placed makes a difference. For example $\sqrt{\int_a^b x^2 dx} \ne \int_a^b \sqrt{x^2}dx$, which can be verified by carrying out each integration.
In 2, you have the same mistake as in #1, of the misplaced square root, and the integration is incorrect. $\int \cos(n\pi x)dx \ne \sin(n\pi x) + C$. You can verify this by differentiating $\sin(n\pi x)$ and seeing that you don't get $\cos(n\pi x)$. You have also lost the c that's on the left side.

Thanks, this was critical! Note that c appears at the end in that fractional expression, which should be $\sqrt{c}$.

 

[Mark44]

My advice is to stop what you're doing, and go back and review first-year calculus.

Thanks for this fantastically stimulating comment!

A more apt description would be "realistic."
How can you expect to learn Quantum Mechanics without the requisite mathematics knowledge?

 

[WWGD]

Hi, after reading on boundedness and unboundedness, I like to prove that an operator, S, is unbounded. However I am not sure this is good enough?
\begin{equation}
||S\psi|| \leqslant c ||\psi||
\end{equation}
\begin{equation}
||S\psi|| = \bigg( \int_a^b (S \psi)^2 dx \bigg)^{1/2}
\end{equation}
\begin{equation}
\bigg( \int_a^b (S \psi)^2 dx \bigg)^{1/2} \leqslant c ||\psi||
\end{equation}
\begin{equation}
\int_a^b (S \psi)^2 dx \leqslant (c ||\psi||)^{2}
\end{equation}
Now comes something which I am not sure on, I cancel the integral by derivating on both sides.
\begin{equation}
(S \psi)^2 \leqslant c^2 \frac{d}{dx} ||\psi||^2 dx
\end{equation}
\begin{equation}
||\psi||^2 = \int_a^b |\psi|^2 dx
\end{equation}
Which finally gives:
\begin{equation}
(S \psi)^2 \leqslant c^2 \frac{d}{dx} \int_a^b |\psi|^2 dx
\end{equation}
which gives:
\begin{equation}
(S \psi)^2 \leqslant c^2 |\psi|^2
\end{equation}
Something is not right here, because it seems too simple integrals vanishing and derivatives canceling intergrals, and it doesn't really say whether S is unbounded, as long as $\psi$ is kept general. I can of course test this on some function, but that would be no proof!

Is this possible to be used as a basis for proving unboundedness? If it was, the final answer does not say anything, because the operators action on the square of $\psi$ still does not guarantee that it is smaller than c.

Thanks!

You may also want to specify the spaces: domain and target , so that we know what norms to use for each.