Are bounded operators bounded indepedently on the function?

[SeM]

Hi thanks to George, I found the following criteria for boundedness:

\begin{equation}
\frac{||Bf(x)||}{||f(x)||} < ||Bf(x)||
\end{equation}

If one takes f(x) = x, and consider B = (h/id/dx - g), where g is some constant, then B is bounded in the interval 0-$\pi$. However, given that I am new to operator algebra, I am not sure whether this means that B is ALWAYS bounded for ANY f(x)?

Thanks!

 

[mathman]

Your inequality doesn't make sense. For example if ||f(x)||=1/2, then the right hand side needs a coefficient of 2.

To answer the general questions, the operator is presumed to be an operator on some function space. Bounded operator means ||Bf(x)|| < C||f(x)|| where C is a constant independent of f(x) for all functions in the space.

 

[SeM]

Your inequality doesn't make sense. For example if ||f(x)||=1/2, then the right hand side needs a coefficient of 2.

To answer the general questions, the operator is presumed to be an operator on some function space. Bounded operator means ||Bf(x)|| < C||f(x)|| where C is a constant independent of f(x) for all functions in the space.

Hi Mathman, thanks for this.I made a typo there, the criterion for boundedness of an operator T should be:

\begin{equation}
\sqrt {\frac{||Tf(x)||}{||f(x)||}} < ||Tf(x)||
\end{equation}

http://www.fjordforsk.no/Unititled.jpg

 

[mathman]

Your revised definition is the same (wrong) as the original. All you did was replace the letter B by the letter T, but otherwise left the expression unchanged.

 

[jim mcnamara]

@mathman Would it help if you posted latex for what it ought to be?

 

[SeM]

Mathman, As you can see from the attached image, it should be as given with the square root. I can't really find any other form, than the given one in the book, unless Kreyszig was wrong there ( in addition to the typo on the upper limit of the integral which should be 1/n thanks to some other guys answering on another thread).

 

[mathman]

@mathman Would it help if you posted latex for what it ought to be?

My first response to the statement has it. An operator T is bounded if $$||Tf(x)||\le C||f(x)||$$ for all f(x) in the function space of interest, where C does not depend on f(x).

 

[SeM]

This inequality is also the book. So why are there two inequalities for boundedness and is there a typo in the page-scan?

 

[FactChecker]

This inequality is also the book. So why are there two inequalities for boundedness and is there a typo in the page-scan?

You don't show the scan of the book definition, but @mathman 's definition is correct. (see https://en.wikipedia.org/wiki/Bounded_operator ). The definitions that you are quoting do not make sense.

 

[SeM]

You don't show the scan of the book definition, but @mathman 's definition is correct. (see https://en.wikipedia.org/wiki/Bounded_operator ). The definitions that you are quoting do not make sense.

Is the scan non visible?

Please see again:

note a typo in the upper limit, 1 should be 1/n. except for this, the last part, the inequality seems correct.

 

[FactChecker]

Is the scan non visible?

Please see again:

View attachment 217369

note a typo in the upper limit, 1 should be 1/n. except for this, the last part, the inequality seems correct.

It's perfectly readable. No equation on that page is the same as equation (2) of post #3. Nothing on that page is the definition of a bounded operator. The last equation is related to the definition only to show that the example is not bounded.