Functional Analysis - Operator Norm

I need to show that the operator norm of

\ell_{\varphi}:C^0[a,b]\rightarrow\mathbb{R}

,

\displaystyle \ell_{\varphi}(f) = \int_a^b \varphi(t)f(t)\;dt

is

\displaystyle \|\ell_{\varphi}\|_{\text{op}} = \int_a^b |\varphi(t)|\;dt

where

\varphi \in C^0[a,b]

.

It's easy enough to see that the norm's bounded above by this, but I need to find a function

f

such that this norm is attained. The obvious choice is

f(t) = \text{sgn}(\varphi(t))

, but this isn't continuous. If

\varphi

has a finite number of zeros then it can just be approximated by joining the discontinuities in some neighbourhood

(-\varepsilon, \varepsilon)

of each zero. However, if

\varphi

is nowhere constant and its zeros are, for example, the Cantor set which is nowhere dense and contains no isolated points, I'm not sure how to construct an approximation.

Any ideas?

Reply 1

mathprof

The proof involves a trick:

Unparseable latex formula:

[br]\begin{array}{ccl}[br]\int_a^b|\phi(t)|dt &=&\displaystyle{ \int_a^b\frac{|\phi(t)|+n|\phi(t)|^2}{1+n|\phi(t)|}}dt }\\[br]&=&\displaystyle{ \int_a^b\frac{|\phi(t)|}{1+n| \phi (t)|}dt}+\displaystyle{\ell_\phi \Big(\frac{n|\phi(t)|}{1+n|\phi(t)|}\Big) \\[br]&\le&\displaystyle{\frac{1}{n}}+\|\ell_\phi\|.}[br]\end{array}[br]

Reply 2

IrrationalNumber

12

Original post by mathprof

The proof involves a trick:

Unparseable latex formula:

[br]\begin{array}{ccl}[br]\int_a^b|\phi(t)|dt &=&\displaystyle{ \int_a^b\frac{|\phi(t)|+n|\phi(t)|^2}{1+n|\phi(t)|}}dt }\\[br]&=&\displaystyle{ \int_a^b\frac{|\phi(t)|}{1+n| \phi (t)|}dt}+\displaystyle{\ell_\phi \Big(\frac{n|\phi(t)|}{1+n|\phi(t)|}\Big) \\[br]&\le&\displaystyle{\frac{1}{n}}+\|\ell_\phi\|.}[br]\end{array}[br]

Presumably you mean

\displaystyle{\ell_\phi \Big(\frac{n\phi(t)}{1+n|\phi(t)|}\Big)}

?

That's a really nice method! Is this alright as well?

Spoiler

(edited 13 years ago)

Reply 3

mathprof

Yes, I meant it without the absoute value. I don't see a difference however since both are continuous and have the same value.

By alright do you mean is it a valid proof? Can you see something wrong with it?

(edited 13 years ago)

Reply 4

IrrationalNumber

12

Original post by mathprof

Yes, I meant it without the absoute value. I don't see a difference however since both are continuous and have the same value.

Well if you remove the absolute value they don't have the same value,

Unparseable latex formula:

l_\phi(|\phi|) \neq \l_\phi(\phi)

.

By alright do you mean is it a valid proof? Can you see something wrong with it?

Erm, I don't know? I am worried about assuming that the dense subsets I'm using can be bounded by the functions they are converging to but other than that I don't see anything wrong... I figured I'd ask? I mean obviously your proof is far simpler but I wanted to check I understood properly...

Reply 5

.matt

OP

1

Original post by mathprof

The proof involves a trick:

Unparseable latex formula:

[br]\begin{array}{ccl}[br]\int_a^b|\phi(t)|dt &=&\displaystyle{ \int_a^b\frac{|\phi(t)|+n|\phi(t)|^2}{1+n|\phi(t)|}}dt }\\[br]&=&\displaystyle{ \int_a^b\frac{|\phi(t)|}{1+n| \phi (t)|}dt}+\displaystyle{\ell_\phi \Big(\frac{n|\phi(t)|}{1+n|\phi(t)|}\Big) \\[br]&\le&\displaystyle{\frac{1}{n}}+\|\ell_\phi\|.}[br]\end{array}[br]

Wow, I really like this proof, thank you! :biggrin:

I think IrrationalNumber's right about the thing in the bracket though, as

\phi(t)|\phi(t)| \neq \phi(t)^2

in general. I'm guessing the first term on the last line is meant to be (b-a)/n as well? (not that it affects the result in any way)

Original post by IrrationalNumber

Presumably you mean

\displaystyle{\ell_\phi \Big(\frac{n\phi(t)}{1+n|\phi(t)|}\Big)}

?

That's a really nice method! Is this alright as well?

Spoiler

That seems to work too, and I'm guessing more along the lines of what they'd expect us to do in an exam. I don't see any problems with boundedness given that

\phi

is bounded? (continuous and defined on a closed interval)

Reply 6

mathprof

Original post by IrrationalNumber

Well if you remove the absolute value they don't have the same value,

Unparseable latex formula:

l_\phi(|\phi|) \neq \l_\phi(\phi)

.

I beg your pardon, I confused

\phi |\phi|

with

\phi^2

, as has already been pointed out.

Erm, I don't know? I am worried about assuming that the dense subsets I'm using can be bounded by the functions they are converging to but other than that I don't see anything wrong... I figured I'd ask? I mean obviously your proof is far simpler but I wanted to check I understood properly...