Equivalent definitions for the norm of a linear functional

[AxiomOfChoice]

Can someone please explain why the following three definitions for the norm of a bounded linear functional are equivalent?

<br /> \| f \| = \sup_{0 &lt; \|x\| &lt; 1} \frac{|f(x)|}{\| x \|},<br />

and

<br /> \| f \| = \sup_{0 &lt; \| x \| \leq 1} \frac{|f(x)|}{\| x \|},<br />

and

<br /> \| f \| = \sup_{\| x \| = 1} \frac{|f(x)|}{\| x \|} = \sup_{\| x \| = 1} |f(x)|.<br />

(Thanks to micromass for reminding me about the last equality.) Every book I have just asserts their equivalence but provides no explanation. Thanks!

 

[micromass]

It is evident that

\sup_{0&lt;\|x\|&lt;1}{\frac{|f(x)|}{\|x\|}}\leq \sup_{0&lt;\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}We will now prove:

\sup_{0&lt;\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}\leq \sup_{\|x\|=1}{|f(x)|}

Let's assume that (xn) is a sequence with 0&lt;\|x_n\|\leq 1, and such that

\frac{|f(x_n)|}{\|x_n\|}\rightarrow \sup_{0&lt;\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}

Now, we put y_n=x_n/\|x_n\|. Then \|y_n\|=1. Furthermore:

|f(y_n)|=\frac{|f(x_n)|}{\|x_n\|}\rightarrow \sup_{0&lt;\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}

This proves that the inequality must hold.

We will now prove

\sup_{\|x\|=1}{|f(x)|}\leq \sup_{0&lt;\|x\|&lt;1}{\frac{|f(x)|}{\|x\|}}

Assume that (xn) is a sequence such that \|x_n\|=1 and such that

|f(x_n)|\rightarrow \sup_{\|x\|=1}{|f(x)|}

Now, we put y_n=\frac{n-1}{n}x_n, then 0&lt;\|y_n\|&lt;1, and

\frac{|f(y_n)|}{\|y_n\|}=|f(x_n)|\rightarrow \sup_{\|x\|=1}{|f(x)|}

This proves that this inequality must also hold.

 

[spamiam]

Either your book has a different definition of the operator norm, or there are some typos in the first post. The norm I'm familiar with can be found here: http://en.wikipedia.org/wiki/Operator_norm#Equivalent_definitions.

Let's call the norms

(1) \sup\{\|f(x)\| : \|x\| \leq 1 \}

(2) \sup\{\|f(x)\| : \|x\| = 1 \}

(3) \sup \left\{\frac{\|f(x)\|}{\|x\|} : x \neq 0\right\}

(1) is equivalent to (2) because a linear functional on will attain its maximum on the boundary of the set, i.e., when \|x\|=1.

(2) is equivalent to (3) because f is linear, so cf(x) = f(cx). Then if x is nonzero, we have

<br /> \frac{\|f(x)\|}{\|x\|} = \left\|f\left(\frac{x}{\|x\|}\right)\right\|<br />

and \frac{x}{\|x\|} has norm 1.

 

[mathwonk]

these all look trivially the same just from the definition of linear.

i.e. if x is any non zero vector and c is any non zero scalar then f(cx) = cf(x),

so |f(x)|/||x|| = (c/c)(|f(x)|/||x||) = |f(cx)|/||cx||.

Thus |f(x)|/||x|| is constant on lines through the origin.

This proves immediately that all limits in all posts above are the same, since the sets of

quotients are all the same.

 

[AxiomOfChoice]

It is evident that

\sup_{0&lt;\|x\|&lt;1}{\frac{|f(x)|}{\|x\|}}\leq \sup_{0&lt;\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}


We will now prove:

\sup_{0&lt;\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}\leq \sup_{\|x\|=1}{|f(x)|}

Let's assume that (xn) is a sequence with 0&lt;\|x_n\|\leq 1, and such that

\frac{|f(x_n)|}{\|x_n\|}\rightarrow \sup_{0&lt;\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}

Now, we put y_n=x_n/\|x_n\|. Then \|y_n\|=1. Furthermore:

|f(y_n)|=\frac{|f(x_n)|}{\|x_n\|}\rightarrow \sup_{0&lt;\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}

This proves that the inequality must hold.

We will now prove

\sup_{\|x\|=1}{|f(x)|}\leq \sup_{0&lt;\|x\|&lt;1}{\frac{|f(x)|}{\|x\|}}

Assume that (xn) is a sequence such that \|x_n\|=1 and such that

|f(x_n)|\rightarrow \sup_{\|x\|=1}{|f(x)|}

Now, we put y_n=\frac{n-1}{n}x_n, then 0&lt;\|y_n\|&lt;1, and

\frac{|f(y_n)|}{\|y_n\|}=|f(x_n)|\rightarrow \sup_{\|x\|=1}{|f(x)|}

This proves that this inequality must also hold.

Thanks; I like this proof a lot. And I think I understand the argument, provided I'm correct about this detail: "If A and B are sets, and there is a sequence of points in B that converge to \sup A, then we have \sup A \leq \sup B." That's correct, right? Since \alpha = \sup B only if there is a sequence of points in B converging to \alpha.

 

[micromass]

Thanks; I like this proof a lot. And I think I understand the argument, provided I'm correct about this detail: "If A and B are sets, and there is a sequence of points in B that converge to \sup A, then we have \sup A \leq \sup B." That's correct, right? Since \alpha = \sup B only if there is a sequence of points in B converging to \alpha.

Yes, that is basically the idea of what I was trying to do

 

[mathwonk]

if two sets if numbers are the same numbers , then their sups are also the same. done.

 

[AxiomOfChoice]

Incidentally, my professor claims that if we define (Tx)(t)= tx(t) (just multiplication by the independent variable) on L^2[0,1] (square-integrable functions on [0,1]), we have \| T \| = 1. I get why we have \| T \| \leq 1, but I don't see how we can get \| T \| \geq 1. Does anyone see this?

 

[Fredrik]

The trick to prove \|T\|\leq\text{something} is usually to find an upper bound of the set for which \|T\| is the least upper bound. The trick to prove \|T\|\geq\text{something} is usually to use that \|T\| is \geq every member of that set. So you should look for a specific member of that set that solves the problem. (I haven't actually done it for this problem).

By the way, my answer to the question in #1 is that if T:X\rightarrow Y is a bounded linear operator, and we define

\begin{align*}<br /> A &amp;=\bigg\{\frac{\|Tx\|}{\|x\|}\,\bigg|\, x\in X,\ x\neq 0\bigg\}\\<br /> B &amp;=\bigg\{\frac{\|Tx\|}{\|x\|}\,\bigg|\, x\in X,\ 0&lt;\|x\|\leq 1\bigg\}\\<br /> C &amp;=\Big\{\|Tx\|\,\Big|\, x\in X, \|x\|=1\Big\},<br /> \end{align*}

then A=B=C. So don't worry about the supremums. Just show that the sets are the same.

I realize that the question had already been answered (and that my answer is identical to mathwonk's), but I had this already LaTeXed in my personal notes.

 

[AxiomOfChoice]

The trick to prove \|T\|\leq\text{something} is usually to find an upper bound of the set for which \|T\| is the least upper bound. The trick to prove \|T\|\geq\text{something} is usually to use that \|T\| is \geq every member of that set. So you should look for a specific member of that set that solves the problem. (I haven't actually done it for this problem).

Right; that's what I'm trying to do (so far, without success).

 

[AxiomOfChoice]

The trick to prove \|T\|\leq\text{something} is usually to find an upper bound of the set for which \|T\| is the least upper bound. The trick to prove \|T\|\geq\text{something} is usually to use that \|T\| is \geq every member of that set. So you should look for a specific member of that set that solves the problem. (I haven't actually done it for this problem).

By the way, my answer to the question in #1 is that if T:X\rightarrow Y is a bounded linear operator, and we define

\begin{align*}<br /> A &amp;=\bigg\{\frac{\|Tx\|}{\|x\|}\,\bigg|\, x\in X,\ x\neq 0\bigg\}\\<br /> B &amp;=\bigg\{\frac{\|Tx\|}{\|x\|}\,\bigg|\, x\in X,\ 0&lt;\|x\|\leq 1\bigg\}\\<br /> C &amp;=\Big\{\|Tx\|\,\Big|\, x\in X, \|x\|=1\Big\},<br /> \end{align*}

then A=B=C. So don't worry about the supremums. Just show that the sets are the same.

I realize that the question had already been answered, but I had this already LaTeXed in my personal notes.

Thanks

 

[AxiomOfChoice]

Ok, I'm confused. If

<br /> \| T \| = \sup_{x\neq 0} \frac{\| Tx \|}{\| x \|},<br />

then to show we have \| T \| \geq 1, we need to show that there is a nonzero x such that the expression above is equal to 1. But this means

<br /> \| Tx \| = \sqrt{\int_0^1 t^2 |x(t)|^2 dt} = \sqrt{\int_0^1 |x(t)|^2 dt} = \| x \|,<br />

which implies

<br /> \int_0^1 |x(t)|^2 (t^2 - 1)dt = 0.<br />

But this can't be! On [0,1], (t^2 - 1) is negative, but |x(t)|^2 is positive! And there's no way the integral can be zero unless the product is positive in some places and negative in others! Am I wrong here?

My professor stated \| T \| = 1 without proof, so I can't imagine it's this hard...

 

[AxiomOfChoice]

...actually, now that I think about it, I guess it would be sufficient to come up with a sequence x_n(t) \in L^2[0,1] such that

<br /> \frac{\int_0^1 t^2 |x_n(t)|^2 dt}{\int_0^1 |x_n(t)|^2 dt} \to 1.<br />

Can anyone think of such a sequence? I can't really do it...

 

[Landau]

Note that L^2[0,1] contains more than the continuous functions!

For every 0&lt;\epsilon&lt;1, let A_{\epsilon}:=\{t\in[0,1]\ |\ t\geq 1-\epsilon\}. It has finite Lebesgue measure \mu(A_\epsilon)=\epsilon. Then, for f=\chi_{A_\epsilon}\in L^2[0,1] the characteristic function of this set, we have

\|f\|^2=\int_{1-\epsilon}^1 dt=\epsilon;
\|Tf\|^2=\int_{1-\epsilon}^1 |t|^2dt\geq \epsilon (1-\epsilon)^2.

Hence

\|T\|=\sup_{g} \frac{\|Tg\|}{\|g}\geq \frac{\|Tf\|}{\|f}\geq 1-\epsilon.

But 0<epsilon<1 was arbitrary, so \|T\|=1.

This argument can be generalized to prove that the multiplication operator

T_{\phi}:L^2[0,1]\to L^2[0,1]
f\mapsto f\phi

has norm equal to the essential supremum of \phi. (Indeed, \phi(t)=t has (essential) supremum on [0,1] equal to 1.)

 

[AxiomOfChoice]

Note that L^2[0,1] contains more than the continuous functions!

For every 0&lt;\epsilon&lt;1, let A_{\epsilon}:=\{t\in[0,1]\ |\ t\geq 1-\epsilon\}. It has finite Lebesgue measure \mu(A_\epsilon)=\epsilon. Then, for f=\chi_{A_\epsilon}\in L^2[0,1] the characteristic function of this set, we have

\|f\|^2=\int_{1-\epsilon}^1 dt=\epsilon;
\|Tf\|^2=\int_{1-\epsilon}^1 |t|^2dt\geq \epsilon (1-\epsilon)^2.

Hence

\|T\|=\sup_{g} \frac{\|Tg\|}{\|g}\geq \frac{\|Tf\|}{\|f}\geq 1-\epsilon.

But 0<epsilon<1 was arbitrary, so \|T\|=1.

This argument can be generalized to prove that the multiplication operator

T_{\phi}:L^2[0,1]\to L^2[0,1]
f\mapsto f\phi

has norm equal to the essential supremum of \phi. (Indeed, \phi(t)=t has (essential) supremum on [0,1] equal to 1.)

You're right; this works. Thanks a lot Landau! I've also managed to think of the following example: If you look at x_n = \sqrt n \chi_{[1-1/n,1]}, I think this sequence has the property that \| Tx_n \| / \| x_n \| \to 1.

 

[Landau]

Well, sure. It's basically what I just wrote, with epsilon=1/n, and a factor sqrt{n} (which seems unnecessary) in front of it.