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?

$$\| f \| = \sup_{0 < \|x\| < 1} \frac{|f(x)|}{\| x \|},$$

and

$$\| f \| = \sup_{0 < \| x \| \leq 1} \frac{|f(x)|}{\| x \|},$$

and

$$\| f \| = \sup_{\| x \| = 1} \frac{|f(x)|}{\| x \|} = \sup_{\| x \| = 1} |f(x)|.$$

(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

$$\sup_{0<\|x\|<1}{\frac{|f(x)|}{\|x\|}}\leq \sup_{0<\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}$$We will now prove:

$$\sup_{0<\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}\leq \sup_{\|x\|=1}{|f(x)|}$$

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

$$\frac{|f(x_n)|}{\|x_n\|}\rightarrow \sup_{0<\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}$$

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

$$|f(y_n)|=\frac{|f(x_n)|}{\|x_n\|}\rightarrow \sup_{0<\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}$$

This proves that the inequality must hold.

We will now prove

$$\sup_{\|x\|=1}{|f(x)|}\leq \sup_{0<\|x\|<1}{\frac{|f(x)|}{\|x\|}}$$

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 $$0<\|y_n\|<1$$, 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

$$\frac{\|f(x)\|}{\|x\|} = \left\|f\left(\frac{x}{\|x\|}\right)\right\|$$

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

$$\sup_{0<\|x\|<1}{\frac{|f(x)|}{\|x\|}}\leq \sup_{0<\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}$$


We will now prove:

$$\sup_{0<\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}\leq \sup_{\|x\|=1}{|f(x)|}$$

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

$$\frac{|f(x_n)|}{\|x_n\|}\rightarrow \sup_{0<\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}$$

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

$$|f(y_n)|=\frac{|f(x_n)|}{\|x_n\|}\rightarrow \sup_{0<\|x\|\leq 1}{\frac{|f(x)|}{\|x\|}}$$

This proves that the inequality must hold.

We will now prove

$$\sup_{\|x\|=1}{|f(x)|}\leq \sup_{0<\|x\|<1}{\frac{|f(x)|}{\|x\|}}$$

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 $$0<\|y_n\|<1$$, 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

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

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

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

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

$$\| T \| = \sup_{x\neq 0} \frac{\| Tx \|}{\| x \|},$$

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

$$\| Tx \| = \sqrt{\int_0^1 t^2 |x(t)|^2 dt} = \sqrt{\int_0^1 |x(t)|^2 dt} = \| x \|,$$

which implies

$$\int_0^1 |x(t)|^2 (t^2 - 1)dt = 0.$$

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

$$\frac{\int_0^1 t^2 |x_n(t)|^2 dt}{\int_0^1 |x_n(t)|^2 dt} \to 1.$$

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 $0<\epsilon<1$, 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 $0<\epsilon<1$, 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.