MHB Norm of a Bounded Linear Functional

[Sudharaka]

Hi everyone, :)

Here's a question with my answer, but I just want to confirm whether this is correct. The answer seems so obvious that I just thought that maybe this is not what the question asks for. Anyway, hope you can give some ideas on this one.

Problem:

Let \(X\) be a finite dimensional linear space. Let \(x_1,\,\cdots,\,x_n\) be a basis of \(X\). Define the norm,

\[\|x\|_p=\left\|\sum_{i=1}^{n}a_i x_i\right\|_p=\|(a_1,\,\cdots,\,a_n)\|_p\]

If \(f\) is a bounded linear functional on \(X\), find the norm \(\|f\|\).

My Answer:

We have been given the following theorem;

Let \((X,\,\|\cdot\|_X)\) and \((Y,\,\|\cdot\|_{Y})\) be two normed linear spaces over \(F\) and \(B(X,\,Y)\) denote the set of all bounded linear functions from \(X\) to \(Y\). Then the function \(\|\cdot\|:\,B(X,Y)\rightarrow\mathbb{R}\) defined by,

\[\|T\|=\sup_{x\in X,\, \|x\|_X\neq 0}\frac{\|T(x)\|_Y}{\|x\|_X}\]

for \(T\in B(X,\,Y)\) is a norm on \(B(X,\,Y)\).

From the above theorem we know that the set of all bounded linear functionals, \(B(X,\,\mathbb{R})\) has the norm,

\[\|f\|=\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|f(x)|}{\|x\|_p}=\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|f(x)|}{\left(\sum_{i=1}^{n}|a_i|^p\right)^{1/p}}\]

 

[Evgeny.Makarov]

From the above theorem we know that the set of all bounded linear functionals, \(B(X,\,\mathbb{R})\) has the norm,

\[\|f\|=\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|f(x)|}{\|x\|_p}=\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|f(x)|}{\left(\sum_{i=1}^{n}|a_i|^p\right)^{1/p}}\]

That's the definition of $\|f\|$. You probably have to express it in terms of $f$ and $x_1,\dots,x_n$,

 

[Sudharaka]

That's the definition of $\|f\|$. You probably have to express it in terms of $f$ and $x_1,\dots,x_n$,

Thanks very much for the reply. Well I can substitute \(x=a_1 x_1+\cdots+a_n x_n\) and get,

\[\|f\|=\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|f(x)|}{\left(\sum_{i=1}^{n}|a_i|^p\right) ^{1/p}}=\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|a_1 f(x_1)+\cdots+a_n f(x_n)|}{\left(\sum_{i=1}^{n}|a_i|^p\right) ^{1/p}}\]

But can I simplify any more? I don't think so. Correct? :)

 

[Evgeny.Makarov]

You may because you need to find not what this fraction equals, but what its supremum is. Could it be that $\|f\|=\max(\|f(x_1)\|,\dots,\|f(x_n)\|)$? I just don't remember this stuff very well.

 

[Opalg]

At a first glance, without thinking about it carefully, I would assume that the dual of a $p$-norm ought to be a $q$-norm, where $$\frac1p + \frac1q = 1.$$ So my guess is that $$\|f\| = \Bigl(\sum_{i=1}^n|f(x_i)|^q\Bigr)^{\!1/q}.$$

 

[Sudharaka]

At a first glance, without thinking about it carefully, I would assume that the dual of a $p$-norm ought to be a $q$-norm, where $$\frac1p + \frac1q = 1.$$ So my guess is that $$\|f\| = \Bigl(\sum_{i=1}^n|f(x_i)|^q\Bigr)^{\!1/q}.$$

I don't know if there's a false in this argument and if there is please let me know,

\begin{eqnarray}

\|f\|&=&\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|f(x)|}{\|x\|_p}\\

&=&\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|f(x)|}{\left(\sum_{i=1}^{n}|a_i|^p\right) ^{1/p}}\\

&=&\sup_{x\in X,\, \|x\|_p\neq 0}\frac{\left|\sum_{i=1}^{n}a_i f(x_i)\right|}{\left(\sum_{i=1}^{n}|a_i|^p\right) ^{1/p}}

\end{eqnarray}

By the Holder's inequality we get,

\[\left|\sum_{i=1}^{n}a_i f(x_i)\right|\leq\sum_{i=1}^{n}\left|a_i f(x_i)\right|\leq \left(\sum_{i=1}^{n}\left|a_i\right|^p\right)^{1/p}\left(\sum_{i=1}^{n}\left|f(x_i)\right|^q\right)^{1/q}\]

So we see that for a proper choice of \(x\) we can make,

\[\left|\sum_{i=1}^{n}a_i f(x_i)\right|=\left(\sum_{i=1}^{n}\left|a_i\right|^p\right)^{1/p}\left(\sum_{i=1}^{n}\left|f(x_i)\right|^q\right)^{1/q}\]

Therefore,

\[\|f\|=\left(\sum_{i=1}^{n}\left|f(x_i)\right|^q \right)^{1/q}\]

 

[Opalg]

I don't know if there's a false in this argument and if there is please let me know,

\begin{eqnarray}

\|f\|&=&\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|f(x)|}{\|x\|_p}\\

&=&\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|f(x)|}{\left(\sum_{i=1}^{n}|a_i|^p\right) ^{1/p}}\\

&=&\sup_{x\in X,\, \|x\|_p\neq 0}\frac{\left|\sum_{i=1}^{n}a_i f(x_i)\right|}{\left(\sum_{i=1}^{n}|a_i|^p\right) ^{1/p}}

\end{eqnarray}

By the Holder's inequality we get,

\[\left|\sum_{i=1}^{n}a_i f(x_i)\right|\leq\sum_{i=1}^{n}\left|a_i f(x_i)\right|\leq \left(\sum_{i=1}^{n}\left|a_i\right|^p\right)^{1/p}\left(\sum_{i=1}^{n}\left|f(x_i)\right|^q\right)^{1/q}\]

So we see that for a proper choice of \(x\) we can make,

\[\left|\sum_{i=1}^{n}a_i f(x_i)\right|=\left(\sum_{i=1}^{n}\left|a_i\right|^p\right)^{1/p}\left(\sum_{i=1}^{n}\left|f(x_i)\right|^q\right)^{1/q}\]

Therefore,

\[\|f\|=\left(\sum_{i=1}^{n}\left|f(x_i)\right|^q \right)^{1/q}\]

Yes, that looks good – except that you have swept some of the messy detail under the carpet by saying "for a proper choice of \(x\)...". (Wink)

 

[I like Serena]

As an observation, instead of writing $||x||_p \ne 0$, you can also write $x \ne 0$.
It is an axiom of a norm that these two are identical.

 

[Sudharaka]

Yes, that looks good – except that you have swept some of the messy detail under the carpet by saying "for a proper choice of \(x\)...". (Wink)

:p Yeah, need to find the method to solve the problem quickly without going into too much detail. Grad studies is a race against time; I always find myself trying hard to do all the problems they give us, all throughout the week, but until the last moment I cannot complete them. Thank you very much for all your help. I really appreciate it. :)

As an observation, instead of writing $||x||_p \ne 0$, you can also write $x \ne 0$.
It is an axiom of a norm that these two are identical.

Thanks very much, I missed this little point. :)