# [SOLVED]Norm of a Bounded Linear Functional

#### Sudharaka

##### Well-known member
MHB Math Helper
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 $$\|f\|$$.

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

##### Well-known member
MHB Math Scholar
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

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Scholar
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

##### MHB Oldtimer
Staff member
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 $$\displaystyle \frac1p + \frac1q = 1.$$ So my guess is that $$\displaystyle \|f\| = \Bigl(\sum_{i=1}^n|f(x_i)|^q\Bigr)^{\!1/q}.$$

#### Sudharaka

##### Well-known member
MHB Math Helper
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 $$\displaystyle \frac1p + \frac1q = 1.$$ So my guess is that $$\displaystyle \|f\| = \Bigl(\sum_{i=1}^n|f(x_i)|^q\Bigr)^{\!1/q}.$$
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

##### MHB Oldtimer
Staff member
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 $$x$$...".

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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

##### Well-known member
MHB Math Helper
Yes, that looks good – except that you have swept some of the messy detail under the carpet by saying "for a proper choice of $$x$$...".
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.