Find an example of a linear functional with some properties

[docnet]

Homework Statement

Let $X$ be a Banach space and $X^\ast$ its dual space. Find an $f\in X^\ast$ such that $$\sup_{||x||\leq 1}||f||=1$$ and $f(x)\neq 1$ whenever $||x||=1$.

Relevant Equations

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I considered $X=\mathbb{R}^n$ and quickly realized any linear functional like $f=a_1x_1+\cdots a_nx_n$ would attain a maximum on the boundary. I regret to say that my knowledge of topology is still very limited, and did a lot of experimenting with a pen and paper without fruitful results. And I did some researching to find a possible example.

Would $X=l^1$ and $f=(0.9, 0.99, 0.999, \cdots ) \in l^\infty$ work? It says that no element in $X$ would give $f(x) = \sum_i f_i x_i = 1$, and $||f||=1$ in the sup norm. My understanding is still hazy and isn't quite clear to me what about makes this all work, and obviously I have a lot of new learning to do.

Thank you (excited to learn another challenging math topic).

 

[nuuskur]

This is a corollary of Hahn-Banach (cf. Prop 6.6, for example). We can fix a proper closed subspace $X_0\subseteq X$, then there exists $x^*$ with norm $1$ such that it vanishes on $X_0$. So if $S_X\subseteq X_0$, the requirement of $x^*(x)\neq 1$ is met for all $x\in S_X$.

Your condition should read
<br /> \sup _{\|x\|\leqslant 1} \|f(x)\| = 1. \quad (\text{or } |f(x)|)<br />
The LHS expression is precisely the norm of $f$ and it suffices to consider the supremum over the unit sphere.

 

[docnet]

This is a corollary of Hahn-Banach (cf. Prop 6.6, for example). We can fix a proper closed subspace $X_0\subseteq X$, then there exists $x^*$ with norm $1$ such that it vanishes on $X_0$. So if $S_X\subseteq X_0$, the requirement of $x^*(x)\neq 1$ is met for all $x\in S_X$.

This might be an obvious question, but if $X_0$ corresponds to the closed unit ball, and $x^*$ vanishes on $X_0$, i.e., $x^*(x)=0$ for all $x\in X_0$, isn't it a contradiction to $\sup_{||x||\leq 1}x*(x)=1$?

 

[WWGD]

Do you distinguish here between the Continuous dual and the Algebraic dual?

 

[docnet]

Do you distinguish here between the Continuous dual and the Algebraic dual?

I don't know much about the distinction between the two types of duals, but I assume it's the case where $f$ is a linear functional in the continuous dual. If $f$ is allowed to be discontinuous, would it be more trivial to find a discontinuous linear functional to satisfy $||f||=1$ and $f(x)\neq 1$ with $||x||=1$?

 

[nuuskur]

Might be mechanical thinking on my part - whenever I see Banach spaces and duals mentioned, I switch my brain to "Hahn-Banach mode" and assume we are talking about topological dual (continuous linear functionals).

If it is discontinuous, then you find your subspace where something behaves the way you want and then extend its basis to a basis on the entire space.

This might be an obvious question, but if $X_0$ corresponds to the closed unit ball, and $x^*$ vanishes on $X_0$, i.e., $x^*(x)=0$ for all $x\in X_0$, isn't it a contradiction to $\sup_{||x||\leq 1}x*(x)=1$?

Careful, $X_0$ is a closed vector subspace. In the one dimensional case, there is no proper nonzero example. As soon as $0\neq a\in X_0$ we have $X_0=\mathbb R$. In multidimensional case, the proper subspaces could correspond to lines and (hyper)planes, for instance.

edit: Whenever a subspace contains the unit ball, it is the entire space $X$, simply because $X$ is a countable union of scaled unit balls.

I'm not 100% sure right now what happens when $X_0$ only contains the unit sphere. I think it's still the entire space because we can express any $x\in X$ as $\lambda z\in \lambda S_X$ for some $\lambda$.

What a clever question!

 

[nuuskur]

Here's another idea. A linear map $f:X\to Y$ between normed spaces is continuous, surjective and has unit norm if $f$ maps the unit sphere to the unit sphere. So, can we have a functional that maps $S_X$ to $-1$ or $\{ \exp (it)\}\setminus \{1\}$, (depending on whether it's over $\mathbb R$ or $\mathbb C$)?

 

[WWGD]

Yes, I guess you're right. It's most likely the continuous dual.

 

[nuuskur]

We'd be hard pressed to find such an example from #7 over $\mathbb R$ due to linearity: if $f(x)=-1$, then $f(-x)=-f(x)=1$ for any $x\in S_X$.