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Is there a $X$ norm space such that we can find a linear continuous functional $f$ that holds the following properties:

$\|f\|=1$ but there is no $x \in X$ such that $\|x\|=1$ and $f(x)=1$

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Let $X=\ell^1$. Let $a_n=1-2^{-n}$ for $n\in\Bbb N$, and let $a=\langle a_n:n\in\Bbb N\rangle\in\ell^\infty$. For $x=\langle x_n:n\in\Bbb N\rangle\in\ell^1$ let $f(x)=\sum_{n\in\Bbb N}a_nx_n$. Then $\|f\|=1$, but if $0\ne x\in\ell^1$, then

$$|f(x)|=\left|\sum_{n\in\Bbb N}a_nx_n\right|\le\sum_{n\in\Bbb N}a_n|x_n|<\|x\|\;.$$

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I think is easier to prove continuity if you take $a_n = 2^{-n}$ the rest is the same thanks alot –  Maria Sep 20 '12 at 7:36
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Another classic example is: take $X = C([0,2])$, and consider the linear functional $f(x) = \frac{1}{2}\int_0^1 x(t)\,dt - \frac{1}{2}\int_1^2 x(t)\,dt$. Then $\| f\| = 1$ but the norm is not achieved in the unit ball. Intuitively, the $x$ which achieved the norm would have to be $x = 1_{[0,1]} - 1_{[1,2]}$ but this is not a continuous function.

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Yes. You want a normed linear space $X$ in which the closed unit ball is not weakly compact. It so happens that a normed linear space has weakly compact closed unit ball iff it is reflexive, so any nonreflexive space will give you a counterexample.

One example is $\ell_\infty$, the set of bounded sequences with norm $\|(x_n)\|_\infty=\sup_n |x_n|$. This has a basis $\{e_n\}$ (indexing from $1$) where $e_n$ is the sequence with the $n^{th}$ term equal to $1$ and all other terms $0$. We can define a functional $f$ on $\ell_\infty$ by $f(e_n)=2^{-n}$. This has norm $\|f\|=1$ but the norm is not achieved on the closed unit ball.

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I don't think your $f$ is well defined on $\ell_\infty$. What is $f$ of the sequence which is identically 1? –  Nate Eldredge Sep 20 '12 at 3:49
    
@NateEldredge Thanks, I was thinking $\ell_1$ when defining $f$. –  Alex Becker Sep 20 '12 at 3:52
    
Also your first paragraph is not true. For instance, if $X$ is an infinite-dimensional Hilbert space, the closed unit ball is not compact, but every continuous linear functional achieves its norm on the unit ball (by the Riesz representation theorem). –  Nate Eldredge Sep 20 '12 at 3:58
    
But it's easy to fix the first paragraph: "You want a normed linear space X in which the closed unit ball is not weakly compact. It so happens that a normed linear space has weakly compact closed unit ball iff it is reflexive, so any non-reflexive space will give you a counterexample." –  user31373 Sep 20 '12 at 18:42
    
@LVK Very good point, I forgot that I was using weak compactness rather than compactness. –  Alex Becker Sep 20 '12 at 18:44
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Let $X=\ell^1$. Let $a_n=2^{−n}$ for $n∈N$. For $x=⟨x_n:n \in N⟩∈ℓ1$.

Let $f(x)=\sum_{n\in\Bbb N}a_nx_n$.

$f$ is clearly lineal, thus continuous because is obviously continuous on $0$.

Then $\|f\|=1$, if $0≠x \in \ell^1$ we have:

$|f(x)|=\left|\sum_{n\in\Bbb N}a_nx_n\right|\le\sum_{n\in\Bbb N}a_n|x_n|<\|x\|\;.$

thanks a lot for your support:D

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The norm of your $f$ is not $1$, it's $1/2$. –  Martin Argerami Sep 21 '12 at 4:05
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