Linear continuous function on a norm space 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 $f(x)=1$
 A: 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.
A: 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 
$$|f(x)|=\left|\sum_{n\in\Bbb N}a_nx_n\right|\le\sum_{n\in\Bbb N}a_n|x_n|<\|x\|\;.$$
A: 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.
A: 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
