Restriction of bounded linear functionals attaining their norm Let $X$ be a (real) normed space and let $Y$ be a closed subspace of $X$.
Suppose that the bounded linear functional $f\in X^\ast$ attains its norm on the closed unit ball of $X$. Must $f\vert_Y \in Y^\ast$ attain its norm on the closed unit ball of $Y$?
Addendum. Suppose that every bounded linear functional $f\in X^\ast$ attains its norm on the closed unit ball of $X$. Must $g \in Y^\ast$ attain its norm on the closed unit ball of $Y$?
 A: For the original question, we can construct easy counterexamples by considering a product, $X = Y \times \mathbb{R}$, endow it with the sum norm of its two factors, and take a linear functional $\lambda$ on $Y$ that doesn't attain its norm (if that is possible, of course), and extend it by $\Lambda((x,t)) = \lambda(x) + C\cdot t$ for some $C > \lVert\lambda\rVert$. Then $\lVert\Lambda\rVert = C$, and $\Lambda$ attains its norm in $(0,1)$.
For the addendum, note that by James' theorem, if $X$ is a Banach space so that every continuous linear functional on $X$ attains its norm [on the closed unit ball], then $X$ is reflexive, and thus $Y$ as a closed subspace of a reflexive space is also reflexive, hence every continuous linear functional on $Y$ also attains its norm [on the closed unit ball].
If $X$ is an incomplete normed space such that every continuous linear functional on $X$ attains its norm, then the completion of $X$ is reflexive, hence the restriction to the closed subspace $Y \subset X$ attains its norm certainly if $Y$ is a complete subspace (because then $Y$ is reflexive). If $Y$ is not complete, I expect that a continuous linear functional on $Y$ need not attain its norm, but I can't offer an example.
A: James theorem gives the complete solution to the question. The reflexive case is as Daniel Fischer explained. If $Z$ is not reflexive, then there exists $f\in B_{Z^*}$ not attaining its norm at $B_{Z}$. But $f$ attains it norm at $B_{Z^{**}}$. Take $X=Z^{**}$, $Y=Z$ and $f$ the functional. 
For a concrete example in this situation, think at $c_0$ as subspace of $l_\infty$. Take $f=(f_n)_n\in B_{l_1}$ with $f_n>0$ for all $n\in\mathbb{N}$. Then, $f$ attains its norm (which is the same looked as a vector in $c_0^*=l_1$ or in $l_\infty^*$) in a unique point in $l_\infty$, mamely $(1,1,\ldots,1)$. 
Edited: I read properly David Fincher answer and it also contains the complete solution simply "adding to Y one more dimension", roughly speaking. Anyway I believe the example with $c_0$ is quite natural and illustrative. This argument also gives an example for "David Finscher situation". Take $X=Y=\varphi\subset l_2$ ($\varphi$ is the space formed by the sequences which coordinates are zero but finitely many). $\overline{X}^{\|\cdot\|}=l_2$,  but a functional $f=(f_n)_n\in B_{l_2}$, $f_n>0$ for all $n$, only attains its maximum at $(f_n)_n$.
