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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 $Y$?

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  • $\begingroup$ No. Consider $X = Y \times \mathbb{R}$ to construct easy counterexamples. $\endgroup$ – Daniel Fischer Oct 25 '13 at 17:18
  • $\begingroup$ What if we assume that every bounded linear functional on $X$ attains its norm? Must a bounded linear functional $g\in Y^\ast$ attain its norm? $\endgroup$ – user103232 Oct 25 '13 at 17:24
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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.

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  • $\begingroup$ So if we were to prove a version of the Hahn-Banach theorem for extending bounded linear functionals not attaining their norm, where would the usual proof fail? $\endgroup$ – user103232 Oct 25 '13 at 19:34
  • $\begingroup$ Do you mean trying to prove that a continuous linear functional that doesn't attain its norm can be extended to a continuous linear functional with the same norm that doesn't attain its norm? Not sure where the proof would fail, but I think in the step where you show you can extend the functional from $M$ to $M \oplus \mathbb{R}\cdot x_0$, you can't show that the extension doesn't attain its norm. $\endgroup$ – Daniel Fischer Oct 25 '13 at 19:44
  • $\begingroup$ Yes, indeed. In such a proof, in order to extend a functional $f\in Y^\ast$ to a normed space $X$ (where $\dim (X/Y) = 1$), we would like to find some $c \in \mathbb{R}$ such that $-f(y) - \Vert z + y \Vert < c < -f(y') + \Vert z + y' \Vert$ for all $y,y'\in Y$ (and some fixed $z \in X\setminus Y$) knowing that $-f(y) - \Vert z + y \Vert < -f(y') + \Vert z + y' \Vert$ is true for all $y,y'\in Y$ (and this is not always possible). $\endgroup$ – user103232 Oct 25 '13 at 20:07
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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 $(f_n)_n$.

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  • $\begingroup$ Bad example my last one. I believe that with $X=\varphi\oplus span\{a\}$, perhaps it works. $\endgroup$ – Enrique Oct 26 '13 at 13:01
  • $\begingroup$ If Z is not reflexive, then there need not be a functional not attaining its norm, unless Z is Banach. $\endgroup$ – user103232 Oct 26 '13 at 18:01
  • $\begingroup$ You are right, let say, if the completion of Y is not reflexive, then always exists X containing Y as a subspace and a functional norm attaining on $B_X$ but which restriction to $Y$ is not. The $X$ above is an idea (I believe it works) of a superspace for $Y$ which a functional ($a$) norm attaining in $B_X$ but not in $B_Y$. I suppose you would like a general proof that any $Y$ whose completion is reflexive is contained in a superspace $X$ admiting a functional norm attaining at $B_X$ but not at $B_Y$. $\endgroup$ – Enrique Oct 27 '13 at 10:54

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