1
$\begingroup$

Let $Y$ be a Banach space and $g\in Y''$ with $\|g\|=1$. Consider $\{\phi_n\}_{n=1}^\infty\subset Y'$ such that $\|\phi_n\|=1$ for each $n\in\mathbb{N}$. How do we prove that there exists $y_n\in Y$ such that $$\phi_i(y_n)=g(\phi_i)$$ for $i=1,2,\dots,n$, and $\|y_n\|\leq 1+1/n$ ?

My attempt: I managed to prove the case for $n=1$, but can't extend it to $n$ greater than 1.

Let $0<|g(\phi_1)|\leq\|g\|\|\phi_1\|=1$. So $\frac 12|g(\phi_1)|\leq\frac 12$.

By definition of $\|\phi_1\|=\sup_{\|y\|\leq 1}|\phi_1(y)|=1$, there exists $y$ such that $\|y\|\leq 1$ and $\frac 12|g(\phi_1)|<|\phi_1(y)|\leq 1$.

Consider $$y'=y\times\frac{1/2|g(\phi_1)|}{|\phi_1y|}$$. Then $\|y'\|\leq 1$ and $|\phi_1(y')|=\frac 12|g(\phi_1)|$.

Then $|\phi_1(2y')=|g(\phi_1)|$, so we may take $y_1=2y'$ or $y_1=-2y'$ depending on sign.

$\endgroup$
5
  • 1
    $\begingroup$ Are you sure is $\|y_n\|\le 1+\frac{1}{n}$? $\endgroup$
    – sinbadh
    Mar 16, 2016 at 9:41
  • $\begingroup$ Yes, I typed it correctly. What do you have in mind? $\endgroup$
    – yoyostein
    Mar 16, 2016 at 9:43
  • 1
    $\begingroup$ I think you can do this with the Principle of Local Reflexivity (which gives this exact result when applied to the subspace span$g$ and span$y_1,\dots,y_n$ with a certain $\epsilon$). There may also be a proof with Goldstine's theorem and weak star density, but it's not so fun to think about (I think this works out similarly if one tracks through the details of the PLR proof.) $\endgroup$ Mar 16, 2016 at 9:58
  • 1
    $\begingroup$ You can find the PLR in, eg, Albiac and Kalton Chapter 11, section 2. $\endgroup$ Mar 16, 2016 at 9:59
  • $\begingroup$ Is there an easier way to prove it than PLR? $\endgroup$
    – yoyostein
    Mar 16, 2016 at 13:26

1 Answer 1

1
$\begingroup$

This is quite a nice question and I think I've found a way of doing it without PLR (although I wouldn't stake myself on this being correct):

Without loss of generality the $\phi_n$'s are linearly independent. If they were linearly dependent, then in what follows you simply ignore the linearly dependent term, I'll show that you can take $\|y_n\| \leq 1+\eta$ for any positive $\eta$ you care to choose.

Fix some $k$. Since they are linearly independent, for $j=1,\dots,k$ we can pick some $x_i$ of norm 1 such that $\phi_i(x_i) > 0$ but $\phi_j(x_i) = 0$ for each $j \neq i$ (this follows from the classic lemma that $\cap \ker f_i \supset \ker g$ if and only if $g$ is a linear combination of the $f$'s). Set $\epsilon$ to be the minimum of the $\phi_i(x_i)$'s.

Using Goldstine's theorem we can find some $x$ such that $|\phi_i(x) - g(\phi_i)| < \delta$ for each $i$, where we determine $\delta$ later.

Now, consider $x - \sum_i c_i \cdot x_i $, where $c_i$ are some constants. Applying $\phi_j$ to this gives $\phi_j(x) - c_i \phi_j(x_j)$. Choosing $c_i = \frac{\phi_j(x) - g(\phi_j)}{\phi_j(x_j)}$ gives that this is equal to $g(\phi_j)$. So $|c_i| \leq \frac{\delta}{\epsilon}$.

By the triangle inequality we have that $\|x - \sum c_i \cdot x_i\| \leq 1 + \frac{\delta k}{\epsilon}$. So, if $\frac{\delta k}{\epsilon} < \eta$ we are done, but we had free choice of $\delta$.

$\endgroup$
8
  • 1
    $\begingroup$ I think you can actually improve on this. Your condition on the $x_i$ implies they are linearly independent. If you take $V$ to be the linear span, the restrictions of the $\varphi_i$ to $V$ are in $V^*$. Since $V$ is finite dimensional, it is reflexive and so there is an $x\in V$ which (after mapping it into $V^{**}$) equals the restriction of $g$ to $V^*$. This should be the sought $x$. $\endgroup$ Mar 16, 2016 at 19:10
  • 1
    $\begingroup$ So in this case, we even have $\|x\|\le 1$. Hence I'm not completely sure if the above argument is correct. $\endgroup$ Mar 16, 2016 at 19:11
  • $\begingroup$ Our lecturer gave a hint of considering a map $y$ to $(\phi_1(y),\dots,\phi_n(y))$, and then using Open mapping theorem and Hahn-Banach. Any idea using this approach? $\endgroup$
    – yoyostein
    Mar 17, 2016 at 3:42
  • $\begingroup$ That hint replicates the 'standard' proof of the Principle of Local Reflexivity. I don't think Vincents comments are true: take $Y = c_0$, $Y^* = \ell_1$, $Y^{**} = \ell_\infty$ and $g = (1,1,1,\dots)$. Then if we take $\phi_1 = (1/2,1/4,1/8,\dots)$ and $\phi_2 = (1,0,\dots)$ then $x_1 = (0,1,0,\dots)$ and $x_2 = (1/2,-1,0,\dots)$ are such that $\phi_1(x_1) = 1/4$, $\phi_2(x_2) = 1/2$ and $\phi_1(x_2) = \phi_2(x_1) = 0$. But then the action of $g$ is replicated by the point $2x_2 + 4x_1 = (1,2,0,\dots)$ (checking, $\phi_1(2x_2+4x_1) = 1/2 + 2*1/4 = 1$ and $\phi_2(2x_2 + 4x_1) = 1$). $\endgroup$ Mar 17, 2016 at 8:29
  • $\begingroup$ But the constructed point has norm 2. $\endgroup$ Mar 17, 2016 at 8:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .