Dual Banach Space question 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.
 A: 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 $\delta$.
