About a property in a reflexive Banach space

Question

Let $E$ be a reflexive Banach space. Let $\{x_n\}_n$ be a bounded sequence of linearly independent elements of $E$. Does there exist a sequence $\{\phi_n\}_n$ of elements of $E^*$ (the dual of $E$) such that $\langle x_n,\phi_m\rangle=\delta_{mn}$ and $\sum\|\phi_n\|<\infty$.

As Nate Eldredge's argument below shows, there is not such sequence $\{\phi_n\}$. Indeed I have a bounded sequence $\{y_n\}$ and I want to know if there is a bounded operator $T\in\mathcal L(E)$ such that $Tx_n=y_n$ for a bounded sequence $\{x_n\}$ in $E$. I thought if $\{x_n\}$ is a bounded linearly independent sequence, then one can define $T=\sum y_n\odot \phi_n$ which $\{\phi_n\}$ is as above. So my main question is that if there is such an operator $T$?

Answer 1

Certainly "linearly independent" is not good enough.

Example. Let $e_n$ be an orthonormal basis in a Hilbert space. Take $$ x_n = \frac{1}{\sqrt{n}}(e_1+\dots +e_n) $$ Coefficients are chosen so that $\|x_n\|=1$. These vectors are linearly independent. What can $\phi_n$ be, so that $\phi_m(x_n) = \delta_{nm}$? Well, if $\phi_n$ is orthogonal to $u_1,\cdots,u_{n-1}$, then $\phi_n$ is orthogonal to all linear combinations of these, so it is orthogonal to $e_1, e_2,\cdots, e_{n-1}$. But then $$ 1 = \phi_n(x_n) = \phi_n\left(\frac{1}{\sqrt{n}} e_n\right) $$ and therefore $\|\phi_n\| \ge \sqrt{n}$. Not bounded.

Next question. Is there a bounded linear operator such that $T(x_n) = e_n$? Now we get $$ \langle T(e_n), e_n\rangle = \sqrt{n} $$ so that operator is not bounded. More detail: $T(x_1), \dots, T(x_{n-1})$ are all in the span of $\{e_1,\dots,e_{n-1}\}$, so $T$ of any linear combination of $x_1,\cdots,x_{n-1}$ is also in that span, and therefore is orthogonal to $e_n$. For $1 \le m < 1$, $e_m$ is a linear combination of $x_1,\cdots, x_{n-1}$. So $\langle T(e_m),e_n\rangle = 0$. Thus $$ \langle T(e_n),e_n \rangle = \langle T(e_1+\cdots+e_n),e_n\rangle = \langle T(\sqrt{n} x_n), e_n\rangle = \sqrt{n}\langle e_n,e_n\rangle = \sqrt{n} $$