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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 an arbitrary 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 $T$?

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 $T$?

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 want to have a bounded operator $T\in\mathcal L(E)$ such that $Tx_n=y_n$ for an arbitrary bounded sequence $\{y_n\}$ in $E$. I thought I can define $T=\sum y_n\odot \phi_n$ which $\{\phi_n\}$ is as above. So my main question is that if such an operator $T$ can be defined?

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 an arbitrary 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 $T$?

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 $\sum\|\phi_n\|<\infty$.

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 want to have a bounded operator $T\in\mathcal L(E)$ such that $Tx_n=y_n$ for an arbitrary bounded sequence $\{y_n\}$ in $E$. I thought I can define $T=\sum y_n\odot \phi_n$ which $\{\phi_n\}$ is as above. So my main question is that if such an operator $T$ can be defined?