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Let $(\phi_n)$ be a sequence of continuous linear functionals from $X$ - Banach space to $\mathbb{R}$ such that $(\phi_n(x))$ is convergent for all $x\in X$. Show that if a sequence $(x_n)$ is convergent in X norm, then $(\phi_n(x_n))$ is convergent.

My attempt:

All assumptions of Banach-Steinhaus are satisfied. So we have $\sup_n\|\phi_n\|\le M, \:\:M>0$. So we have $|\phi_n (x_n)|\le M\|x_n\|$ for all $n$. If $(x_n)$ is convergent, so is $(\phi_n (x_n))$.

Is this ok?

Edit: Let $x\in X$ be a limit of $(x_n)$ and $\phi(x)$ be a limit of $(\phi_n(x))$.

$|\phi_n (x_n-x)|=|\phi_n (x_n)-\phi_n(x)|\le M\|x_n-x\| \rightarrow 0 \:\:\:\:$ letting $n$ go to $\infty$

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    $\begingroup$ The uniform bound is good, but your final sentence could use some epsilon work. $\endgroup$ – Ravi Jun 14 '14 at 6:51
  • $\begingroup$ I dont know about epsilon work, but I edited it. Could you check it? $\endgroup$ – luka5z Jun 14 '14 at 6:59
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    $\begingroup$ I am not sure I follow your inequalities. I was thinking you would show that your sequence $(\phi_n(x_n))$ is Cauchy. Start with $|\phi_m(m)-\phi_n(n)|$ and use things you know (Including $x_n\to x$ and $||\phi_n||\leq M$) to show this is less than $\epsilon$. Since we are in a Banach space, Cauchy implies Convergent. $\endgroup$ – Ravi Jun 14 '14 at 7:18
  • $\begingroup$ Why is your first term $|\phi_n(x_n-x)|$? I dont see why this is particularly relevant. $\endgroup$ – Ravi Jun 14 '14 at 7:43
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Concerning your last edit, you're almost there. You don't have to show anything is a Cauchy sequence when you already know the limit: $$ \begin{align} |\phi_{n}(x_{n})-\lim_{n}\phi_{n}(x)| & \le |\phi_{n}(x_{n}-x)|+|\phi_{n}(x)-\lim_{n}\phi_{n}(x)| \\ & \le M\|x_{n}-x\|+|\phi_{n}(x)-\lim_{n}\phi_{n}(x)| \end{align} $$ Each term on the right tends to $0$ as $n\rightarrow 0$. A few simple details and you're there.

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