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Erwin Kreyzsig's Introduction to Functional Analysis , Section 4.8- Problem2:

Let $X,Y$ be normed spaces, $ T\in B(X,Y)$ and $(x_n)$ a sequence in X. Prove that if $ \forall f\in X' \quad f(x_n) \rightarrow f(x_0)$, then $ \forall g \in Y' \quad g(Tx_n) \rightarrow g(Tx_0).$

I attempted as follows: Consider $$ |g(Tx_n) - g(Tx_0) | \leq ||g||.||T||.|| x_n-x_0|| $$ I thought to replace $g$ with $T, $ which would give me

$$ |g(Tx_n) - g(Tx_0) | =| T(gx_n) - T(gx_0)| \leq ||T||.|g(x_n)-g(x_0)| $$ and conclusion directly follows. But I could not assure that what I did is correct.

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  • $\begingroup$ Writing $g(x_n)$ doesn't make sense in general since $g$ is a functional defined on $Y$ whilst $x_n$ is a sequence in $X$ so you can't swap $T$ and $g$ as you have done. $\endgroup$ – Rhys Steele Apr 22 '17 at 10:53
  • $\begingroup$ I can't directly do that. But, as @Kenny have pointed out, If I show $g\circ T$ is a linear and bounded operator from $X $ to $Y$. Then the result is immediate. $\endgroup$ – Quantes Apr 22 '17 at 11:00
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Do $X'$ and $Y'$ denote the bounded linear functionals on $X$ and $Y$ respectively?

If so, I think the way to proceed is to show that if $g$ is a bounded linear functional on $Y$ and $T$ is a bounded operator from $X$ to $Y$, then $g \circ T$ is a bounded linear functional on $X$. Once you've done this, the result follows immediatedly by taking $f = g \circ T$.

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    $\begingroup$ Yes, exactly. I can show that $goT \in X'$ since $g$ and $T$ is linear bounded. $\endgroup$ – Quantes Apr 22 '17 at 11:05

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