Let $X$ and $Y$ be Banach space. If $T: X\to Y$ is a linear map so that $f(T)\in X^*$ (dual space of $X$) for every $f\in Y^*$, then $T$ is bounded.
Here is the proof by contradiction of this question, but I am confused where is the contradiction.
Let $x_n\to x$ in $X$ and $Tx_n\to y$ in $Y$. It is enough to show that $Tx=y$ by closed mapping theorem.
If not, $Tx\neq y$. Then by a corollary of the Hahn-Banach theorem,
Let $Y$ be a closed subspace of a normed space $X$, $u\in X\setminus Y$ and $\rho=dist(u,Y)$. Then there exists a linear functional $f\in X^*$ so that $f(x)=1$ and $f|Y=0$, and $\|f\|=\rho^{-1}$.
we take $Y=\{0\}$ and then there exists $f\in Y^*$ so that $f(Tx-y)=1$, where is a contradiction.
I am confused why that is a contradiction.
Best Answer
By assumption, $Tx_n-y\to 0$, and since $f$ is continuous, $f(Tx_n-y)\to f(0)=0$. On the other hand, for each $n$ we have:
$f(Tx_n-y)=fT(x_n)-f(y)\to fT(x)-f(y)=f(Tx-y)=1,$
where we used the assumption that $fT$ is continuous. So we got $0=1$, a contradiction.