# Composition involving bounded linear operators

I recently come across the following statement mentioned in a proof:

Let $X,Y$ be normed linear spaces and $T:X \rightarrow Y$ be a linear operator. if for every bounded linear functional $U: Y \rightarrow \mathbb{R}$ , $UT$ is bounded, then $T$ is also bounded.

How can we justify this statement?

• Your $U$ is not a linear functional, as $U(\lambda y) = |\lambda| U(y)$ which shows a violation for negative $\lambda$. – Roland Feb 4 '16 at 21:59

I'm going to revise the notation to make things easier to follow.

Say $X^*$ is the space of bounded linear functionals on $X$, and similarly for $Y^*$. I'm going to write $x$ and $y$ for elements of $X$ and $Y$ and I'm going to write $x^*$ and $y^*$ for elements of $X^*$ and $Y^*$.

Define the adjoint $T^*:Y^*\to X^*$ as usual: $$T^*y^*=y^*T.$$We can use the Closed Graph Theorem to show that $T^*$ is bounded (note that $X^*$ and $Y^*$ are Banach spaces). Assume that $y_n^*\to y^*$ and $T^*y_n^*\to x^*$; we need to show that $x^*=T^*y^*$. By definition that means we need to show that $x^*(x)=T^*y^*(x)$ for every $x\in X$. But since norm convergence in the dual implies pointwise convergence, $$x^*(x)=\lim T^*y_n^*(x)=\lim y_n^*(Tx)=y^*(Tx)=T^*y^*(x).$$

So $T^*$ is bounded. This implies that $T$ is bounded. NOTE that the Hahn-Banach theorem applies to incomplete normed vector spaces, in particular if $y\in Y$ then $$||y||=\sup_{||y^*||=1}|y^*(y)|.$$So for $x\in X$ we have $$||Tx||=\sup_{||y^*||=1}|y^*Tx|=\sup_{||y||=1}|T^*y^*x| \le \sup_{||y||=1}||T^*y^*||\,||x||\le||T^*||\,||x||.$$

We use the Closed Graph Theorem. Assume $x_n \to x$ and $Tx_n \to y$. We need to prove that $Tx = y$. By Hahn Banach this is true if $f(y) = f(Tx)$ for all $f \in Y^*$. But $f(Tx) = \lim_n f(Tx_n) = f(y)$. So we are done.

(In case $X,Y$ are Banach.)

• I thought that this only applies to Banach spaces. Only normed linear space is assumed. – erik Feb 4 '16 at 21:56
• Do you think that my above argument is correct? – erik Feb 4 '16 at 21:56
• Ok, I see. Your operator $U$ is not linear and indeed my argument only works for Banach spaces. – user42761 Feb 4 '16 at 22:00