# $T$ is bounded iff $S\circ T$ is bounded.

Let $X , Y, Z$ are Banach spaces. $T:X\to Y$ be a linear map and $S:Y\to Z$ be a one-one linear map which is bounded. Show that $T$ is bounded iff $S\circ T$ is bounded.

One side is trivial that I know. But how to prove that if $S\circ T$ is bounded then $T$ is bounded too.

I think I got this one.

Since $X, Y, Z$ are Banach spaces, so we can use the Closed Graph theorem. We want to show $T$ is bounded. We prove $Gr(T)$ is closed. So take $(x_{n}, T(x_{n})) \to (x,y)$. That means $x_{n} \to x$ and $T(x_{n})\to y$. We need to prove $T(x)=y$.

Now we have since $S\circ T$ is bounded $S\circ T(x_{n}) \to S\circ T(x)$. Again since $S$ is bounded we have $S(T(x_{n})\to S(y)$. So by uniqueness of limit we have $S(T(x))=S(y)$. Since $S$ is one-one $T(x)=y$.

Hope this is fine!

The result is true if $S$ is bijective, otherwise it is false, $S$ is invertible and bounded implies that $S$ is open by the open mapping theorem. We deduce that $S^{-1}$ is continuous, hence bounded. Write $U=S\circ T$, and suppose that $U$ is bounded, we have $T=S^{-1}\circ U$ is bounded as product of two bounded maps.

• I don't understand why $S$ is invertible. The image of $S$ does not need to be closed. – Severin Schraven Jul 9 '17 at 18:33
• Saying that $S$ is invertible is different from saying that $S$ is one to one. We cannot guarantee that $S$ has a bounded two-sided inverse. – Omnomnomnom Jul 9 '17 at 18:33
• The open mapin theorem says that the image of a surjective map is open, this implies that the inverse is continuous. – Tsemo Aristide Jul 9 '17 at 18:34
• @TsemoAristide the word surjective is key. $S$ here is injective, and not necessarily surjective. – Omnomnomnom Jul 9 '17 at 18:35
• @TsemoAristide For many, one-to-one (or 1-1 or one-one) means injective. It is not clear what the asker means. – Omnomnomnom Jul 9 '17 at 18:38