# $T\in\mathcal{L}(X,Y)$ maps closed bounded subsets onto closed subsets $\implies$ Range $T$ is closed.

Given two normed spaces $X$ and $Y$ and let $T$ be a bounded linear operator $T:X\to Y$. Assume that $T$ maps bounded and closed subsets of $X$ onto closed subsets of $Y$. Show that the range of $T$ is closed.

Can someone give me a hint. That would be nice.

Cheers

• Are you allowed that bounded linear operators are continuous? – Joe Johnson 126 Sep 3 '14 at 10:26
• Sure. However, that would be very easy to show. – Peter Sep 3 '14 at 10:27
• Possible duplicate: math.stackexchange.com/questions/183146/… – Joe Johnson 126 Sep 3 '14 at 10:48
• @Joe Johnson: Note that T is not injective in my case! – Peter Sep 3 '14 at 10:49

First observe that the proof given in Bounded linear operator maps norm-bounded, closed sets to closed sets. Implies closed range? does not use the fact that $X,Y$ are assumed to be complete.

It hence remains to show that we can reduce to the case of an injective operator. To this end, we "factor out the kernel":

As $T$ is bounded, the kernel $Z := \mathrm{ker}(T)$ is a closed subspace of $X$. This shows that the norm

$$\Vert x + Z \Vert := \mathrm{d}(x, Z) = \inf \{\Vert x - z \Vert \mid z \in Z\}$$

defines a norm on the quotient space $X/Z$ (check this, closedness is used to establish definiteness of the norm).

Let $\pi : X \to X/Z, x \mapsto x + Z$ be the canonical projection. Note that $\pi$ is bounded. It is now easy to see that the map

$$\widetilde{T} : X/Z \to Y, x+Z \mapsto Tx$$

is well-defined and bounded with $T = \widetilde{T} \circ \pi$. Hence, $\mathrm{range}\widetilde{T} = \mathrm{range}T$, so that it suffices to show the claim for the injective(!) map $\widetilde{T}$.

Hence, we only have to verify that $\widetilde{T}$ fulfills the requirements. To this end, let $M \subset X/Z$ be bounded and closed, e.g. $M \subset B_R (0)$ for some $R > 0$. I claim that

$$\widetilde{T}(M) = T(\pi^{-1}(M) \cap \overline{B_R}(0)).$$

As $\pi^{-1}(M) \cap \overline{B_R}(0)$ is closed and bounded, this suffices.

The inclusion "$\supset$" above is clear because for $x \in \pi^{-1}(M)$, we have $Tx = \widetilde{T}(\pi x) \in T(M)$. For the inclusion "$\subset$", let $x +Z \in M$ be arbitrary. Because of $M \subset B_R(0)$, the definition of the norm on $X/Z$ shows that there is some $z \in Z$ with $\Vert x - z \Vert < R$.

But then $\pi (x-z) = x+Z \in M$, i.e. $x-z \in \pi^{-1}(M) \cap \overline{B_R}(0)$ and

$$\widetilde{T}(x+Z) = \widetilde{T}(\pi x) = Tx = T(x-z) \in T(\pi^{-1}(M) \cap \overline{B_R}(0)).$$

As $x+Z \in M$ was arbitrary, this completes the proof.