TITLE: Is duality an exact functor on Banach spaces or Hilbert spaces?
QUESTION [8 upvotes]: Let $V,V',V''$ and $W$ be vector spaces over $k$. Then, it is known that $\operatorname{Hom}(\cdot,V)$ is a contravariant exact functor, i.e. for each exact sequence
$0\to V'\to V\to V'' \to 0$,
and each $W$, the induced sequence
$0\to \operatorname{Hom}(V'',W)\to\operatorname{Hom}(V,W)\to\operatorname{Hom}(V',W)\to 0$ 
is exact.
But what if all spaces involved are Banach spaces (over $\mathbb{C}$ or $\mathbb{R}$) and if we replace $\operatorname{Hom}(\cdot,V)$ by $\operatorname{L}(\cdot,V)$, i.e. the continuous linear maps with domain $V$? Is this functor still exact? What if we restict ourselves even stronger to Hilbert spaces? I'm particularly interested in the specialization of $W=\mathbb{R}$ or $W=\mathbb{C}$, where we get the dual spaces (again, only continuous maps considered in the Banach- or Hilbert case).
I'm learning for an exam and the books I've been reading say nothing about it...
Thank you!

REPLY [2 votes]: I'm going to adjust your notation, because $V'$ looks too much like a dual space to me.
In the category of Banach spaces, where the morphisms are the continuous linear maps, one should perhaps interpret "image" in the categorical sense, as the closure of the image.  (See the comments on Martin's answer.)  If we interpret "exact" according to this sense, then  a sequence $$0 \to X_1 \overset{S}{\to} X_2 \overset{T}{\to} X_3 \to 0$$ is exact iff $\ker S = 0$, $\overline{S(X_1)} = \ker T$, and $T(X_2)$ is dense in $X_3$.
In this setting, the functor $L(\cdot, W)$ is not exact, not even when $W = \mathbb{R}$.  For instance, let's take $X_1 = \ell^1$, $X_2 = \ell^2$, $S$ the inclusion map, and $X_3 = 0$.  Since $S$ is injective and has dense image, this sequence is exact.  If $L(\cdot, \mathbb{R})$ were an exact functor, then $S^* : (\ell^2)^* \to (\ell_1)^*$ should also be injective and have dense image.  $S^*$ is injective,  but since it is a map from the separable Banach space $(\ell^2)^* = \ell^2$ to the non-separable Banach space $(\ell^1)^* = \ell^\infty$, it cannot have dense image.  (In fact, $S^*$ is just the inclusion map $\ell^2 \to \ell^\infty$ and the closure of $\ell^2$ in $\ell^\infty $ is $c_0$, the sequences which vanish at infinity.)
If you work in the category of reflexive Banach spaces, then $L(\cdot, \mathbb{R})$ is an exact functor; this is a fairly straightforward diagram chase using the Hahn-Banach theorem.