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Let $L$ be a linear operator from Banach space $X$ to $Y$. Is the dimension of the kernel of the adjoint of $L$ the same as the dimension of the cokernel? The cokernel is $Y/(Im L)$.

Also, is the index of operators for which this quantity is defined on, same some subspace of the bounded linear operators from $X$ to $Y$, continuous on this set? This is the case for Fredholm operators, but I was wondering if it was true in more general settings.

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In the "Encyclopedic Dictionary of Mathematics" it says that this is true if the range is closed and the domain is dense. But it says this under the heading of Fredholm operators, so I don't know if this is true without the Fredholm hypothesis. –  Q-the-curvature Nov 17 '11 at 3:27
    
So, you want unbounded operators? You should say so, there's nothing in your question that hints at that. Be that as it may, it is completely wrong if the range isn't closed, even for bounded operators: In fact for bounded operators you have $\ker{L^\ast} = (\operatorname{im}{L})^\perp$, so in particular if the range is dense you have that $\ker{L^\ast} = 0$. If the range is dense the image must have infinite co-dimension, so the cokernel of $L$ is infinite-dimensional. On the other hand, the best you can hope for in the second question is upper semi-continuity: perturb the identity with ... –  t.b. Nov 17 '11 at 5:36
    
... the orthogonal projection $P$ onto a subspace of a Hilbert space, so consider $I + tP$. For real $t \neq -1$ this operator is self-adjoint and invertible, so has zero cokernel, while for $t = -1$ you get a cokernel isomorphic to the range of $P$ which can be of arbitrary dimension. –  t.b. Nov 17 '11 at 5:38

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