# Kernel of continuously varying family of operators

Let $X,Y$ be Banach spaces, $F_t:X \to Y, t\in \mathbb{R}$ be a family of bounded linear operators that are continuously varying, in the sense that $F_t(x)$ is a continuous function of $t$ for each $x \in X$. I am interested in the following question:

When is $\dim\left(\ker\left(F_t\right)\right)$ a locally non-increasing function of $t$?

I believe that this is true when $X,Y$ are finite dimensional (please feel free to correct me if I am wrong on this point). But to what extent is it true for infinite-dimensional spaces? In particular, if I add one or more of the following conditions, do we have some known results?

1. $X=Y$,
2. $X,Y$ are Fréchet or Hilbert spaces
3. $\dim\left(\ker\left(F_t\right)\right)$ is known to be finite for all $t$.
4. $F_t$ is equicontinuous (in some meaning of the word)

The reason I am interested in this question is because a paper I recently read proves that the Betti numbers of a compact smooth manifold is locally non-increasing with respect to a deformation of its differentiable structure, by noting that $b_k=\dim \left(\ker\left(\Delta_t \right) \right)$ and remarking that the Laplacian operator $\Delta_t$ is smoothly varying, thus its kernel has locally non-increasing dimension 'by the theory of spectrum of operators'. I am trying to fill out the details of this argument and also trying to see if this argument can be generalized to other cohomology theories, e.g. Dolbeault cohomology.

I apologize if this happens to be a very easy/very nontrivial question, since I am not very experienced with operator theory. Thank you in advance!

Even in finite dimensional case, dim(ker$F_t$) may not be locally non-increasing. For example, take $X=Y=\mathbb{R}$ with usual norm and define $F_t(x)=tx$ for $t,x\in \mathbb{R}$.

Then, dim(ker$F_t)=0$ when $t\neq0$ and $1$ when $t=0$ which is not non-increasing in the neighbourhood of $t=0$.

• Why is it not non-increasing in a nbd of 0? It is 1 at 0 and 0 at any other nearby point. Maybe you are confused by my terminology, by 'locally nonincreasing', I mean upper semicontinuous. – Chi Cheuk Tsang Jul 6 '17 at 14:07