# Fredholm Index/ Laplacian / inverse function theorem

I'm studying an application of inverse function theorem for Banach spaces.

The theorem I would like to prove is:

$X$ is a compact manifold.

And $p:(a,b) \rightarrow (c,d)$ is an increasing function. ($p'>0$)

For any function $g \in C^{ \infty }(X)$ with $c < g < d$ there is a unique $f \in C^{\infty}(X)$ with $a < f < b$ solving the equation:

$p(f) - \Delta f = g$.

The function that should be defined is:

$P: C^{ 2 + \alpha} \rightarrow C^{\alpha}$

$~~~~~~ f \mapsto p(f) - \Delta f$

$DP(f)h = p'(f)h - \Delta h$

I would like to prove that $DP(f)$ is invertible. The Hamilton's article about Nash Moser theorem, page 111 (where I've found this example) asserts that since laplaian is selfadjoint $DP(f)$ has index is zero.

I don't understand why $DP(f)$ has index zero. I just can see it is a Fredholm operator.

Another possible problem is that laplacian is selfadjoint in $L^2(X)$ but maybe not in these other spaces.

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 The index of a Fredholm operator $F$ is usually defined to be $dim(ker(F)) - dim(coker(F))$, but you can just as well define it to be $dim(ker(F)) - dim(ker(F^*))$. This makes it clear that the index of $F$ is $0$ if $F$ is self-adjoint, as your operator $DP(f)$ appears to be. This is why one often formulates index problems for graded self-adjoint operators; the overall operator has index $0$, but its even/odd and odd/even parts may have nonzero index. – Paul Siegel Jul 12 '12 at 4:15