Showing that a nonlinear operator over function spaces is differentiable and locally invertible?

Question

I am aware that the implicit and inverse function theorems can be generalized to infinite dimensional cases, but I am having difficulty in applying it to a specific calculation.

Let $C^1_{\mathbb{R}}[0,1]$ be the space of real-valued $C^1$ functions on the interval $[0,1]$. If we impose the following norm: $$ \begin{equation} \lVert f \rVert := \lVert f \rVert_{\sup}+\lVert f' \rVert_{\sup} \end{equation} $$ for $f \in C^1_{\mathbb{R}}[0,1]$, it is clear that $C^1_{\mathbb{R}}[0,1]$ is a Banach space over $\mathbb{R}$.
Similarly, $C_{\mathbb{R}}[0,1]$ is the Banach space of real-valued continuous functions on $[0,1]$ with the supremum norm.

Then, we can think of the operator $F:C_{\mathbb{R}}^1[0,1] \to C_{\mathbb{R}}[0,1]$ defined by $$ \begin{equation} F(f):=\sinh(f)+(f')^2 \end{equation} $$ How do I show that (or is it indeed true that) $F$ is strongly $C^1$ and the derivative at each point is a linear isomorphism? More generally, how about the case $f \to G(f,f')$ for any smooth function $G : \mathbb{R}^2 \to \mathbb{R}$ whose Jacobian determinant never vanishes?

I appreciate any help.

Answer 1

The derivative of $F$ at $f \in C^1[0,1])$ is by definition the linear operator $F'(f): C^1[0,1] \rightarrow C[0,1]$ given by $$ F'(f)\dot{f} = \lim_{t\rightarrow 0} \frac{F(f+t\dot{f}) - F(f)}{t}. $$ Here, you get $$ F'(f) = (\cosh f)\dot{f} + 2f'\dot{f}'. $$ To apply the inverse function theorem, you need a linear operator $G(f): C[0,1] \rightarrow C^1[0,1]$ such that $F'(f)G(f)g = g$. In other words, given any $g in C[0,1]$, you need to be able to solve for $\dot{f} \in C^1[0,1]$ so that $$ (\cosh f)\dot{f} + 2f'\dot{f}' = g. $$ It's easy to check that if $f = 0$, then this is not possible. $F'(0)$ does have a right inverse, but it is not a bounded linear map from $C[0,1]$ to $C^1[0,1]$. So you cannot invert $F$ near $f = 0$ using the inverse function theorem for Banach spaces. In order to use the inverse function theorem, you need to "regain" the derivative you lost because $F'(f)$ is a first order differential operator.

So you want to restrict to $f$ such that the linear ODE above is a nondegenerate one, i.e., you can write it in the form $$ \dot{f}' = \dots, $$ where the right side is continuous on $[0,1]$. Next, you show that, for a given $f$, this ODE always has a solution $\dot{f} \in C^1[0,1]$. You also need to show that the map from $g$ to $\dot{f}$ can be chosen to be a bounded linear map from $C[0,1]$ to $C^1[0,1]$. This is all straightforward using the existence and uniqueness theorem for ODEs.

Offhand, I don't see how to show that $F$ is locally invertible on a neighborhood of $0$ in $C^1[0,1]$. However, you can show that it is locally invertible in a neighborhood of $0$ in $C^k[0,1]$ for $k$ sufficiently large (including $k = \infty$) by using the Nash-Moser inverse function theorem, as mentioned by @DCM.

Answer 2

If $F(f) = G(f,f')$ for some $G\in C^\infty(\mathbb{R}^2)$ then I think you can establish that

$$ F'(f)u = (\partial_1 G)(f,f')u + (\partial_2 G)(f,f')u' $$

holds in much the same way as if you were dealing with functions of real variables. Writing $D:C^1 \to C$ for differentiation and $I:C^1\to C$ for the inclusion map,

$$ F'(f) = (\partial_1 G)(f,f')I + (\partial_2 G)(f,f')D $$

is:

• Fredholm if $(\partial_2 G)(f,f')$ is everywhere nonzero; and
• Compact if $(\partial_2 G)(f,f')$ is everywhere zero

In the second case, $F'(f)$ can't be an isomorphism.

For the example you give (i.e. $G(x,y) = \sinh(x)+y^2$), $F'(0)$ is the inclusion of $C^1$ into $C$, which can't be an isomorphism for the simpler reason that not all continuous functions are $C^1$.