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

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.

• Isn't $F'(0)$ just the inclusion map sending $C^1([0,1])$ into $C([0,1])$? (or have I just displayed an embarrassing inability to differentiate ;))
– DCM
Dec 29, 2021 at 22:22
• Just in case you're not aware of it already: there are lots of nice examples of this sort of thing in R. S. Hamilton's "The Inverse Function Theorem of Nash and Moser".
– DCM
Dec 29, 2021 at 22:55

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.

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 $$C^1$$.