dimension codimension problem for a perturbation operator 
Let $H$ be a separable Hilbert space and let $L:H\to H$ be a compact
self-adjoint bounded linear operator . Let $\lambda_0$ be a simple
characteristic value and let $\phi_0$ be an eigenvector such that
$\lambda_0L\phi_0=\phi_0$ and $\langle L\phi_0,\phi_0\rangle=1$ .
Let $A:H\to H$ be a compact map such that $A(0)=0$ and the following
property holds : There exists a constant $C>0$ if $||u||\leq r$ and
$||v||\leq r$ for $u,v\in H$ , we have $||A(u)-A(v)||\leq Cr^2||u-v||$
.
Define a map $f:H\times\mathbb{R}\to H$ by $f(u,\lambda)=u-\lambda Lu+A(u)$
for all $(u,\lambda)\in H\times\mathbb{R}$ . We can assume $f$ is $C^p$ for some $p\geq2$ .

Now I want to show two things :
$(1)$ $\ker(\partial_uf(0,\lambda_0))$ is a one-dimensional subspace of $H$ .
$(2)$ $\text{range}(\partial_uf(0,\lambda_0))$ has codimension $1$ .
What I tried so far :
$(1)$ It's not hard to see $\partial_uf(0,\lambda_0)=I-\lambda_0L+A$ and hence $$\partial_uf(0,\lambda_0)\phi_0=\phi_0-\lambda_0L\phi_0+A\phi_0=A\phi_0$$ but I am getting nowhere to show that $\text{span}\{\phi_0\}=\ker(\partial_uf(0,\lambda_0))$ .
$(2)$ It is enough to show there exists some $h^*\in H^*\setminus\{0\}$ such that $\ker(h^*)=\text{range}(\partial_uf(0,\lambda_0))$ , i.e. we need to find such $h^*$ so that $$h^*(u-\lambda_0Lu+A(u))=0 \ \ \ \text{for all} \ \ u\in H$$ But I am unable to find such $h^*$ .
I want to verify the approach of solutions I am trying to make . If my approach is misleading , then any alternative method of approach would be appreciated . Regards .
Edit : I attached screenshots related to the problem .

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Of course , I wanted to check hypothesis $(ii),(iii)$ of theorem $4.3.3$ as : "It is trivial to check that all the hypothesis of theorem 4.3.3. are verified..." .
 A: I think your post missed out on some motivation. Probably, you tried to isolate the problem (for example, there is no motivation of the constants $C$ and $r$ and you also do not explain why we can assume that $f$ is in $C^p$) but the assumptions you present here are not sufficient. Maybe try to post the full problem.

It's not hard to see $\partial_uf(0,\lambda_0)=I-\lambda_0L+A$ ...

This is actually wrong. Since $A$ is generally not linear, $\partial_uf(0,\lambda_0)$ is generally not a linear operator are therefore not the derivative on $f$ with respect to $u$. One rather has $$\partial_uf(0,\lambda_0) = I - \lambda_0 L + DA(0),$$ where $DA(0) \colon H \to H$ is the Fréchet derivative of $H$ which is by definition linear.
However, there is a counter example to the statements you try to prove even in finite dimensions. Indeed, consider the separable Hilbert space $H = \mathbb R^2$ and the linear operator $L \colon H \to H$ given by the matrix
$$ L := \begin{pmatrix} 1 & 0 \\ 0 & 2\end{pmatrix}.$$
Then $L$ is compact since $H$ is finite-dimensional and self-adjoint since its eigenvalues $1$ and $2$ are real (actually it is even positive in the Hilbert-space sense and also positive in the sense of Banach lattices). Moreover, both eigenvalues $1$ and $2$ are simple, i.e., their respective eigenspaces are one-dimensional.
Moreover, set $A(x) := Lx$ for all $x \in H$. Then $A(0) = 0$ and $A$ satisfies the growth condition since
\begin{align*}
\lVert A(x) - A(y) \rVert_2^2 &= \lvert x_1 - y_1 \rvert^2 + 4 \lvert x_2 - y_2 \rvert^2 \\
&\leq 4 (\lvert x_1 - y_1 \rvert^2 + \lvert x_2 - y_2 \rvert^2) = 4 \lVert x - y\rVert_2^2
\end{align*}
for all $x, \, y \in H$. Just pick $C = 2$ and $r = 1$ e.g. but actually the growth condition we showed above is stronger than yours since our estimation does not depend on any $r$ whatsoever.
Moreover, $A$ is a compact map, i.e., if $(x_n)_{n \in \mathbb N}$ is bounded in $H$, then $(A(x_n))_{n \in \mathbb N}$ has a convergent subsequence. Namely, $(A(x_n))_{n \in \mathbb N}$ is bounded if $(x_n)_{n \in \mathbb N}$ is bounded due to the growth condition. Since $H$ is finite-dimensional, the Bolzano-Weierstrass theorem then implies that $(A(x_n))_{n \in \mathbb N}$ has a convergent subsequence. Note that this argument works for all functions $A : H \to H$ satisfying the growth condition.
Now let's come to your two assertions: Clearly, we have $DA(0) = L$ and thus $$\partial_uf(0, 1) = I - \lambda_0 L + DA(0) = I$$, where $\lambda_0 = 1$. Therefore,
\begin{align*}
\ker \partial_uf(0, 1) & = \ker I = \{0\}, \\
\operatorname{codim}\partial_uf(0, 1) H &= \dim(H/\partial_uf(0, 1) H) = \dim (H/H) = \{0\}
\end{align*}
which shows that neither $\ker \partial_uf(0, 1)$ nor $\operatorname{codim}\partial_uf(0, 1) H$ is one-dimensional. However, I would really like to see the full problem!
EDIT: I just want to point out two things about Theorem 4.3.3 that really irritate me. I stress that this might just be a result of me being part of a different mathematical community.
However, eigenvalues are usually defined to be $\lambda \in \mathbb C$ such that $L \psi_0 = \lambda \psi_0$ and usually one considers eigenvectors that are normalized, i.e., $\langle \psi_0, \psi_0 \rangle = 1$, and not such that $\langle L\psi_0, \psi_0 \rangle = 1$.
Moreover, it is bad style to assume that $f$ is in $\mathcal C^p$ instead of assuming that $A$ is in $\mathcal C^p$. That would imply the former statement but yielding a weaker theorem. Maybe I am just not aware of the applications but I would bet that you would usally check that $f$ is in $\mathcal C^p$ by proving that $A$ is in $\mathcal C^p$. Surely, if you have some kind of cancelations going on you do not need that $A$ is in $\mathcal C^p$. However, I highly doubt that you can expect that from non-toy examples.
I would like to suggest that maybe your reference is flawed. That would not be unusual. If someone else can spot a flawed argument in my post I would be glad to know.
