Linearization of operator Suppose $K:X\to Y$ is nonlinear operator between two Banach spaces. That is $f\mapsto K(f)$. The linearized operator is written as $$Kf_0 + K'f_0(f-f_0)$$ I would like to ask is it correct to say that $K'$ is a Frechet derivative? Frechet derivative is a linear operator, right? What does $K'f_0(f-f_0)$ mean? Is it a product of $K'(f_0)$ and $f-f_0$ or it is an operator $K'$ of $f_0(f-f_0)$?
 A: If $K'$ exists at $f_0$, then indeed $K'$ is the Frechet derivative. We can consider $K'$ as a map
$$ K': V \rightarrow B(V,W),$$
where $B(V,W)$ is the space of bounded linear operators from $V$ to $W$. Thus $K'(f_0)$ is an element of $B(V,W)$, so it acts on $(f - f_0)$ as suggested by the expression you wrote out. One should avoid calling this a "product", since it's really an operator acting on a vector. 
A: As other mentioned, $K'$ maps the points of $V$ to a bounded, linear operator $V \to W$.
In order that you have a Frechet derivative in $f_0 \in V$, you need to verify
$$\lim_{\lVert h\rVert_V \to 0} \frac{\lVert K(f_0 + h) - K(f_0) - K'(f_0)\,h\rVert_W}{\lVert h\rVert_V} = 0.$$
Now, let me answer the question you asked in a comment: Let $G : \mathbb{R} \to \mathbb{R}$ be differentiable and not affine. Let $V = L^p(\mu)$, $W = \mathbb{R}$.
Then, $K = \int G(f(x)) \, d\mu(x)$ is only Frechet differentiable if $p > 1$.
In any case, you need also some additional conditions on $G$, see Goldberg, Kampowski, Tröltzsch: On Nemytskii operators in Lp-spaces of abstract functions,
Math. Nachrichten 155 (1992), 127-140.
edit: If $K$ is differentiable (Gateaux or Frechet), you get the derivative
$$K'(f_0) \, h = \int G'(f_0(x)) \, h(x) \, d\mu(x),$$
as expected.
Note that Gateaux differentiability may also hold for $p = 1$.
A: $K'(f_0)$ is a linear transformation.  So we are talking about a linear operator operating on a vector here.
