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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 $f_0(f-f_0)$?

Bayes
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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.

Christopher A. Wong
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  • Thank you! If I consider a simple example. Let $Kf = \int G(f(x))dx$, where $G$ is some continuous function. Then $K'f = \int f(x)G'(f(x))dx$? How can I write $K'f_0(f-f_0)$? – Bayes Mar 28 '13 at 00:59
  • Forgot to mention that $G$ is nonlinear. – Bayes Mar 28 '13 at 01:05
  • Would it be $K'f_0(f-f_0) = \int (f(x)-f_0(x))G'(f(x)-f_0(x))dx$? – Bayes Mar 28 '13 at 01:14
  • I'm actually not exactly sure about your example. What type of object is $Kf$? – Christopher A. Wong Mar 28 '13 at 01:40
  • @ChristopherA.Wong: You are wrong. In order that $K'$ is a Frechet derivative, you need an additional property. – gerw Mar 28 '13 at 09:03
  • That property is implicit in the existence of the linearization, typically. Otherwise the linearization would have unbounded error in a neighborhood of the point. – Christopher A. Wong Mar 28 '13 at 10:07
  • @ChristopherA.Wong: But there are many ways of "linearize" (Frechet, Gateaux, Hadamard,...). E.g. obtaining the Gateaux-derivative by directional derivatives. This is also a linearization. – gerw Mar 28 '13 at 10:54
  • @gerw, I suppose I agree, there are some technical delicacies. I think that for this problem though, the notation indicates that $K'f_0$ is to be taken as a linear operator. – Christopher A. Wong Mar 28 '13 at 19:17
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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 $p = 1$.

gerw
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  • Thank you all for answers. I will upvote your posts once my reputation will be high enough and I will be able to do that. @gerw Suppose that conditions on $G$ are satisfied. How would $K'f_0(f-f_0)$ look like? I'm having problems with very basic undertanding. – Bayes Mar 28 '13 at 09:35
  • @Bayes: I will add that to my answer. – gerw Mar 28 '13 at 10:03
  • @gerw Please will you upload that paper you cited. It is not available here. – C_Guy Oct 06 '14 at 15:53
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$K'(f_0)$ is a linear transformation. So we are talking about a linear operator operating on a vector here.

ncmathsadist
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