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If I consider the sequence $\{x_n\}\in L^2(\Omega)$ such that: $$ x_n \rightarrow x $$ We know that $x\in L^2(\Omega)$ because we're in a Banach space. So I can say that $||x_n-x||_{L^2(\Omega)}\rightarrow 0$ when $n\rightarrow\infty$ ? I know it is a really stupid question, but I'd like to be sure about this because I'm having some troubles with Gateaux derivative. If I can find the limit for: $$ \lim_{t\rightarrow 0}\frac{F(x_0+th)-F(x_0)}{t}=Ah $$ where $F:X\rightarrow Y$, with $X,Y$ Banach, and $A\in\mathcal{L}(X,Y)$. Can i say without problems that $A$ is the Gateaux-derivative for the operator $F$ at the point $x_0$? Or should I do something else?

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  • $\begingroup$ Out of curiosity, what do you mean by $x_n \to x$ if not $\|x_n - x\| \to 0$? Also, how do you define the Gateaux derivative (and once you answer this, does it not answer your question)? $\endgroup$
    – Tom
    Jun 3, 2014 at 15:47
  • $\begingroup$ I mean that it is impossibile to obtain $||(F(x_0+th)-F(x_0))/t -Ah||_y \nrightarrow 0$ because it is the same thing written in two different ways. That's what I mean^^ $\endgroup$
    – rusca91
    Jun 3, 2014 at 16:01

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First of all: $x_n \to x$ does not mean nothing if you don't specify the topology (or the norm, if you are not comfortable with topology). In some cases, this topology is implicitly given: for example with $x_n\in L^2(\Omega)$, the statement "$x_n\to x$" will be understood as $\lim_{n\to{+\infty}} \|x_n -x\| = 0$.

So, yes! You can say it since it's the definition! Moreover $x\in L^2$ since $L^2$ is closed, the completness of $L^2$ is not relevant here.

For your question about Gateaux derivation: looking at the wikipedia page of "Gateaux derivation", A is not the "Gateaux derivative of $F$", but more precisely the "Gateaux derivative of $F$ at the point $x_0$" (denoted $dF(x_0;\cdot)$ in wiki).

By the way, $A$ may depend on $x_0$.

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  • $\begingroup$ Yes, you're right. I missed that it is "at the point $x_0$. $\endgroup$
    – rusca91
    Jun 3, 2014 at 16:06

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