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?
2026-03-25 15:11:12.1774451472
Simple question about convergence and Gateaux derivative
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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 $x_0$.