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3

Hint: take your favourite non-Hausdorff topological space $X$, and let $S$ consist of two points that do not have disjoint neighbourhoods.


2

$$a_n-a_m =\langle x_n, y_n\rangle -\langle x_m, y_m\rangle$$ $$= \langle x_n, y_n\rangle - \langle x_m, y_n\rangle +\langle x_m, y_n\rangle -\langle x_m, y_m\rangle$$ $$= \langle x_n-x_m, y_n\rangle +\langle x_m, y_n-y_m\rangle$$ Hence $$\left|a_n-a_m\right| \leq \left| \langle x_n-x_m, y_n\rangle +\langle x_m, y_n-y_m\rangle \right|$$ ...


2

Not if $X$ is non-trivial. Consider any $x \in X$ with $\|x\| = 1$. We note that for any $\delta > 0$ and $n \in \Bbb N$, we have $$ \langle x + \delta x,nx\rangle = \langle x,nx\rangle + n\delta \langle x,x\rangle = \langle x,nx\rangle + n\delta $$ so that $$ |\langle x + \delta x,nx\rangle - \langle x,nx\rangle| = n \delta $$ the conclusion follows. ...


1

One of the main tools on infinite matrices and Hilbert spaces operators is the so-called Schur's test. This is Exercise 45 in Halmos' A Hilbert Space Problem Book. Schur's test. Let $A=[a_{ij}]_{i,j\in\mathbb{N}}$ be an infinite matrix. Suppose that there exist positive numbers $p_i>0$, $q_j>0$ ($i,j\in\mathbb{N}$), $\beta>0$ and $\gamma>0$ ...


1

First you have to find a candidate for $L(0,0)$, given by the partial derivatives. This gives you the $0$ matrix. Then observe that $f(0,0)=0$. So $r(0;h)$ is equal to $f(h1,h2)$.


1

Your first line is indeed a definition of $r$: \begin{equation*} r(x_0; h) = F(x_0 + h) - F(x_0) - L(x_0) \, h. \end{equation*} As mentioned by lalala, the typical road for proving Fr├ęchet differentiability is the following: Find a candidate for $L(x_0)$. This can be done by formal calculations or by considering directional derivatives. Define the ...



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