Convergence in norm implication

Question

Consider $(x_n)$ as sequence in a Hilbert space.

Does $\vert\vert x-x_n \vert\vert^2 \rightarrow 0$ implies $x_n \rightarrow x$? If yes, why :)?

Answer 1

$$||x-x_n ||=\sqrt{||x-x_n ||^2}\to \sqrt{0}=0 $$

so the answer is YES.

Answer 2

If $H$ is a Hilbert space, $(x_n)$ a sequence in $H$ and $x \in H$, then by definition(!!):

$$ x_n \to x \iff ||x_n-x|| \to 0.$$

Furthermore we have for a real sequencc $(a_n)$:

$$ a_n \to 0 \iff a_n^2 \to 0.$$

Answer 3

I'd go like this

$$ ||x-x_n|| \rightarrow 0 \Leftrightarrow \forall \epsilon > 0 \;\exists n_0 \; \text{s.t.}\; \forall n \geq n_0 \Rightarrow ||x - x_n|| < \epsilon \Leftrightarrow\\ \forall \epsilon > 0 \;\exists n_0 \; \text{s.t.}\; \forall n \geq n_0 \Rightarrow \sqrt{||x - x_n||^2} < \epsilon \Leftrightarrow \\ \forall \epsilon > 0 \;\exists n_0 \; \text{s.t.}\; \forall n \geq n_0 \Rightarrow d(x-x_n,0) < \epsilon \Leftrightarrow \\ \forall \epsilon > 0 \;\exists n_0 \; \text{s.t.}\; \forall n \geq n_0 \Rightarrow d(x,x_n) < \epsilon \Leftrightarrow x_n \rightarrow x $$