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If $H$ is a Hilbert space and $x \in H$ then does it follow that $||x|| < \infty$?

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depends on what you mean by the norm. for instance $\ell^2$ is the space of sequences $a=(a_n)$, $a_n\in\mathbb{C}$ with $||a||:=\sqrt{\sum_n|a_n|^2}<\infty$. –  yoyo Feb 25 '11 at 18:27
    
$||x||^2=<x,x>$, im not looking for a particular hilbert space, but rather i want to know if this is true for any hilbert space –  jack Feb 25 '11 at 18:29
    
every (infinite dimensional) seperable hilbert space is isomorphic to $\ell^2$. the sequence is just the coefficients of some orthonormal basis. –  yoyo Feb 25 '11 at 18:31

3 Answers 3

up vote 3 down vote accepted

Short answer (with some extra text to fill it out): Yes.

:)

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On any real or complex vector space $X$ for which a norm $\|\cdot\|$ is defined, part of the definition is that $\|x\|$ is a real number for each $x\in X$. The norm on a real or complex inner product space $H$ fits into this context, because part of the definition of the inner product is that $\langle x,y\rangle$ is a real or complex number for each $x$ and $y$ in $H$, and that $\langle x,x\rangle$ is nonnegative for each $x\in H$, and hence $\langle x,x\rangle$ is a nonnegative real number (excluding the possibility of $\langle x,x\rangle=\infty$).

In some contexts there is notational abuse of $\|\cdot\|$, which may be the source of the question here. For example, suppose that $\mu$ is a positive measure on $X$, and $1\leq p\lt \infty$. Some authors will say that for a measurable real or complex-valued function $f$ on $X$, $\|f\|_p$ is defined to be the $p^\text{th}$ root of $\int_X |f|^pd\mu$, before defining $L^p(\mu)$ to be the set of such $f$ for which $\|f\|_p$ is finite. With this convention, $\|\cdot\|_p$ is a norm when restricted to $L^p(\mu)$, but the extended notation allows $\|f\|_p=\infty$ to also be a meaningful statement; it is equivalent to saying that $f$ is not in $L^p(\mu)$. So for example, $\|f\|_2$ can be infinite for some measurable $f$, but $\|\cdot\|_2$ is a norm on the Hilbert space $L^2(\mu)$, meaning in part that $\|f\|_2$ is a nonnegative real number for all $f\in L^2(\mu)$.

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Pick a general vector space $V$ over $\mathbb{R}$ or $\mathbb{C}$. A norm on $V$ is a function $N: V \rightarrow \mathbb{R}$ such that

(i) $N(v) \geq 0$

(ii) $N(kv) = |k|N(v)$

(iii) $N(v) = 0$ iff $v = 0$

(iv) $N$ satisfies the triangle inequality.

Such an $N$ is a generalization of the standard Euclidean norm on $\mathbb{R}^{n}$ (the norm derived from Euclidean distance). For this reason we usually write $\| v \|$ instead of something like $N(v)$.

So by definition, all norms are positive. A Hilbert space is, again by definition, a normed vector space with some additional properties. In particular the norm is induced by an inner product $<,>$ in the obvious way (again a generalization of Euclidean space). Note that $\| \cdot \|$ = $\sqrt{<,>}$ can't be negative, but if it takes complex values then the function $\| \cdot \|$ isn't a norm for trivial reasons, so your space isn't a Hilbert space (again for trivial reasons -- the inner product must induce a norm for it to be a Hilbert space).

Edit: I really need to start checking the dates when these things are written...

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