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
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$||x||^2=<x,x>$, 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
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2every (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
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@yoyo We don't really need separability here. Every Hilbert space $\mathcal H$ has an orthonormal basis ${e_\alpha}{\alpha\in A}$. There is a unitary map: $\mathcal H\to l^2(A)$ defined by $\mathcal H\ni x\mapsto \hat x\in l^2(A)$ with the $\alpha$'s coordinate of $\hat x$ is $\hat x(\alpha)=\langle x,e\alpha\rangle$. The linearity is easy to check. Isometry follows from Parseval's identity. And isometry also implies injection. It remains to show it is surjective. Note that every square summable sequence naturally corresponds to a Cauchy sequence in $\mathcal H$ by Pythagorean theorem. – Bach Apr 14 '20 at 03:29
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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 $f\in L^2(\mu)$.
Jonas Meyer
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Can you please clarify why we need to exclude the possibility of $\langle x,x\rangle=\infty$ ? – texmex Aug 08 '23 at 14:12
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1@texmex: it's a consequence of the definition of the inner product, its values are complex numbers (or real numbers in the real case) unless you are using some nonstandard definition in which case it would hopefully be called out in context. – Jonas Meyer Oct 23 '23 at 21:17
