is $x \in H$ then does it follow that $||x|| < \infty$?
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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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