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Let $p, q \in {]1,\infty[}$ where $\frac{1}{p}+\frac{1}{q}=1$ and define

$J\colon L^{q} \to (L^{p})^{*}, f \mapsto \ell_{f}: L^{p} \ni g\mapsto \ell_{f}(g)=\int_{X}fg\,d\mu$

My question is related in general to the notion of well definedness for operators. In particular, why is it sufficient for an operator to be considered well-defined, when we can show that for any $f \in L^{p}$ that $\|\ell_{f}\|_{*}<\infty$. In my previous, elementary versions of well-definedness, we defined it as for any $f \in L^{q}$ there exists exactly one $\ell\in (L^{p})^{*}$ so that $Jf=\ell$. Now showing that $\| Jf\|_{*}<\infty$ definitely shows that $Jf \in (L^{p})^{*}$ but it does not show that there is a unique $\ell \in (L^{p})^{*}$ so that $Jf=\ell$.

In other words, there may exist another $m \in (L^{p})^{*}$ where $m\neq \ell$ so that $m = Jf$ and $\ell = Jf$. I thought that this was the point of proving well-definition.

Why is it always reduced to the idea that well-definition depends only on whether $\|\ell_{f}\|_{*}<\infty$?

Any clarity on this topic will help me greatly.

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    $\begingroup$ What's the definition of "well-defined operator" that you're using? From what I know, the phrase "well-defined" is an abuse of language—when we say "$J$ is a well-defined operator", we don't actually mean "$J$ is an operator, and it has the property of being well-defined", we mean "the given definition of $J$, which is purported to be the definition of an operator, actually is the definition of an operator". $\endgroup$ – Tanner Swett Jun 21 at 22:01
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There are two important notes here to take:

  1. The space $L^p(X,\mu)$ is actually a quotient space (by the subspace of almost everywhere zero functions), and the formal definition $g\mapsto \int_X|fg|\,d\mu$ of the mapping $\ell_f$ refers to actual functions and not their cosets in the quotient.
    So, well-definedness here effectively means that if $g$ is almost everywhere zero, then $\ell_f(g)=0$.

  2. Boundedness of a linear operator is rather connected to the fact that it is everywhere defined.
    For a typical example, it's usual to consider the differentiation operator $f\mapsto f'$ in e.g. $L^2(\Bbb R)$ or $C(\Bbb R)$, which is not bounded, and is defined only on a dense subspace.

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