# Why is boundedness sufficient for well-definedness

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.

• 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". – Tanner Swett Jun 21 at 22:01

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.