2
$\begingroup$

I) Let $X$ be a Banach space. In his book Brézis introduces two definitions (section II.5):

1) If $M \subset X$ then he defines $$M^\perp=\{f \in X' \quad f(x)=0 \quad \forall x \in M\}$$ 2) If $N \subset X'$ then he defines $$N^\perp=\{x \in X \quad f(x)=0 \quad \forall f \in N\}$$ Isn't it absurd since, if $X$ is a Banach space then so is its dual $X'$ and therefore, for $N \subset X'$ the definition 1) gives $$N^\perp=\{f \in X'' \quad f(x)=0 \quad \forall x \in N\}$$ which does not correspond to the definition 2) unless $X$ is reflexive (which is not assumed there, this notion is introduced later in the book).

II) I also have another question. Assume now that $X$ is a Hilbert space. Suppose that, for whatever reason, I don't want to identify $X'$ with $X$ nor $X''$ with $X$. Then, how do you call the following subspace: $$\{x \in X \quad \langle x,m \rangle_H=0 \quad \forall m \in M\}$$ where $M$ is a subset of $X$ and $\langle x,m \rangle_H$ denotes the inner product of $X$ between the two elements $x,m \in X$.

$\endgroup$
1
$\begingroup$

The asymmetry in the definition is purposeful. The definition of $N^{\perp}$ requires that you know that you're working in the dual of a Banach space.This means that you will get different spaces if you consider $l^{\infty}$ as a Banach space by itself or as the dual of $l^1$. It will always be clear from context.

You will notice this asymmetry again on page 13 when he introduces the convex conjugate, but from the examples that follow it turns out this actually contains the orthogonal complement as a special case. It will become more clear why this is done with the Frenchel-Moreau theorem and when you get to weak topologies in chapter three.

I suppose you would still call it the orthogonal complement, although I don't know that I've ever seen anyone use the more general definition when working in a Hilbert space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.