2
$\begingroup$

Theorem. Let $X$ be a normed space and $Y$ be a Banach space. Then the set of continuous linear maps $L(X,Y)$ is a Banach space (with the operator norm).

From the well-known theorem above, we get an immediate consequence: Dual spaces $X^*$ are always Banach spaces.

Why should that be true intuitively? (I am not looking for a proof.)

I'd like to think that somehow, the space $L(X,Y)$ plays more on $Y$ than on $X$, so it inherites being Banach from $Y$. Though, I can't quite make out a clear intuition for all this.

$\endgroup$
0
$\begingroup$

If you have a sequence of functionals $f_n:X\to\mathbb R$ which is Cauchy in the operator norm, then in particular this implies that for any $x\in X$ the sequence $f_n(x)$ is Cauchy in $\mathbb R$, which means that it has a limit, which we denote by $f(x)$. This gives us a function $f:X\to\mathbb R$ such that $f_n$ converges to $f$ pointwise. Intuitively one might expect that $f$ is then in fact the limit of $f_n$ in $X^*$ - at the very least $f$ is a candidate to be one, and we just need to show that it works.

From there it's actually easy to get a complete proof - it's pretty clear that $f$ is a linear functional, so we just would need to check it's bounded and $f_n$ in fact converge to $f$, which is just a matter of writing down an estimate uniform in $X$.

$\endgroup$
  • $\begingroup$ Thank you. This is almost the entire proof, though unfortunately not what I'm looking for as intuition - it doesn't give me the deeper reason I seek. The convergence of $f_n(x)$ is connected to what I wrote about things "playing in $Y$ rather than in $X$", though - so I feel my hunch is close to some intuition. $\endgroup$ – Qi Zhu Feb 3 at 20:48
  • 1
    $\begingroup$ I agree with you. In fact, I think it should pretty clear (imo even intuitively so) that things will be playing in $Y$ - after all, distance between operators is defined in terms of their values, i.e. operators are close if their values are. $\endgroup$ – Wojowu Feb 3 at 20:59
  • $\begingroup$ Oh yes, that is great! That (pretty clear fact as you said) makes "things playing in $Y$" clearer. $\endgroup$ – Qi Zhu Feb 3 at 21:02
0
$\begingroup$

The lemma stated doesn't have to really do with the Theorem you mentioned, but rather has to do with $\mathbb R$ being a complete normed space. Noting that $X$ is Banach and $\mathbb R$ is Banach too, it's easy to show that for $x \in X$, the sequence $f_n(x)$ would be Cauchy in $\mathbb R$, which falls down to showing that it converges into it, thus the dual $X^*$ will be complete as well and thus Banach.

$\endgroup$
  • 2
    $\begingroup$ I don't really why it doesn't have to do with the mentioned theorem - $\mathbb{R}$ being Banach is simply a special case of $Y$ being Banach, is it not? $\endgroup$ – Qi Zhu Feb 3 at 20:44
  • $\begingroup$ @Kezer The dual space $X^*$ of $X$ is the space of all the linear functionals, such that $f: X \to \mathbb R$. Thus, it is pretty straightforward to study it individually, as it is trivial to prove that $\mathbb R$ is Banach. $\endgroup$ – Rebellos Feb 3 at 20:46
  • $\begingroup$ I agree one can study it individually - but didn't you yourself say that the crucial property is that $\mathbb{R}$ is complete? That for me says that the critical property, in this case, is completeness, in other words, it falls down to $Y$ being Banach. So for me, this is either a natural generalization of the lemma or an immediate consequence of the theorem - which (again, for me) means that they are connected. At least, this is my philosophy of this. $\endgroup$ – Qi Zhu Feb 3 at 20:52

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.