# Intuition: Dual Space is always Banach

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.

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 $$X$$.

• 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. – Kezer Feb 3 at 20:48
• 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. – Wojowu Feb 3 at 20:59
• Oh yes, that is great! That (pretty clear fact as you said) makes "things playing in $Y$" clearer. – Kezer Feb 3 at 21:02

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.

• 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? – Kezer Feb 3 at 20:44
• @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. – Rebellos Feb 3 at 20:46
• 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. – Kezer Feb 3 at 20:52