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 $X$.

  • $\begingroup$ 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
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    $\begingroup$ 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

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.

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    $\begingroup$ 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

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