2
$\begingroup$

So, let $X$ be a Banach space under norm $||\cdot||_X$ and $K\subset X$ a compact subset and a vector subspace of $X$. If we now restrict the norm from $X$ to $K$ (i.e. we take for $x \in K$, $||x||_K=||x||_X$ and by this define norm in $K$), will $K$ together with this norm be a Banach space?

What I think is that this should be positive, as $X$ is Banach space, it is before all, vector space, and $K$ , as a vector space, is vector space itself. Also, $||\cdot||_K$ is obviously a norm, and if sequence $\{x_n\}$ is Cauchy in K, it is as well Cauchy in $X$, because of how norms are defined. As $X$ is Banach, there is a $x = \lim_{n\to \infty} x_n$ and $x \in X$. But, there is a sequence in $K$ converging to that $x$ meaning $x$ is a point of accumulation of $K$. As $K$ is compact, it has to be closed, and hence contain all of its accumulation points, $x$ included. So, a random Cauchy sequence $x_n$ in $K$ converges to a point $x \in K$, and $K$ is Banach as well.

Am I right?? Because this is simple and important and still not in any textbook or wikipedia, and this shall be among the first things proven about Banach spaces??

2 Answers 2

2

The only compact subspace is the trivial one $K = \{ 0 \}$. Yes, this is a Banach space, but not a particularly interesting one.

  • 1
    To add to this: the usual "closed and bounded" notion of compactness from the reals kind of applies in more generality. Particularly, every compact subset of a Banach space is closed and bounded but those conditions are not enough to determine compactness. They are necessary but far from sufficient in general. Pairing this with your setup, it is easy to see that the only case is the trivial subspace. – 2017-01-23
1

Your (correct) proof shows: if $K$ is a compact subset of a Banach space, the $K$ is, with the metric $d(x,y)=||x-y||_X$, a complete metric space.

$K$ is not a Banach space, since $K$ is not a (vector) subspace of $X$

  • 0
    It is assumed that it is a subspace in the post. – 2017-01-23
  • 0
    @ Cameron William: then we have $K=\{0\}$. – 2017-01-23