# An infinite dimensional closed subspace of $C[0,1]$

I've been studying Functional Analysis for a test and this was one of the questions in one of the previous tests:

Construct an infinite-dimensional closed subspace of $$C[0,1]$$ with respect to the supremum norm.

I know every infinite-dimensional Banach Space has a subspace which is not closed but I cannot find a closed one. Also, why does a supremum norm matter?

• How about $C[0,1]$ itself? – Severin Schraven Jul 27 at 23:13
• Well, the norm does matter because apriori $C[0,1]$ is just a set. We want to speak about closed sets and thus we need a topology. The topology is induced by the norm. – Severin Schraven Jul 27 at 23:15
• If you want some more examples you might want to consider the spaces$$A_y:=\{ f\in C[0,1] \ \vert \ \forall x\in [0,y] : f(x)=0\}$$ where $y\in [0,1)$. – Severin Schraven Jul 27 at 23:20
• @SeverinSchraven I meant proper subspace, sorry Edit: Also, I meant how does the answer differ if we use any other norm. – Abhimanyu Swami Jul 27 at 23:47
• The $A_y$ are all proper subspaces. In general you can use the following idea. Just take $Ker(\varphi)$ for any $\varphi: X \rightarrow \mathbb{R}$ continuous linear functional. This will give you a closed infinite-dimensional subspace of $X$. If $\varphi$ is not identically zero, then $Ker(\varphi)$ is also proper. – Severin Schraven Jul 28 at 15:42

As mentioned earlier, $$C([0,1])$$ is of course a closed subspace of itself and you can verify that it has infinite dimension. Regardless, if you meant to ask for a proper subspace, then $$C_0([0,1]) = \{ f\in C([0,1]): f(0)=f(1)=0\}$$ works. I'll leave it to you to show why.
• Does this use Stone–Weierstrass theorem, with the constant term and the sum of coefficients of the polynomials being equal to $0$? – Abhimanyu Swami Jul 27 at 23:45
• @AbhimanyuSwami No, we don't need anything. You just need to show that it cannot be finite dimensional and that it is closed in $C([0,1])$. – BigbearZzz Jul 28 at 9:25
We can use some abstract machinery to generalize to any infinite-dimensional normed space $$X$$. Pick some $$0\neq v\in X$$ and define the linear function $$f: \mathbb{R} v \rightarrow \mathbb{R}, \ rv \mapsto r$$. By the Hahn-Banach Theorem there exists a continuous linear function $$\varphi: X\rightarrow \mathbb{R}$$, such that $$\varphi(rv)=r$$. Thus, $$Ker(\varphi)$$ is a closed infinite-dimensional subspace of $$X$$. As $$\varphi$$ is not identically zero we get that $$Ker (\varphi)$$ is also proper.