Closed infinite dimensional linear subspace of $L^p((0,1))$ consisting of continuous functions

Question

Let $1 \le p < +\infty$. Construct a closed infinite-dimensional linear subspace of $L^p((0,1))$ consisting of continuous functions.

I tried the set of all polynomials on $(0,1)$, but polynomials are dense in $L^p$, so it's not closed. Any other ideas?

An idea: try the space of continuous functions which are piecewise linear on each interval $[1/(n+1), 1/n]$. — Nate Eldredge, Apr 12 '17 at 20:49
Note that by http://math.stackexchange.com/questions/727274/finite-dimensional-subspace-of-c0-1 it is impossible for this subspace to consist entirely of bounded functions. — Nate Eldredge, Apr 12 '17 at 21:04
@Nate, in your linked question, the domain is $[0,1]$, and the hint I'm given is the open property of domain is important. — , Apr 12 '17 at 22:08

user251257 · Accepted Answer · 2017-04-13T10:14:45.290

2

Let $h: \mathbb R\to \mathbb R$ be the hat function on $[0,1]$, that's is $h(x) = 2\max(0, 1 - |2x-1|)$. For $n\in \mathbb N$ let $$ h_n(x) = 2^n h\left( 2^{n}x-1 \right). $$ Then, consider the subspace $$ S = \left\{ \sum_{n=1}^\infty a_nh_n \mid a \in \ell^1 \right\}.$$

Actually $S$ is isometric isomorph to $\ell^1$.

edited Apr 13 '17 at 10:14

answered Apr 12 '17 at 21:04

user251257

9,417

Sorry, what is hut function, I can't find it by google. How is $h$ defined on $(0,1/2)$ and $(1/2,1)$? – Apr 12 '17 at 22:11
@Philo it was a typo. It should be hat function. Sorry. – user251257 Apr 12 '17 at 22:17
@Philo: Presumably one assumes it will be linear on those intervals, but it really doesn't matter that much. The main point is that it's zero outside $[0,1]$. – Nate Eldredge Apr 12 '17 at 23:48
@Philo I add the definition of the hat function. I hope it is more clear now. – user251257 Apr 13 '17 at 14:47

Closed infinite dimensional linear subspace of $L^p((0,1))$ consisting of continuous functions

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