Closed subspaces of $L^2(0,1)$

Question

I would like to prove that the almost-everywhere constant functions, and the functions whose integral is 0 are closed subspaces of $L^2(0,1)$. It's readily seen that they are subspaces. I'm finding some difficulties trying to show that they are closed.

For example, with the constants:

I take a convergent sequence of almost-everywhere constant functions $f_n$. It converges to a function $y$. I know that for each $\epsilon>0$ i can find a $n_\epsilon$ such that for $n>n_\epsilon$ I have

$\int\limits_{0}^{1}|f_n(x)-y(x)|^2 dx<\epsilon$

Well It's obvious that $y$ is a constant a.e. but how can I show it formally? Same problem for the convergent sequences of functions $u_n$ such that $\int\limits_{0}^{1}u_n(x)dx=0.$

Any help is appreciated.

Answer 1

1. For the constants: prove that the subspace is complete. In an Hilbert space, this is equivalent to being closed, and in this case it is easier.
2. For the null integrals subspace: show that $\int f_n\to \int f$.

Answer 2

For the first problem, use that any $L^2$-convergent sequence has a subsequence that converges almost everywhere.

Alternatively, you can use that any finite-dimensional subspace of any normed vector space is closed. This essentially follows (in your case), because

$$\Bbb{R} \rightarrow \Bbb{R}_+, x \mapsto \Vert x \cdot \chi_{[0,1]} \Vert_{L^2}$$

gives a (necessarily equivalent) norm on $\Bbb{R}$.

For the second part, use that the functional

$$L^2 \rightarrow \Bbb{R}, f \mapsto \int_0^1 f(x) \,dx$$

is bounded and hence continuous (use Cauchy-Schwarz with the constant $1$ function), so that the kernel is a closed subspace.