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Consider the set

$E := \{f \in C[a,b] , \int_a^b f(t) dt = 0 \}$

Show that E is a closed linear subspace of $C[a,b]$.

I know that a subset $E$ is closed if $\overline{E} \subseteq E$ (where $\overline{E}$ is the closure of $E$).

I don't know how I can prove this - any ideas where to start?

Mittens
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binaryBigInt
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    The set $E$ pretty much hints at a solution. – AlvinL Dec 31 '22 at 15:22
  • Presumably $C([a,b])$ is equipped with the uniform norm. If that is so, then things follows by taking a sequence $f_n\in E$ that converges (in uniform norm to $f$) – Mittens Dec 31 '22 at 17:06

1 Answers1

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Whether $E$ is closed or not depends on the norm on $C[a,b]$. (Ofcourse to talk about closedness you need to specify the topology on that space . If you view it as a normed space then you need to specify the norm on it)

For example if it is the sup norm $||\cdot||_{\infty}$ ,then if $f_{n}$ be a sequence in $E$ such that $||f_{n}-f||_{\infty}\to 0$ Then $f_{n}$ converges to $f$ uniformly and hence $$0=\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dx =\int_{a}^{b}\lim_{n\to\infty}f_{n}(x)\,dx =\int_{a}^{b}f(x)\,dx$$

Hence $f\in E$ and hence $E$ is closed.

But with some other norm say $L^{1}$ , $E$ need not be closed in $L^1[a,b]$ which is the completion of $C[a,b]$ with the $L^{1}$ norm. This is not to be confused with the fact that $E$ is closed in $C[a,b]$ equipped with the $L^{1}$ norm (i.e. wrt subspace topology).

Let $$f_{n}=\begin{cases} 1\,, x\in [\frac{1}{n},1]\\ -1 \,, x\in [-1,-\frac{1}{n}]\\ nx\,, x\in[-\frac{1}{n},\frac{1}{n}]\end{cases}$$

Then $\int_{-1}^{1}f_{n}\,dx=0$ and

$$\int_{-1}^{1}|f_{n}(x)-\text{sgn}(x)|\,dx =\int_{\frac{-1}{n}}^{\frac{1}{n}}|nx-\text{sgn}(x)|=\frac{1}{n}\to 0 $$ .

and hence $f_{n}\xrightarrow{L^{1}} \text{sgn}(x)$.

But quite clearly $\text{sgn}(x)$ does not even lie in $C[a,b]$ let alone in $E$.

So $E$ (and also $C[a,b]$) cannot be closed in $L^1[a,b]$).

But $E$ will always be closed as a subspace of $C[a,b]$ for any norm on $C[a,b]$ which makes the linear functional $T:C[a,b]\to \Bbb{R}$ such that $T(f)=\int_{a}^{b}f(x)\,dx $ continuous as then $E$ is the kernel .

So in particular $E$ is closed in $(C[a,b],||\cdot||_{L^{1}})$ as $|T(f)|=|\int_{a}^{b}f\,dx|\leq (b-a)||f||_{L^{1}}$ and hence $T$ is continuous.

Mr.Gandalf Sauron
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  • +1. With the $\sup$ norm we can also observe that the function $\psi(f)=\int_a^b f(x)dx$ is continuous because it is Lipschitz-continuous because $|\psi(f)-\psi(g)|\le |b-a|\cdot |f-g|$. So $E=\psi^{-1}{0}$ is closed because ${0}$ is closed. – DanielWainfleet Dec 31 '22 at 17:20
  • @DanielWainfleet I have added that at the end of my answer. – Mr.Gandalf Sauron Dec 31 '22 at 17:23
  • I find the second example slightly misleading since $E$ is closed in $C[a,b]$ w.r.t. $L^1$-norm, and indeed any norm dominated by $|\cdot|_\infty$. Nowhere did you claim otherwise but still... – mechanodroid Dec 31 '22 at 18:19
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    @mechanodroid Okay I agree. Maybe I should make it more prominent that I am talking about closedness in the complete space $L^{1}$ rather than with respect to the subspace topology $C[a,b]$ . – Mr.Gandalf Sauron Dec 31 '22 at 18:25
  • @Mr.GandalfSauron Yeah, it confused me at first. Interestingly, I find it actually quite difficult to find an explicit example of a norm on $C[a,b]$ such that $E$ is not closed. It is possible that the axiom of choice is required to construct such a norm. – mechanodroid Dec 31 '22 at 18:28
  • I think it is better to think about a norm which makes $f\mapsto \int_{a}^{b}f$ an unbounded/discontinuous functional . Ofcourse the kernel $E$ is closed iff the above map is continuous. – Mr.Gandalf Sauron Dec 31 '22 at 18:39
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    As I suspected, you won't be able to explicitly state a norm on $C[a,b]$ such that this functional is discontinuous (I just asked a question about it: https://math.stackexchange.com/q/4609350/144766). – mechanodroid Jan 01 '23 at 13:54