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