Closure of set in $C[0,1]$ with $L_1$ norm

Question

Let $C[0,1]$ be endowed with the $L_1$ norm. I am trying to prove/disprove that $S=\{f\in C[0,1]:\;f\left(\frac{1}{2}\right)=0\}$ is closed.

I am pretty sure it is, so I considered a convergent sequence $f_n\in S$ and I am trying to show that $f$ must also be in $S$. Intuitively I can see that we cannot escape $S$ without violating continuity, but I am having trouble putting this into a proof.

If $f(\frac{1}{2})>0$ then we can find $|f(\frac{1}{2})-f(x)|<\epsilon$ for $x$ in an interval around $\frac{1}{2}$, and let $\lambda=\inf |f(\frac{1}{2}-f(x)|$ on that interval. Similarly we can find say $|f_n(x)|<\lambda/2$ when $x$ on an interval $I_n$ around $\frac{1}{2}$. The trouble is I can't see why $f_n$ couldn't "cancel" most of $f$ on the interval so that $||f_n-f||_1<\epsilon$ too.

Could anyone help with this?

Answer 1

$S$ is not closed. Consider $$f_n(x)=\min\left(\left\lvert n\left(x-\frac12\right)\right\rvert,1\right)$$

$f_n\in C[0,1]$ and $$\forall n\ge 2,\quad\int_0^1 \lvert 1-f_n(x)\rvert\,dx =\frac1n$$

So $f_n\to 1$ in $L^1$ norm, but $1\notin S$.

It is actually dense: you can show that, for all $g\in C[0,1]$, $f_n\cdot g\to g$ in $L^1$ norm (recall that $\lVert g\rVert_\infty$ is finite) and $f_n\cdot g\in S$.

Answer 2

Better might be to show that every $f \not \in S$ has an open neighborhood $U_f$ where $U_f \cap S = \varnothing$. Then since the complement of $S$ (the union of all the $U_f$) is open, $S$ is closed. You're approximately there, I'm just not sure you're seeing your sentences as going in this direction.

Answer 3

If $f_n(x) =|x-1/2|^{1/n},$ then each $f_n \in S,$ but $f_n \to 1$ in the $L^1$ norm.