Why are the continuous functions not dense in $L^\infty$?

I mean both concretely (i.e. a counter example) and intuitively why is this the case.


Consider $$f(x)=\begin{cases}0&\text{if }x<0\\1&\text{if }x\ge 0.\end{cases}$$ Any continuous $g$ with $\|f-g\|_\infty<\frac 13$ must have $g(x)<f(x)+\frac13=\frac13$ for all $x<0$. By continuity, $g(0)\le \frac13$, contradicting $g(0)>f(0)-\frac13=\frac 23$.

Even if we only require $|f(x)-g(x)|<\frac13$ for almost all $x$, the argument above still holds (with using continuity on the right as well).

Intuitively, the continuous $g$ cannot do the jump at once, it needs some "preparation" and "relaxation".

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    $\begingroup$ In other words, the topology on $L^{\infty}$ is strictly finer than its restriction to $C(\mathbb{R},\mathbb{R})$? $\endgroup$ – AIM_BLB Mar 18 '20 at 15:37

If a convergent sequence of continuous functions converges uniformly then the limiting function is continuous. Consequently, the only functions which can be approximated to arbitrary accuracy in $L^\infty$ by continuous functions are the continuous functions. In other words, the subspace of $L^\infty$ consisting of continuous functions is closed.

Of course, there are many functions in $L^\infty$ which are not continuous!


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