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.
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)
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".
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!