Operator $Tf=f(0)$ in $L^\infty[0,1]$

Question

This question for $$T:C[0,1]\to \mathbb{R}$$ such that $Tf = f(0)$ is treated here, where we conclude the operator is bounded. Now, I read this doesn't hold if $T:L^\infty[0,1]\to\mathbb{R}$ (either the operator is not linear or is not bounded, I am not sure).

I was wondering how to prove that. Since $||f||_\infty$ is defined as the smallest $M$ such that $|f(x)|\le M$ a.e is sufficient to consider a function such that $f(0) = \infty$ and $f(x) = 0$ for $x\in(0,1]$ ? or in general, a function that take big values in a set of zero measure (something like a family $f_n$ such that $f_n(0)=n$). Thanks in advance!

Answer 1

This is not well defined function so its continuity does not make sense. Elements of $L^{\infty}$ are equivalence classes of functions so changing the value at $0$ does not change the element.

If $f(x)=0$ for all $x$ and $g(x)=0$ for $x \neq 0, g(0)=1$ then $f=g$ as elements of $L^{\infty}$ but they have different values at $0$.

Based on a comment below I will show discontinuity of $T$ when viewed as a map from $C[0,1]$ with the $L^{2}$ norm. Let $f_n(x)=1-nx$ for $0 \leq x \leq \frac 1 n$ and $0$ for $x >\frac 1 n$. Then $f_n \to 0$ in $L^{2}$ norm but $T(f_n)=1$ for all $n$. However, $L^{\infty}$ norm on $C[0,1]$ is same as the usual sup norm so $T$ is continuous for this norm.