# Doubt in Example realted to dual of $L^{\infty}$ is strictly bigger than $L^1$

I was reading Brezis I stuck at following Example

I do not know how contradication occur. First we take functional of compact supported space then we extend than function to whole $$L^{\infty}$$ this I understand

But I do not understand why to take f(0)=0 and proceed .

Any Help will be appreciated

Of course when it is assumed that $$\displaystyle\int uf=\left<\phi,f\right>=f(0)$$ and $$f(0)=0$$, then $$\displaystyle\int uf=0$$ for all such $$f$$.

The statement that $$\left<\phi,f\right>=0$$ for all $$f\in L^{\infty}$$ entails that $$\left<\phi,f\right>=f(0)=0$$ for $$f\in C_{c}$$ by the extension, this is a contradiction since we can always find $$f\in C_{c}$$ such that $$f(0)\ne 0$$, say, a triangle splines with $$f(0)=1$$.