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

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I was reading Brezis I stuck at following Example enter image description here

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 .

Please can any one give me idea about proof

Any Help will be appreciated

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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 $f(0)=1$.