Let $H=L_2 [0,1]$ and $G=\{f \in L_2[0,1] : \int_{0}^{1}xf(x)dx=0\}$. Find the Hahn - Banach extension of the functional $L:G \rightarrow H$ defined as $$L(f)=\int_{0}^{1}x^2f(x)dx$$
In the proof of the Hahn - Banach theorem, we obtain some possible interval of values of the extension, but I don't know how to use this fact. Then I tried to explore the set $G$ for a certain property but I couldn't find anything.
I will be grateful for any hints or suggestions.
If $G$ is a closed subspace of a Hilbert space $H$ and $g$ is a continuous linear funnctional on $G$ then the Hahn Banach extension $h$ of $g$ is given by $h(x+y)=g(x)$ where $x \in M$ and $y \in M^{\perp}$. In other words $h$ is $0$ on $M^{\perp}$.
In this case $M^{\perp}= \{cx: c \in F\}$. ($F$ being the scalar field). Can you write down the extension explicitly using these facts? [Any $f$ can be decomopsed as $cx +(f(x)-cx)$ where $c=2\int_0^{1} tf(t)dt$. This makes $f(x)-cx \in M^{\perp}$].