In Hilbert space $L_{2}(0,1)$ we have linear functional $f:L_{2}(0,1)\rightarrow \mathbb{R}$ such as $f(x)=\int_{0}^{1}t^{2}[x(t)-x(1-t)]dt$ for any $x\in L_{2}(0,1)$.
First of all, I want to find $Kerf$ but I don't really know where to start. After that I have to compute $d(t^{2},Kerf)$ which would be easier if I knew the $Kerf$ representation. Definition that I know is:
$Kerf:=\{x\in X: f(x)=0\}$
After that, I want to check the norm of linear functional $||f||$. What I'm supposed to do here? I don't know how exactly should look like $||f||=\sup_{x\neq0}\frac{|f(x)|}{||x||}$.
Hints: I will assume that you are familiar with some basic facts about Hilbert spaces. The first step here is to simplify the definition of $f$. We have $f(x)=\int_0^{1}t^{2} x(t)dt-\int_0^{1}t^{2}x(1-t)dt$. Put $s=1-t$ in the second term (and the write $t$ for $s$) to get $f(x)=\int_0^{1}t^{2} x(t)dt-\int_0^{1} (1-t)^{2} x(t)dt$. By expanding $(1-t)^{2}$ we get $f(x)=\int_0^{1}g(t)x(t)dt$ where $g(t)=2t-1$. Hence $f(x)=0$ iff $\int_0^{1}g(t)x(t)dt=0$ which means that $ker (f)$ is the orthogonal complement of $g$. It is easy to show that the distance from any point $h$ to the orthogonal complement of $g$ is $\frac {|\langle g , h \rangle|} {\|g\|}$. You can now compute this distance from $t^{2}$ to $ker (f)$ using this formula.
Finding the norm of $f$ is very easy. We have $f(x) =\langle x , g \rangle $ and the norm of this is just $\|g\|$. Hence $\|f\|=\sqrt {\int_0^{1} (2t-1)^{2}dt}$.