We have $X=\{f\in L_{2}[0,1]:\int_{0}^{1}f(t)dt=0\}$ - subspace of Hilbert space. I have managed to find orthogonal complement of X which is:
$X^{\perp}=\{g\in L_{2}[0,1]: g \textrm{ is constant a.e.}\}$
Now I have to find the distance:
a) $d(t^{2},X)$
b) $d(t^{2},X^{\perp})$
For part a) I tried this aproach:
$||t^2-f(t)||^{2}=\int^{1}_{0}|t^2-f(t)|^{2}dt \geq \int^{1}_{0}(t^2-|f(t)|)^{2}dt=\int^{1}_{0}(\frac{1}{3}+[t^2-\frac{1}{3}-|f(t)|])^{2}dt=\int^{1}_{0}(\frac{1}{3})^2dt+2\int^{1}_{0}(t^2-\frac{1}{3}-|f(t)|)dt+\int^{1}_{0}(t^2-\frac{1}{3}-|f(t)|)^{2}dt$
Second integral seems fine for me, if $\int fdt=0 $ then $ \int|f|dt=0$ which I need for calculations.
But in the third one I don't know how to integrate for example $t^{2}*|f(t)|$. I'm stuck at this part. Maybe there is another way to calculate the distance between function and subspace? ($d(x,L):=inf_{y\in L}||x-y||$). I'm not so good at functional-analysis so I would appreciate for some explanations.
For part b) we should just find the infimum of $||t^{2}-c||$, since $g(t)=c$ is constant?
The distance of an element $\ f\ $ of a Hilbert space $\ \mathbb{H}\ $ from a subspace $\ X\ $ is its distance from its perpendicular projection onto that subspace. When, as in case a) here, $\ X^\perp = \{c h\,|\, c\in \mathbb{C}\}\ $, with $\ 0\ne h\in \mathbb{H}\ $, is one-dimensional, the perpendicular projection of $\ f\ $ onto $\ X\ $ is $$ f-\frac{\langle f,h\rangle}{\|h\|^2} h\ , $$ and $$ d(f,X)=\left|\left\langle f,\frac{h} {\|h\|}\right\rangle\right|\ . $$ In your case you can choose $\ h(t)\equiv 1\ $, for which $\ \|h\|\ =1\ $.
Your proposed method for solving b) should work, but you could also use the same idea as given for part a) above, the perpendicular projection of $\ f\ $ onto $\ X^\perp\ $ being $\ \frac{\langle f,h\rangle}{\|h\|^2} h\ $ and $\ d(f,X^\perp)=\sqrt{\|f\|^2- \left\langle f,\frac{h} {\|h\|}\right\rangle^2}\ $.