Let $L^2[-1, 1]$ be the Hilbert space of real valued square integrable functions on $[-1, 1]$ equipped with the norm $$\|f\|_2=\big(\int_{-1}^1|f(x)|^2\, dx\big)^{\frac{1}{2}}.$$ Consider the subspace $M=\{f\in L^2[-1, 1]\ |\ \int_{-1}^1f(x)\, dx=0\}.$
For $f(x)=x^2$, define $d=\inf\{\|f-g\|:g\in M\}$. Then find $d$
My effort: I note that $d=d(f, M)$, i.e. distance of $f$ from $M$.
Let $g\in M^{\bot}$ (orthogonal complement of $M$). This means that $\langle g, f\rangle=0$ for all $f\in M$. That is $$\int_{-1}^1 f(x)g(x)\, dx=0.$$ Since we know that $\int_{-1}^1f(x)\, dx=0$, we conclude that $g(x)=c$ almost everywhere.
Let $g(x)\equiv h=1$ which means $M^{\bot}=\{r:r\in \mathbb{R}\}$. Now, we use the result that the distance of an element $f$ of a Hilbert space from a subspace $M$ is its distance from its perpendicular projection onto that subspace.
In our case, the perpendicular projection of $ f$ onto $M$ is $$ f-\frac{\langle f,h\rangle}{\|h\|^2} h\ , \ \text{and}\ d(f,M)=\left|\left\langle f,\frac{h} {\|h\|}\right\rangle\right|\ . $$ In our case, we have $ h(t)\equiv 1 $, for which $ \|h\|\ =1$. Therefore, $d=d(f,M)=\int_{-1}^1x^2=\frac23.$ But the answer is $\frac{\sqrt{2}}{3}$.
Please tell me where I am wrong.
I think $\|h\|=1$ is not correct: $$\|h\| = \sqrt{\int_{-1}^1 h^2\, dt} = \sqrt{\int_{-1}^1 1\, dt} = \sqrt{2}.$$ If you plug this in, the expected result follows.