Let $H$ be Hilbert space and $\left(\cdot,\cdot\right)$ be it's inner product.
And let $S$ be a closed subspace in $H$.
If there is a closest element $g_{0}$ in $S$ to $f\in H$, I want to prove $\left(f-g_{0},g\right)=0$ for any $g\in S$.
My attempt : By assumption, $\|f-g_{0}\|\leq\|f-g\|$ for all $g\in S$. So $$ \|f-g_{0}\|^{2}\leq\|f-g\|^{2}. $$ And I expand this. But there is nothing I can obtain. Anyone can help me?
On
By a well known theorem there are $P:H\to S$, $Q\to S^{\perp}$ linear maps such that for all $x\in H,$ $x=Px+Qx$ (unique decomposition) and $Px$, $Qx$ the closet elements to $S$ and $S^{\perp}$ respectively. In our case, $$f=Pf+Qf=g_0+\underbrace{Qf}_{\in S^{\perp}}\Rightarrow f-g_0\in S^{\perp}\Rightarrow\langle f-g_0,g\rangle=0,\quad \forall g\in S.$$
On
Ok from what you wrote I guess we are supposing that $g_0$ is existing. If such an element exists, then let's take $y$ in S, and $\theta$ in $\mathbb{R}$.
Then, $\theta y + g_0$ is an element of the subspace S, and it's further from f than $g_0$ (this is how we defined $g_0$ after all). So, $$||f-g_0||^2 \le ||f-g_0-\theta y||^2$$
By rewriting RHS:
$$||f-g_0||^2 \le ||f-g_0||^2 - 2 \theta(f-g_0,y)+\theta^2||y||^2$$
Then you only have to subtract $||f-g_0||^2$ from each side. You then have:
$$\theta(f-g_0, y)+\theta^2||y||^2\ge 0$$ and this for any $\theta$. The only way to ensure this is to have $(f-g_0,y)=0$ for any y in S.
Well, anoher way from your starting point. Denote $z=f-g_0$. Without loss of generality we can suppose that $ \left\|{g}\right\|=1$. We have $$\langle z,z\rangle=\left\|{z}\right\|^2\le\left\|{z-\alpha g}\right\|^2=\langle z-\alpha g,z-\alpha g\rangle\text{ for all }\alpha \text{ scalar}. $$ Symplyfing, $0\le -\alpha \langle g,z \rangle-\bar{\alpha}\langle z,g \rangle+\alpha\bar{\alpha}$. Now, for $\alpha=\langle z,g \rangle$ we get $0\le -\left |\langle z,g\rangle\right |^2$, so $$\langle z,g\rangle=\langle f-g_0,g\rangle=0.$$