# Quotient of pre-hilbert space

Let $$H$$ be a pre-hilbert space and let $$F$$ be a complete subspace of $$H$$. Show that $$H/F$$ is a pre-hilbert space.

My attemp:

Consider the following theorem:

$$\it{Theorem}$$: Let $$H$$ a pre-hilbert space and let $$F$$ be a complete subspace of $$H$$, $$F$$ $$\neq$$ $$\emptyset$$ and $$x \in H$$. Then there exist an unique $$f_{0} \in F$$ such that $$||x - f_{0}|| = dist(x, F)$$. This $$f_{0}$$ is characterized by $$x - f_{0} \in F^{\perp}$$.

I defined $$\phi: H/F \to F^{\perp}$$ by $$\phi(\bar{x}) = x - f_{0}$$, where $$f_{0}$$ is given by the the above theorem.

I would like show that $$\phi$$ is linear isometry. I already prove that $$||\phi(x)|| = ||x||$$ and $$\phi$$ is sobrejetive, but i can't show that this mapping is linear.

Hint: if $$x-f_0 \in F^{\perp}$$ and $$y-f_1 \in F^{\perp}$$ with $$f_0,f_1 \in F$$ then $$(ax+by)-(af_0+bf_1) \in F^{\perp}$$ and $$af_0+bf_1 \in F$$ for any scalars $$a$$ and $$b$$. [I have used the elmenetary fact that $$F^{\perp}$$ is a linear space].
• Sorry, but I don't understand. I have $\phi(\bar{x} + \alpha \bar{y})$ = $\phi(\overline{x + \alpha y}) = (x + \alpha y) - f'$. How can i say that $f' = f_{0} + \alpha f_{1}$? Jun 16, 2019 at 23:53
• $f'$ is unique. So if you know that $(ax+by)-(af_0+bf_1) \in F^{\perp}$ and $af_0+bf_1 \in F$ you can conclude that $f'$ must be $af_0+bf_1$. Jun 16, 2019 at 23:55