# Distance to a vector subspace

Let E be the set of real sequences $$(x_n)_{n\in N}$$ such that the series $$\sum u_{n}^2$$ converges.

For $$x=(x_n)_{n \in N}$$and $$y=(y_n)_{n \in N}$$

Inner product :

(x|y) = $$\sum_{n=0}^{+\infty} x_ny_n$$

Let F be the vector subspace of E formed from zero sequences starting from a certain rank.

If $$z=(\frac{1}{n+1})_{n\in N}$$, show that $$z \in E$$and calculate $$d(z,F)$$

Showing that $$z\in E$$ is easy with Riemann. I tried to find an orthogonal base of F to calculate $$d(z,F)$$ but I couldn’t find one. Is there another way to do the calculus of a distance without using an orthogonal projection? Here the spaces considered have infinite dimension so there has to be another way !

• @mathcounterexamples.net I am struggling with Latex as I am new to it and i still don’t know how to make a proper sum and can’t find help anywhere to achieve it. Jan 24 at 17:43
• @mathcounterexamples.net Should I repost my question as it had been downvoted ? Jan 24 at 17:53
• To what kind of Riemann theorem are you referring to? Jan 24 at 19:25
• @mathcounterexamples.net there was a mistake in what i wrote. It should be clearer now Jan 24 at 19:27

The given space $$E$$ is the Hilbert space $$\ell^2(\mathbb R)$$. The space $$F$$ of eventually vanishing sequences is usually denoted by $$c_{00}$$.
$$c_{00}$$ is dense in $$\ell^2(\mathbb R)$$. This is quite easy to prove. Hence $$d(x,F)=0$$ for any $$x \in E$$.
• Could you give a hint to prove that $c_{00}$ is dense in $\ell^2(\mathbb R)$ ? Jan 24 at 20:20
• Look at the distance between an element $x \in E$ and the same element with zeros starting at rank $n$. Jan 24 at 20:26
• For such a sequence of elements of $c00$, the distance intuitively tends to 0, thus proving that $x$ is adherent to $c00$ and that $c00$ is dense in $\ell^2(\mathbb R)$. I just don’t know how to formulate clearly with the inner product that is given Jan 24 at 20:33