Projection in $\ell^2 (\mathbb{N})$

Question

Let $X=\ell^2(\mathbb{N})$ and $C=\left\{x=(x_k) \in X: x_0+x_1+x_2=1\right\}$. I have prooved that $C$ is a non-empty, convex and closed subsect of $X$. Now I want to find the projection of the element $0=(0,0,\dots) \in \ell^2(\mathbb{N})$. Can you help me?

Answer 1

Since the projection in Hilbert space gives the nearest point, you are looking to minimize the quantity $x_0^2+x_1^2+x_2^2$ subject to the condition $x_0+x_1+x_2=1$, and it is not too hard to prove that the minimum occurs when $x_0=x_1=x_2=1/3$.

Answer 2

Any element of $C$ can be written as $$x = x_0, x_1, 1-x_0-x_1,x_3, \dots$$ with $x \in \ell^2(\mathbb N)$.

You now have to find the minimum of $$x_0^2+x_1^2 + (1-x_0-x_1)^2 + x_3^2 + \dots.$$

Obviously you need to have $0=x_3=x_4= \dots$.

And you're left to find the minimum of $$f(x,y)=x^2+y^2+(1-x-y)^2$$ which is obtained for $x=y=1/3$ and therefore the projection is the sequence $a=1/3,1/3,1/3,0,0,\dots$.

Answer 3

You can write $C$ as $$C=\{x\in X: x\cdot y=1\}$$ where $y=(1,1,1,0,0,\ldots)$. Therefore the point of minimum distance will be a multiple of $y$, and the only point satisfying the constraint is $(\tfrac13,\tfrac13,\tfrac13,0,0,\ldots)$.