Is infimum of $\left\{\dfrac{1}{2} \sum_{n=1}^\infty \alpha_n\left( \dfrac{\langle u,e_n\rangle}{\alpha_n}-\langle x,e_n\rangle \right)^2 \right\}=0$?

Question

Let $H$ be a Hilbert space, $\{e_n\}_{n=1}^\infty$ be an orthonormal basis for $H$ and $\{\alpha_n\}_{n=1}^\infty$ be a sequence in $(0,1)$. Let $u\in H$. Are we sure that $$\inf_{x\in H} \left\{\dfrac{1}{2} \sum_{n=1}^\infty \alpha_n \left( \dfrac{\langle u,e_n\rangle}{\alpha_n} - \langle x,e_n\rangle \right)^2 \right\} = 0?$$

We know that the term inside the sum is always nonnegative. Thus, $0$ is a lower bound. But can we be sure that it will not take some positive value? Why or why not? Thanks!

Answer 1

Represent $u$ in Hilbert space's base $\{e_i\}$ as $$u = \sum_{i} u_i e_i.$$

Take $ x $ as $$x = \sum_{i} \frac{u_i}{a_i} e_i.$$

We find $$\frac{\langle u, e_n\rangle}{a_n} = \frac{u_n}{a_n} = \langle x, e_n\rangle.$$

So chosen $x$ is the infimum point.