Inner Product Space $X$ with $x,y \in X$, ${\lVert y \rVert} = 1$, solve for $\lambda \in \mathbb{C}$ that minimizes ${\lVert x - \lambda y \rVert}$

Question

Inner Product Space $X$ with $x,y \in X$, ${\lVert y \rVert} = 1$, solve for $\lambda \in \mathbb{C}$ that minimizes ${\lVert x - \lambda y \rVert}$

\begin{align*} {\lVert x - \lambda y \rVert}^2 &= {\lVert x \rVert}^2 - \lambda {\langle x,y \rangle} - \overline{\lambda {\langle x,y \rangle}} + \lambda \overline{\lambda} \\ \end{align*}

I believe this is minimized when $(x - \lambda y) \perp y$ or $\langle x - \lambda y, y \rangle = 0$, which is $\lambda = {\langle y,x \rangle} = \overline{\langle x,y \rangle}$. How can I demonstrate that this value will minimize ${\lVert x - \lambda y \rVert}^2$?

Answer 1

Tracing through Kreyszig might lead you to Lemma 3.3-2, which says that if $Y$ is a complete subspace of an inner product space $X$, then for any fixed $x\in X$ there is a unique $y\in Y$ such that $\|x-y\|=\inf_{z\in Y}\|x-z\|$, and that $x-y$ is orthogonal to $Y$. For your specific case you can take $Y=\operatorname{span}\{y\}$; then you get from the lemma that the $\lambda$ minimizing $\|x-\lambda y\|$ satisfies $$\langle x-\lambda y,\eta y\rangle=\overline\eta\langle x,y\rangle-\lambda\overline\eta\langle y,y\rangle=0$$ for all $\eta\in\mathbb{C}$, so that $\langle x,y\rangle=\lambda\langle y,y\rangle$. It follows that $\lambda=\langle x,y\rangle/\langle y,y\rangle$ is the minimizer. Since $\|y\|=1$, your conjecture of $\lambda=\langle x,y\rangle$ is correct.