Suppose $K$ is a nonempty closed convex set in a Hilbert space $H$. If $x_0+M\subset K$ for some linear subspace $M$ of $H$, prove that $\langle x_0,y\rangle=0$ for all $y\in M$; in other words, $x_0$ is orthogonal to $M$.(Rudin: Fourier Analysis on Groups)
I do not understand why the statement is true. If we set $H=\mathbb R^2$ with the inner product, $M=(\mathbb R,0)$ is the real line and $K=(\mathbb R,1)$, $x_0=(1,1)$, then $x_0+M\subset K$ and $\langle x_0,(1,0)\rangle=1\ne 0$?
Edit:
$x_0$ is supposed to be the unique element in $K$ with the minimal norm...
It says right before the thing you quoted "Each closed convex set $K$ in a Hilbert space $H$ has a unique element $x_0$ of minimal norm".
Claim: Let $K$ be a closed convex set in a Hilbert space. Let $x_0 \in K$ be such that $||x|| \ge ||x_0||$ for all $x \in K$. Let $M$ be a linear subspace of $H$ such that $x_0+M \subseteq K$. Then $\langle x_0,y\rangle = 0$ for all $y \in M$.
Proof: Take $y \in M$. Fix $t \in \mathbb{R}$. Since $ty \in M$, we have $||x_0||^2 \le ||x_0+ty||^2 = ||x_0||^2+t^2||y||^2+2t\langle x_0,y\rangle$. Looking at $t > 0$ gives $\langle x_0,y\rangle \ge \frac{-t}{2}||y||^2$, and looking at $t < 0$ gives $\langle x_0,y\rangle \le \frac{-t}{2}||y||^2$. So $\langle x_0,y\rangle = 0$.