Inequality of a linear operator on Hilbert space.

Question

Let $T \colon \mathcal H \to \mathcal H$ be a linear operator and let $x,y \in \mathcal H$. We assume that

$$ \forall \ z \in \mathcal H, \ \left \langle y-Tz,x-z \right \rangle \ge 0. $$ Show that $Tx=y$.

Any hints?

Answer 1

Let $z=x-th$, where $t$ is real, then $\langle y-Tx,h \rangle + t \langle Th, h \rangle \ge 0 $ for $h$ and $t$.

In particular, $\langle y-Tx,h \rangle \ge 0$ for all $h$ and choosing $h=y-Tx$ we get $\|y-Tx\|^2 = 0$ or $Tx=y$.

Answer 2

Hint: Fix $h \in \mathcal{H} \setminus \{0\}$ and consider the real quadratic, \begin{align*} \lambda &\mapsto \langle y - T(x + \lambda h), x - (x + \lambda h)\rangle \\ &= \lambda(\langle Th, h \rangle \lambda - \langle y - Tx, h\rangle). \end{align*} This quadratic achieves its minimum at $\lambda = 0$. What does this tell you?