Let $Y$ be a closed subset of a Hilbert space $H$. Let $x \in H, z \in Y$. This equation comes from a proof in Lax' functional analysis book, and the other parts of the proof are not relevant: $$ 2tRe(<v,z>) + t^2||z||^2 \geq 0, $$ where $ <v,x>$ denotes the (complex) scalar product of vectors $v$ and $z$, $||z||$ is the induced norm, and $t \in \mathbf{R}$.
The conclusion from this equation is that $Re(<v,z>) =0,\forall z \in Y$.
I don't see how we can have that conclusion.
$\newcommand{\R}{\mathbb{R}}$It is due to the following fact:
Let $a\ge 0$ and $b\in\mathbb{R}$. If $\color{red}{b\ne 0}$, then there exists $t\in \R$ such that $bt+at^2 < 0$.
From this, you know (by taking contrapositive) that if $a\ge 0$ and $bt+at^2 \ge 0$ for all $t\in \R$ (where $b\in \R$), then $\color{blue}{b=0}$.
To prove that fact above, a hint is to note that $bt+at^2 = t(b+at)$. If $a\ne 0$, you should be able to find the two roots of this quadratic and show that it is negative in between (e.g. at the midpoint of them). If $a=0$, you should be able to show the result quite easily.
Using that limits preserve non-strict inequalities: $$ t>0\implies 2\Re(<v,z>)+t\|z\|^2\ge 0\implies 2\Re(\langle v,z\rangle) = \lim_{t\to 0^+}2\Re(\langle v,z\rangle)+t\|z\|^2\ge 0 $$ $$ t<0\implies 2\Re(<v,z>)+t\|z\|^2\le 0\implies 2\Re(\langle v,z\rangle) = \lim_{t\to 0^-}2\Re(\langle v,z\rangle)+t\|z\|^2\le 0 $$
