Let $X$ be a Hilbert space with $\langle\cdot , \cdot \rangle$, inner product and induced norm $\|\cdot \|$. Denote the unit open ball of $X$ as $B%$. Suppose $z\colon\mathbb{R}^+\to B$ is a possibly discontinuous function defined for all nonnegative time. Suppose further that $x\colon \mathbb{R}^+\cup\{0\}\to X$ is a continuous and differential function such that $\langle x_t(t), x(t)-z(t)\rangle\le 0$ for all time. If I know that $x(0) = 0$ can I conclude that $x(t)\in B$ for all nonnegative $t$? I think I it is true, but I can't prove it to myself.
This is what I have tried. Suppose not, then from the initial condition and continuity there exists a minimum time $t^*\in \mathbb{R}^+$ for which $\|x(t^*)\|=1$ and $\frac{\mathrm{d}}{\mathrm{d}t}\| x\|>0 $, we mean $x$ is crossing the boundary of the unit ball from inside to outside for the first time. Note that from the inner product inequality we have $$ \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \|x\|^2 \le \langle x_t,z\rangle $$
If I could show that $\langle x_t,z\rangle$ is not positive, I would have a contradiction, but I haven't been able to show this. Note that I haven't used that $z(t^*)\in B$,