There is a function $ \phi (x): \mathbb{R}^n \longmapsto \mathbb{R}^n$ which satisfies conditions (1) and (2).
(1): $\langle\phi(x_1)-\phi(x_2),x_1-x_2\rangle \leq \gamma_1 \| x_1-x_2 \|^2$
(2): $ (\phi(x_1)-\phi(x_2))^T (\phi(x_1)-\phi(x_2)) \leq \beta \,\|x_1-x_2\|^2+\rho \langle\phi(x_1)-\phi(x_2),x_1-x_2\rangle$, where $\beta$ and $\rho$ are any real numbers.
$x_1$, and $x_2 \in \mathbb{R}^n $ and are $\in L\infty$.
$\langle \cdot,\cdot \rangle$ is inner product, and $\| \cdot \|$ is the 2-norm. $T$= transpose
What can we say about boundedness($\in L\infty$) and uniformly continuity of $\phi$ ?
What if $\phi:\mathbb{R} \longmapsto \mathbb{R}$ and $x_1, x_2 \in\mathbb{R}$ ?
Here is my solution= I know that since in the worst case, the right-hand side of (2) can be $\infty$, by letting $\phi $=$\infty$ in (2) we have $\infty\le\infty$ which is a mathematical contradiction. Hence $\phi $ should be bounded.is it correct?
about uniform continuity: if $\phi:\mathbb{R} \longmapsto \mathbb{R}$ and $x_1, x_2 \in\mathbb{R}$ from (1) we can write (3)=$(\phi(x_1)-\phi(x_2))\over (x_1-x_2)$$\le \gamma_1$, then since $\phi$ proved to be bounded (in previous part) and $x_1$, $x_2$ are bounded, Hence (3) is bounded. in fact we proved that derivative of $\phi$ is bounded, which is a sufficient for uniform continuity. is it correct?what if $\phi$ and $x_1-x_2$ were vectors?