Using these definitions how to prove $\gamma =\frac{\mu }{L^{2}}$ and $L=\frac{1}{\gamma }$

Question

A mapping $\psi :H\rightarrow H$ is said to strongly monotone if $\exists $ $% \mu >0$ such that

$$ \left\langle \psi \left( u\right) -\psi \left( v\right) ,u-v\right\rangle \geq \mu \left\Vert u-v\right\Vert ^{2} $$

A mapping $\psi :H\rightarrow H$ is said to Lipschitz continuous if $\exists $ $L>0$ such that $$ \left\Vert \psi \left( u\right) -\psi \left( v\right) \right\Vert \leq L\left\Vert u-v\right\Vert $$ A mapping $\psi :H\rightarrow H$ is said to cocoercive if $\exists $ $\gamma >0$ such that $$ \left\langle \psi \left( u\right) -\psi \left( v\right) ,u-v\right\rangle \geq \gamma \left\Vert \psi \left( u\right) -\psi \left( v\right) \right\Vert $$ https://www.sciencedirect.com/science/article/pii/S0022247X13002539

Answer 1

It is easy to follow through these and prove them if:

1. $\mu$ is defined to be the largest such $\mu >0$ such that this holds.$\\$

2. $L$ is defined to be the smallest such $L >0$ such that this holds.$\\$

3. $\gamma$ is defined to be the largest such $\gamma >0$ such that this holds.

This is by noting:

(1) $\left\langle \psi \left( u\right) -\psi \left( v\right) ,u-v\right\rangle \geq \mu \left\Vert u-v\right\Vert ^{2} \geq \left\Vert \psi \left( u\right) -\psi \left( v\right) \right\Vert^2\mu/L^2$

and (2) $\left\langle \psi \left( u\right) -\psi \left( v\right) ,u-v\right\rangle \geq \gamma\left\Vert \psi \left( u\right) -\psi \left( v\right) \right\Vert^2$

In (1), $\mu/L^2$ is the largest constant such that this holds. If it were larger it would contradict the definition of $\mu$ and $L$.

In (2), $\gamma$ is the largest constant such that this holds.

The second statement is the same logic.