Strong convexity and Lipschitz What can you say about the L and $\lambda$ for a $\lambda$-strongly convex differentiable function, if its gradient if L-Lipschitz? 
Also, it is given that $\lVert \nabla f(y) - \nabla f(x)\rVert_2 \ge \lambda\lVert y-x\rVert_2 $
 A: Suppose that $f:\mathbb{R}^n\to\mathbb{R}$ is strongly convex with the modulus $\lambda$ and it is differentiable with its derivative satisfying
$$
\textbf{(I)}\quad\quad\|\nabla f(x)-\nabla f(y)\|\leq L\|x-y\|, \quad \forall x,y\in \mathbb{R}^n.
$$
Then, we have $\lambda \leq L$.
Proof.
Step 1. For all $x,y\in \mathbb{R}^n$
$$
\textbf{(II)}\quad\quad f(x)-f(y)\geq \langle\nabla f(y), x-y\rangle+(\lambda/2)\|x-y\|^2.
$$
By the strong convexity of $f$
$$
f(\alpha x+(1-\alpha)y)\leq \alpha f(x)+(1-\alpha)f(y)-\lambda\alpha(1-\alpha)\|x-y\|^2/2.
$$ 
for all $\alpha\in (0,1)$. It implies that
$$
f(x)-f(y)\geq\frac{f(y+\alpha(x-y))-f(y)}{\alpha}+\lambda(1-\alpha)\|x-y\|^2/2.
$$
Letting $\alpha\to 0^+$, we obtain (I).
Step 2. For all $x,y\in \mathbb{R}^n$ 
$$
\textbf{(III)}\quad\quad\langle \nabla f(x)-\nabla f(x), x-y\rangle\geq \lambda \|x-y\|^2.
$$
Applying inequality (II) we deduce
$$
f(x)-f(y)\geq \langle\nabla f(y), x-y\rangle+(\lambda/2)\|x-y\|^2,
$$
$$
f(y)-f(x)\geq \langle\nabla f(x), y-x\rangle+(\lambda/2)\|x-y\|^2.
$$
Adding two inequalities we get (II).
Step 3. $\quad\lambda\leq L$
Choosing $x,y\in\mathbb{R}^n$ such that $x\ne y$. By (I), (III), and the Cauchy Schwarz
$$
\lambda \|x-y\|^2\leq \langle \nabla f(x)-\nabla f(x), x-y\rangle\leq\|\nabla f(x)-\nabla f(x)\|\|x-y\|\leq L\|x-y\|^2.
$$
Since $x\ne y$, this follows that $\lambda\leq L$.
A: Assuming that the definitions are what I think they are, the two requirements are
$$\begin{aligned}
  \|\nabla f(y)-\nabla f(x)\|&\ge\lambda\|y-z\|&&\text{$f$ is $\lambda$-convex}\\
  \|\nabla f(y)-\nabla f(x)\|&\le L\|y-z\|&&\text{$\nabla f$ is $L$-Lipschitz}
\end{aligned}
$$
for all $y$, $z$.
Really, I insist that you draw the inevitable conclusion without my help.
