Is the map $x \mapsto f(x-\frac{1}{L} \nabla f(x) ) $ coercive provided $f \in C^{\infty}(\mathbb R^n)$ is coercive? Suppose $f: \mathbb R^n \to \mathbb R$ is $C^{\infty}$ smooth and the gradient map is $L$-Lipschitz, i.e., $\|\nabla f(x) - \nabla f(y)\|_2 \le L\|x-y\|_2$. We say a function is coercive if $f(x) \to +\infty$ as $\|x\| \to \infty$. This is equivalent to say that all sublevel sets $\{x: f(x) \le \alpha\}$ is compact for every $\alpha > 0$. I am wondering whether we can prove that the map $F: \mathbb R^n \to \mathbb R$ given by
$$ x \mapsto f\left( x -\frac{1}{L} \nabla f(x) \right)$$
is coercive if $f$ is coercive. The map can be understood as: at any point $x$, we take a gradient descent step, then evaluate the function. 

The answer is no. Take $n=1$ and $f(x)=x^2$. Clearly this is smooth and $\nabla f(x) = 2x$ is two lipschitz. However,
$$ F(x) = f\Big(x- \frac{1}{2} \nabla f(x)\Big) = f(0) $$
is constant and thus not coercive. In fact we can also add some function $\varphi$ is it does not change the lipschitz constant and if it has bounded gradient (then $x- \frac{1}{2} \nabla (f + \varphi)(x)= \frac{1}{2} \nabla \varphi(x)$ remains bounded).
In general you can play the same game by taking $f(x) = \Vert x \Vert_2^2$ and adding some smooth function $\varphi$ which does not change the lipschitz constant of the gradient and such that $\nabla \varphi$ is bounded. Then you get
$$ F(x) = f\big(\nabla \varphi(x)\big). $$
