Prove that $C_b^{2,1}(\mathbb{R}^n \times [0,T])$ is a Banach Space with a weighted maximum norm On the space $C_b^{2,1}(\mathbb{R}^n \times [0,T])$, which is the space of bounded continuous functions with two continuous bounded derivatives in the first variable and (if we can do it) continuous derivative in the second variable, define the weighted sup norm,
$$\|f\|_c:=  \sup\{|f(x,y)e^{cy}|: x \in \mathbb{R}^n, y \in [0,T]\}$$
then define the  norm
$$\Vert f \Vert =  \|f\|_c + \|\nabla f\|_c + \|D^2f\|_c $$ 
or written in full,

$$\|f\| = \sup\{|f(x,y)|e^{cy}: x \in \mathbb{R}^n, y \in [0,T]\} + \sup\{|\nabla f(x,y)|e^{cy}: x \in \mathbb{R}^n, y \in [0,T]\} + \sup\{|D^2 f(x,y)|e^{cy}: x \in \mathbb{R}^n, y \in [0,T]\},$$ 

$c \in \mathbb{R},$ where $\nabla$ is the gradient in the first variable and $D^2$ is the Hessian in the first variable. 
How can I prove that this is a norm and that the space is a Banach space with this norm?
The question is inspired by Prove that $C^1$ is a Banach space with respect to a norm and that it is equivalent to the standard $C^1$ norm
 A: There is no extra work to be done after the answer in the linked question. I quote:

Moreover, since $e^{cy}$ is bounded above and below by positive constants on $[0,T]$, it is equivalent to $\|\cdot\|_{C^1}$ and hence $(\mathcal X,\|\cdot\|)$ is a Banach space. This is again rather obvious, and you can replace $e^{cy}$ with any positive continuous function which is bounded above and below by positive constants.

so since 
$ C^{2,0}_b := \{ f \in C^{2,0}(\mathbb R^n \times [0,T]) : \|f\|_{C^0} + \|\nabla f\|_{C^0} + \|D^2f\|_{C^0}<\infty\}$ is a Banach space with norm $$\|f\|_{C^{2,0}}:=\|f\|_{C^0} + \|\nabla f\|_{C^0} + \|D^2f\|_{C^0}$$
and $\|\cdot\|$ is a norm on this space, the implied inequalities from the quote tell us that this is an equivalent norm for $C^{2,0}_b$.
If you ask for a derivative in $y$, then you're gonna need a different norm. This is also spelled out in the answer you linked to, which I suggest you fully understand first and then come back to this, because its much the same with less notation.
