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?

• I hope you don't mind the notation I introduced – Calvin Khor Mar 9 '18 at 18:44
• That's not a norm, because it's not finite. – David C. Ullrich Mar 9 '18 at 19:33
• @DavidC.Ullrich Sorry, I meant $C^2$ with bounded derivatives. Question edited. – user539115 Mar 9 '18 at 19:42
• And of course somehow you say $C^2$ but you don't ask for derivatives in $y$, else it wouldn't be a Banach space for the same reason $C^1([0,1])$ with $C^0([0,1])$ norm is not Banach. – Calvin Khor Mar 9 '18 at 20:02
• @CalvinKhor I edited the definition of the space to make it clearer. – user539115 Mar 9 '18 at 20:45

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.