On the space $C^1(\mathbb{R}^n \times [0,\infty))$, define
$$\Vert f \Vert = \sup\{|f(x,y)|e^{cy}: x \in \mathbb{R}^n, y \in [0,\infty)\} + \sup\{|\nabla f(x,y)|e^{cy}: x \in \mathbb{R}^n, y \in [0,\infty)\},$$ $c \in \mathbb{R},$ where $\nabla$ is the gradient in the first variable.
How can I prove the following?
(1) This is a norm,
(2) the space is a Banach space with this norm,
(3) this norm is equivalent to the standard norm $$\Vert f\Vert_{C^1} = \sup\{|f(x,y)|: x \in \mathbb{R}^n, y \in [0,\infty)\} + \sup\{|\nabla f(x,y)|: x \in \mathbb{R}^n, y \in [0,\infty)\}.$$
Best Answer
First problem: typically by $C^1(\mathbb R^n\times[0,\infty))$ we mean the space of continuously differentiable functions on $\mathbb R^n\times[0,\infty)$, with no assumption of boundedness. Clearly what you call "the standard norm" is not a norm on this space.
Even if we assume that the space consists of bounded functions with bounded derivatives, it's not a Banach space. Intuitively this is because the $y$-derivative is not involved in the norm. For an explicit counterexample, define
$$\psi_n(t):=\begin{cases} |t|,&|t|>\frac1n,\\ \frac n2t^2+\frac1{2n},&|t|\le\frac1n \end{cases}$$ and let $f_n(x,y)=\varphi(y)\psi_n(1-y)$, where $\varphi$ is a smooth function such that $\varphi=1$ on $[0,2]$ and $\varphi=0$ on $[3,\infty)$. Note that $\psi_n$ is continuously differentiable with $\psi_n(t)\to|t|$ uniformly, so $f_n(x,y)\to f(x,y):=\varphi(y)|1-y|$ uniformly and $\nabla f_n\equiv0$, which implies $\{f_n\}$ is Cauchy. But if $f_n\to g$ in $\|\cdot\|_{C^1}$ then also $f_n\to g$ uniformly, so $g=f\notin C^1$.
So then it seems like we are interested in the space of functions $f(x,y)$ which are bounded, continuous, and have bounded and continuous $x$-derivative $\nabla f(x,y)$. This space looks nothing like what anyone would call $C^1$, so I'm going to call it $\mathcal X$.
Next problem: if $c>0$, then $\|\cdot\|$ is not a norm on $\mathcal X$. Indeed, the constant function $1\in\mathcal X$, and yet $\|1\|=\infty$. If $c\le0$ then $\|\cdot\|$ is indeed a norm - this is easy to prove. However, unless $c=0$ (in which case $\|\cdot\|=\|\cdot\|_{C^1}$), $\mathcal X$ is not a Banach space under this norm. To see this, let $g_n$ be a smooth, non-decreasing function such that $g_n(t)=t$ for $t\le n-\frac1n$ and $g_n(t)=n$ for $t\ge n$, then define $f_n(x,y)=g_n(y)$. Observe that $f_n(x,y)e^{cy}\to ye^{cy}$ uniformly and again $\nabla f_n(c,y)\equiv0$, so $\{f_n\}$ is Cauchy in $(\mathcal X,\|\cdot\|)$. But if $\|f_n-f\|\to0$ then $f_n(x,y)e^{cy}\to f(x,y)e^{cy}$ uniformly, so $f(x,y)=y\notin\mathcal X$.
Basically, the answer to all your questions under any reasonable interpretation is "this is not true". There are two workarounds I can think of. One is to consider the space $\mathcal Y$ of continuous function $f(x,y)$ with continuous $x$-derivative $\nabla f(x,y)$ such that $\|f\|<\infty$. Then indeed we have that $\|\cdot\|$ is a norm on $\mathcal Y$ and $(\mathcal Y,\|\cdot\|)$ is a Banach space. It is a different space from $\mathcal X$, so it doesn't make a lot of sense to talk about the norm being "equivalent" to $\|\cdot\|_{C^1}$. What we have instead is an isometric isomorphism $T:(\mathcal Y,\|\cdot\|)\to(\mathcal X,\|\cdot\|_{C^1})$ given by $Tf(x,y)=f(x,y)e^{cy}$ (this works for any $c\in\mathbb R$, or indeed if you replacing $e^{cy}$ with any positive continuous function). The proof of this is trivial - we basically constructed the space $\mathcal Y$ so that $\|Tf\|_{C^1}=\|f\|$.
The other workaround is to replace $[0,\infty)$ with $[0,T]$. This perhaps makes the most sense, because then $\|\cdot\|$ is a norm on $\mathcal X$. 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.