$X$ is the space of all continuous functions with the norm $||f||=\sup_{t\ge0}e^{kt}|f(t)|<\infty.$ Is $X$ a Banach space? Fix $k\in \mathbb{R}$ and let $X$ be the space of all continuous functions $f:[0,\infty) \to\mathbb{R}$ s.t.
$$\|f\|=\sup_{t\ge0}e^{kt}|f(t)|<\infty.$$
Is $X$ a Banach Space?
I think $X$ is Banach if $k\ge0$ and I had a brief proof. But I have no idea whether $X$ is Banach when $k<0$. Can someone give a brief proof or a counterexample?
 A: Suppose that $\{f_n\}$ is Cauchy. Fix $>0$. Let $n_0$ such that, for all $t$,  $|f_n(t)-f_m(t)|< \,e^{-kt}$ if $n,m\geq n_0$. Now start with $n_0$, and choose $n_{j+1}$ such that $|f_{n_{j+1}}(t)-f_{n_j}(t)|<2^{-j}\, /e^{kt}$. Let 
$$
f(t)=f_{n_0}(t)+\sum_{j=0}^\infty f_{n_{j+1}}(t)-f_{n_j}(t). 
$$ 
This is well-defined for each $t$, since 
$$
\left|\sum_{j=K}^\infty f_{n_{j+1}}(t)-f_{n_j}(t)\right|
\leq \sum_{j=K}^\infty |f_{n_{j+1}}(t)-f_{n_j}(t)|
\leq\sum_{j=K}^\infty 2^{-j}\,\,e^{-kt}=2^{-K+1}\,e^{-kt}. 
$$
Now, since 
$$
f_{n_r}(t)-f(t)=\sum_{j=r}^\infty f_{n_{j+1}}(t)-f_{n_j}(t),
$$
we get that 
$$
e^{kt}|f_{n_r}(t)-f(t)|\leq 2^{-r+1}
$$
So $\|f_{n_r}-f\|\to0$. In particular, 
$$
e^{kt}\,|f(t)|\leq 1 + e^{kt}|f_{n_0}(t)|<\infty. 
$$
So $f\in X$. As the original sequence $\{f_n\}$ was Cauchy, $\|f_n-f\|\to0$. So $f$ is a limit of the sequence, and $X$ is complete. 
A: Unless I'm missing something it seems that this follows pretty easily, for any positive or negative $k$, from completeness of continuous functions in uniform norm.  Following Paul Garrett's suggestion, suppose $f_n$ is Cauchy in $X$ and let $g_n = e^{kt} f_n$.  Then $\|g_n - g_m\|_\infty = \|f_n - f_m\|_X$, so $g_n$ is Cauchy in the space $BC([0,\infty))$ of bounded continuous functions, which is a Banach space, so $g_n$ converges uniformly to some bounded continuous $g$.  Set  $f = e^{-kt} g$, which is in $X$, and now $\|f_n - f\|_X = \|e^{kt} (f_n - f)\|_\infty = \|g_n - g\|_\infty \to 0$.
