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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?

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    $\begingroup$ Perhaps you can share your proof. $\endgroup$
    – Umberto P.
    Commented Feb 9, 2018 at 0:12
  • $\begingroup$ To start, notice that $X$ is the space of continuous functions of exponential order $-k$ $\endgroup$ Commented Feb 9, 2018 at 0:22
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    $\begingroup$ You really should show us that proof, so we can see how $k<0$ changes things... $\endgroup$ Commented Feb 9, 2018 at 0:28
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    $\begingroup$ I don't think the sign of $k$ will end up mattering. If $k<0$, then that just means the functions have to go to zero fast enough to beat the exponential. $\endgroup$ Commented Feb 9, 2018 at 0:37
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    $\begingroup$ Replacing a given $f$ with $f(x)e^{-kx}$ reduces you to the case of bounded continuous functions, which may be less intimidating... $\endgroup$ Commented Feb 9, 2018 at 1:08

2 Answers 2

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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.

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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 $\|f_n - f\|_X = \|e^{kt} (f_n - f)\|_\infty = \|g_n - g\|_\infty \to 0$.

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  • $\begingroup$ Very nice and simple solution, not sure why it had no upvotes. $\endgroup$
    – 2132123
    Commented Jul 7, 2020 at 19:39

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