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Let $g:[0,\infty)\to[1,\infty)$ a continuous strickly increasing function, with $g(0)=1$ and $g(t)\to\infty$, as $t\to\infty$. Let $X=\{x\in C([0,\infty)); \sup_{t\geq 0}{\dfrac{|x(t)|}{g(t)}}<\infty\}$. And if define,

$$\|x\|=\sup_{t\geq 0}{\dfrac{|x(t)|}{g(t)}}\qquad,\forall x\in X$$

Then $(X,\|\cdot\|)$ is a banach space.

My atttemp: Let $(x_n)_n\subset X$ a cauchy sequence, then given $\epsilon>0$, there exist $n_{\epsilon}\in\mathbb{N}$, such that for all $n,m\geq n_{\epsilon}$, we have $$\|x_n-x_m\|<\epsilon\implies \left|\dfrac{x_n(t)}{g(t)}-\dfrac{x_m(t)}{g(t)}\right|<\epsilon$$ Thus, $\left(\frac{x_{n}(t)}{g(t)}\right)_n$ is a cauchy sequence in $\mathbb{K}$. As, $\mathbb{K}$ is a complete space, then $\left(\frac{x_{n}(t)}{g(t)}\right)_n$, converges to $\left(\frac{x(t)}{g(t)}\right)$, when $n\to\infty$. So, we have $$\sup_{t\geq 0}{ \left|\dfrac{x_n(t)}{g(t)}-\dfrac{x(t)}{g(t)}\right| }\to 0$$ Because, $g(t)\to\infty$, when $t\to\infty$. Therefore, for each $\epsilon>0$, exist $n_{\epsilon}\in\mathbb{N}$, such that for all $m\geq n_{\epsilon}$, we have $$\|x-x_m\|<\epsilon$$ So, $x\in X$ and $x_n\to x$.

This proof, is correct? My main problem is I don't be sure of how use the function $g(t)$. Thanks!

1 Answers1

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Try to use existing results rather than jumping into Cauchy sequences. This will give more intuition about what space you are really dealing with.

Let $X_g$ denote the space and norm above with function $g$.

Note that $X_1 $ is just $C[0,\infty)$ with the $\sup$ norm. You know that this is a Banach space.

Note that $f: X_g \to X_1$ defined by$f(x)(t) = {x(t) \over g(t)}$ is a bounded linear map between $X_g$ and $X_1$, and $f$ has a bounded inverse. Hence $X_g$ is a Banach space.

Aside: Note that all we really need here is for $g$ to be continuous and $\inf g > 0$.

copper.hat
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  • Thanks for your answer. Yeah I note that, if $g$ is constant and equal to 1, then $X=C([0,\infty))$, and with the norm of uniform convergence this is a Banach space. But I don't saw how use this argument. A question, what is the inverse of f? Thanks! –  Dec 12 '17 at 22:44
  • Any way, you think my "approach" is correct? –  Dec 12 '17 at 22:45
  • And finally, is correct that, $|f(x)(t)|{\infty}=| x(t)/g(t)|{\infty}=\sup_{t\geq 0}{|x(t)/g(t)|}\leq\sup_{t\geq 0}{|x(t)|/g(t)}=|x|$ –  Dec 12 '17 at 22:58
  • Well, you have the right idea, but some points need to be clarified. You have that ${x_n \over g} $ converges to some $y \in C[0,\infty)$ (I am assuming by $\mathbb{K}$ you mean $C[0,\infty)$ with the $\sup$ norm). Then you define $x=g \cdot y$ (pointwise), note that it is in $X$ and that $x_n \to x$ with the $g$ norm. – copper.hat Dec 13 '17 at 01:08
  • To get the inverse of $f$ note that if $f(x) = y$ then $g(t)y(t) = x(t)$ so we have $f^{-1}(y) = g \cdot y$ (pointwise). – copper.hat Dec 13 '17 at 01:10