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Let $C[0,K]$ be the space of all continuous real valued functions on $[0,K]$ for $K>0$ and $L\geq0$, equipped with the metric $d$ defined by

$$d(f,g)=\sup_{0\leq k\leq K}e^{-Lk}|f(k)-g(k)|.$$

I am trying to show that $(C[0,K],d)$ is a complete metric space.

My attempt:

Let $(f_n)$ be an arbitrary Cauchy sequence in $(C[0,K],d)$. We need to show that it has a limit in $(C[0,K],d)$.

Then for each $\epsilon>0$, there exists $N$ such that $$m,n\geq N\implies d(f_m,f_n)=\sup_{0\leq k\leq K}e^{-Lk}|f_m(k)-f_n(k)|<\epsilon$$

The part that I struggled is the part where we have to show there is an $f\in(C[0,K],d)$ such that $d(f_n,f)<\epsilon$.

Could somebody please give some hints? Thanks!

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  • $\begingroup$ Are there any constraints on $L$? In most cases, the construction of the limit (of the sequence) is somewhat natural, but in this case it seems to be not. My approach would be, as suggested, find an isometry. $\endgroup$ – user305860 Jul 12 '16 at 7:48
  • $\begingroup$ @ErikJoensson I have included that $L$ must be non-negative. $\endgroup$ – user338393 Jul 12 '16 at 8:00
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Hint: Find an isometry between $(C[0,K],d)$ and $C[0,K]$ (with the standard $\sup$ norm).

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  • $\begingroup$ Thanks. I would probably need a bit more hints on constructing the isometry. We need to preserve distance in the isometry, but how could we do that with the factor of $e^{-Lk}$? $\endgroup$ – user338393 Jul 12 '16 at 7:43
  • $\begingroup$ Is there any other way besides constructing an isometry? $\endgroup$ – user338393 Jul 12 '16 at 7:44
  • $\begingroup$ I don't think there is any simpler way (provided you take completness of $C[0,K]$ for granted). Otherwise, more or less exactly the same proof should work. $e^{-Lk}$ is a continuous function, so it is mostly irrelevant. Alternatively, you could just estimate it by the upper bound, but that is more or less the same as writing the isometry and slightly more complicated to write. $\endgroup$ – tomasz Jul 12 '16 at 7:45
  • $\begingroup$ For the isometry: what is the regular sup-distance between $e^{-Lx}f(x)$ and $e^{-Lx}g(x)$? $\endgroup$ – Arthur Jul 12 '16 at 7:57
  • $\begingroup$ @Arthur I don't quite understand what does it mean by sub-distance? Could you please clarify it a bit more? Thanks. $\endgroup$ – user338393 Jul 12 '16 at 8:01
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I think that there is two ways not using the isometry (I agree that this is a very good solution, surely the simplest) to prove the result.

1) Note that your distance is associated with the norm $N(f)=\sup_{0\leq k\leq K}\exp(-kL)|f(k)|$, i.e $d(f,g)=N(f-g)$. If we note by $\|f|_{\infty}$ the usual Sup-norm, use the fact that $\exp(-KL)\leq \exp(-Lk)\leq 1$ to show that $$\exp(-KL)\|f\|_{\infty}\leq N(f)\leq \|f\|_{\infty}$$ ie the two norms on $C([0,K], \mathbb{R})$ are equivalents.

2) Follow the proof for the fact that $C([0,K], \mathbb{R})$ is complete for the $\sup$-norm:

For your Cauchy sequence for the distance $d$,

a) Fix $k$, show that $f_n(k)$ is a Cauchy sequence in $\mathbb{R}$, hence convergent. Now you can define a function $f(k)$ as the pointwise limit of $f_n(k)$

b) For your $\varepsilon$, $N(\varepsilon)$, fix $k$ and in $\exp(-kL)|f_m(k)-f_n(k)|\leq \varepsilon$, let $m\to +\infty$. Deduce that $\exp(-kL)f(k)$ is continuous, hence also $f$.

c) Deduce from what you have done in b) that $d(f,f_n)\leq \varepsilon$ for $n\geq N$ and you are done.

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