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Let X be a Banach space with the norm $||.||$ and f be a vector function which acts from $\mathbb {R}$ into the Banach space X.

Can you please show that the following space is a Banach space ?

$BC^1[\mathbb{R},X]=\{x:\mathbb{R} \rightarrow X: x\in C^1 (\mathbb{R},X), |||x|||=\sup\left\{||x||, ||x'(t)||\}< \infty \right\}$

,i.e., the space of continously differentiable functions on $\mathbb{R}$ and bounded together with their derivatives.

My attempt is as follows:

Let $X$ be a Cauchy sequence in $BC^1[R,X]$ then, $|||f_n−f_m|||<ϵ,$ $n,m>N$ that is, $||f_n(x)−f_m(x)||<ϵ,$ and $||f_n′(x)−f_m′(x)||<ϵ$, ∀ϵ>0. I showed that $f_n(x)\rightarrow f(x)\in C[R,X]$ but i could not showed that $f'_n(x)→g(x)\in C^{1}[R,X].$

Note that this is not a homework problem. I would be appreciate if you could help me.

tuma
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  • It would be better to add your attempt and the exact point where you were stuck, in order to get a more precise stage of support. – BAI Sep 19 '17 at 13:18
  • Can you do the case where $X = \mathbb R$? – GEdgar Sep 19 '17 at 13:45
  • Let $f_n$ be a Cauchy sequence in $BC^{1}[\mathbb{R}, X ]$ then, $|||f_n-f_m|||<\epsilon$, $n,m>N$, that is, $||f_n(x)-f_m(x)||<\epsilon,$ and $||f_n'(x)-f_m'(x)||<\epsilon$, $\forall \epsilon>0$. I show that $f_n(x)\rightarrow f(x)\in BC^{1}[\mathbb{R}, X ]$ but i could not showed that $f_n'(x)\rightarrow g(x)\in BC^{1}[\mathbb{R}, X ] $ – tuma Sep 19 '17 at 14:02
  • How is $x'(t)$ defined in this case? – Daron Sep 19 '17 at 15:05
  • It is a strong derivative – tuma Sep 19 '17 at 15:08

2 Answers2

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Here is an idea that might work. I post it as an answer because it's too long for a comment. Let $(f_n)$ be a Cauchy sequence in $BC^1[\mathbb{R},X]$. As $X$ is Banach, there are functions $f,\ g$ such that $\lim f_n=f$ and $\lim f'_n=g.$ These are pointwise limits. On the other hand, for any $\epsilon>0,\ \||f_m-f_n\||<\epsilon$ for sufficiently large $m,n$ so if we let $m\to \infty$ we see that $f$ is bounded. And since $\|f(t)-f(x)\|\le \|f(t)-f_m(t)\|+\|f_m(t)-f_m(x)\|+\|f_m(x)-f(x)\|,\ f\ $ is continuous. Similar results hold for $g$.

Define for each $f\in BC^1[\mathbb{R},X],\ F(t)=\|f(t)\|.$ The reverse triangle inequality shows that $F_n$ and $F'_n$ are continuous. Therefore, $\int_{a}^{x}F_n'dt=F_n(x)-F_n(a).$ But an application of the Dominated Convergence Theorem gives $\lim \int_{a}^{x}F_n'dt=\int_{a}^{x}\lim F_n'dt=\int_{a}^{x}Gdt$ from which we infer that $F(x)-F(a)=\int_{a}^{x}Gdt$ and so $F'(x)=G(x)$ and therefore $f'=g.$

Matematleta
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  • That's it, the fundamental theorem of calculus makes for the easiest proof. – Giuseppe Negro Sep 19 '17 at 22:15
  • Thank you very much for your solution. – tuma Sep 20 '17 at 12:28
  • Also, in your proof, why you chose $F(t)=|f(t)|$?How can we explain. Is it because of fundamental theorem of calculus in Banach valued functions. Can you give the statement of the fundamental theorem of calculus for real variable Banach valued functions #Chilango Incomprendido – tuma Sep 20 '17 at 13:14
  • $F$ is a real-valued function so the FTC applies, because $F_n,F,G$ are continuous. I think the same argunment will work using the Bochner integral, but I think that's overkill. Anyway, there is still a problem though, because it is not obvious that $F'=G\Rightarrow f'=g.$ I am trying to work this out--- – Matematleta Sep 20 '17 at 14:08
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Let $f_{n}$ be a such Cauchy sequence then $\forall$ $\epsilon$ $>$ 0 $\exists$ M such that $\forall$ m,n $\geq$ M such that $\forall$ t $\in$ $\mathbb{R}$

both $\left \| f_{n}(t)-f_{m}(t) \right \|$ $<$ $\epsilon$ and $\left \| f^{'}_{n}(t)-f^{'}_{m}(t) \right \|$ $<$ $\epsilon$ then there exist two functions a:$\mathbb{R}$ $\rightarrow $ $X$ and b:$\mathbb{R}$ $\rightarrow $ $X$ such that for every t $\in$ $\mathbb{R}$ $f_{n}(t)$ converge to $a(t)$ and $f^{'}_{n}(t)$ converge to $b(t)$ , and because $\forall$ $\epsilon$ $>$ 0 $\exists$ M such that $\forall$ m,n $\geq$ M such that $\forall$ t $\in$ $\mathbb{R}$ both $\left \| f_{n}(t)-f_{m}(t) \right \|$ $<$ $\epsilon$ and $\left \| f^{'}_{n}(t)-f^{'}_{m}(t) \right \|$ $<$ $\epsilon$ we fix n and let m go to $\infty$ so we wille have a and b such that $\forall$$\epsilon$ $>$ 0 $\exists$ N $\forall$ n $\geq$ N $\left \| f^{'}_{n}(t)-b(t) \right \|$ $<$ $\epsilon$ and $\left \| f_{n}(t)-a(t) \right \|$ $<$ $\epsilon$ if we can show that b=$a^{'}$ (i didn't try ) we are done

szw1710
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Abderrahim Abdou
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