One problem in Functional analysis

[Солунац]

Hi to everybody!

I am Serb from Bosnia, and I don`t speak russian ( sorry ), so that`s why I am writing in English. I hope I`ll learn russian someday...

I have a problem that I didn`t solve:

(a) Prove that the space $X = C^n[a,b]$ with norm
$\|f\| = |f(a)|+|f'(a)|+|f''(a)|+|f'''(a)|+…+|f^(^n^-^1^)(a)|+\sup\limits_{t \in[a,b]} |f(t)|$
is Banach`s.

(b) Examen the separability of X.

Russian I could read (but still not write), so u don`t have to write any comments or answers on English ... just on Russian.

Thanks!

Свако добро и све најбоље браћо Руси!
Живјела мајка Србија, живјела прамајка Русија!

[Brukvalub]

В пункте а) нужно просто проверить аксиомы Банахова пространства - это несложно сделать, а в п. b, я думаю, помогут многочлены с рациональными коэффициентами.

[нг]

[mod]Солунац

The formulae on this forum are usually written using $\TeX$ notation. We have an introduction and quick reference, both are unfortunately in Russian.

We will appreciate if you will correct your post. Would you have any problems with rules, language or other, please do not hesitate contact us using PM (ЛС).[/mod]

[Солунац]

@Brukvalub
Thanks.

@нг
Is this OK? I can read Russian (not very good, but stil I can read), so thanks for the introduction and a good will.

+++++++++++++++++++++
I have some further questions...
About (b) - I tried by induction over n.
$C^{n-1}_{[a,b]}\supset C^n_{[a,b]}$
$f\in C^n_{[a,b]} \Rightarrow f '\in C^{n-1}_{[a,b]}$
for arbitrary $\varepsilon > 0$ we could aproximate f ' with a polynomial P (induction hypothesis):
$\|f'-P\|_{C^{n-1}_{[a,b]}}<\frac{\varepsilon}{b-a+1}$
then $Q(t) = f(a)+\int\limits_a^b P(z)dz$ so, it is
$\|f-Q\|_{C^n_{[a,b]}} = |f(a)-f(a)|+|f'(a)-P(a)|+...+|f^{(n-1)}(a)-P^{(n-2)}(a)|+\sup\limits_{[a,b]} |f(t)-Q(t)|$
there is to be done to prove that is
$\|f-Q\| < \varepsilon$
???

Счастливого Рождества!

[lofar]

It seems that $X$ is not complete when $n>1$. Take $f$ and $f_n$ as in the figure below



assume that $f_n$ are from $C^{\infty}[a,b]$, $f_n(t)=0$ for $t\in[a,c]$ ($c$ is the same for all $n$), and $f_n$ converges uniformly to $f$. Sequense $f_n$ is fundamental in $C[a,b]$ thus it is fundamental in $X$ ($||f_n-f_m||_X=||f_n-f_m||_{C[a,b]}$). But $f_n$ has no limit in $X$ since $f$ is not differentiable.

Hint for the (b): Note that $f(t)-Q(t)=\int_a^t(f'(\xi)-P(\xi))\,d\xi$.

[Солунац]

Спасибо, lofar (Христос се роди! )

Still, I'm not quite shore that this $f_n$ is a Cauchy sequence in a space X (of course, it is sad to be like that, but ..., and isn’t it ${\|f_n-f_m\|_X}\geqslant {\|f_n-f_m\|_C_{[a,b]}}$?). Maybe we can assume something like that, but I think it would be better if there is a defined sequence as a proof of incompletes (caunterexample). Sorry if I sad something incorrect ... is there a link on a similar problem (Functional analysis is new to me, and as an exercise book I use the Задачи в упражнения по функциональному анализу, Треногин В. А., Писаревский В. М., Соболева Т. С, so, is there a link or recommendation for a better exercise book which can be download, or this one is a good choice).

Спасибо.

[lofar]

It is counterexample. $f_n(t)=0$ for $t\in[a,c]$ ($a<c<b$). Therefore $0=f_n(a)=f_n'(a)=f_n''(a)=\ldots$ Thus in the definition of $||\cdot||_X$ all the summands except for the last one equal zero.

Such functions ($f_n$'s, i mean) can be constructed using $e^{-1/x^2}$-trick.

[Солунац]

Спасибо, lofar, спасибо.