Kreyszig's FA Prob. $6$ section $2.8$

Question

I'm having a difficult time understand the second part of this question:

The space $C'[a,b]$ or $C^1[a,b]$ is the normed space of all continuously differentiable functions on $J=[a,b]$ with norm defined by:

$||x||=\max_{t\in J}|x(t)|+\max_{t\in J}|x'(t)|$.

Show that the axioms of a norm are satisfied. Show that $f(x)=x'(c)$, $c=(a+b)/2$, defines a bounded linear functional on $C'[a,b]$. Show that $f$ is not bounded, considered as a functional on the subspace of $C[a,b]$ which consists of all continuosly differentiable functions.

For the first part it is easy to show that $f$ is indeed a linear functional and bounded because the elements are continuous and the domain is compact. However I can't understand the last part which states that $f$ is not bounded if we consider it as a functional on the subspace $C^1[a,b]$ of the larger space of continuous functions on $[a,b]$. Maybe $C[a,b]$ is a Banach space (with respect to some norm) and if $f$ was indeed bounded on $C^1[a,b]$ (as a subspace) it would admit a bounded extension $\hat{f}$ defined on $C[a,b]$ and the question is asking me to show that it is impossible? Is the metric induced by the given norm defined $C[a,b]$? I'm really confused, any help is appreciated.

Please make your question accessible without an embedded image. I can barely read that image for its blurriness. On top of which your question should be searchable, which embedded images aren't. — Lee Mosher, Jan 02 '20 at 02:15
I think he means the space $C^1[a, b]$ with the usual supremum norm, i.e., the norm inherited of $C[a, b]$. — Lázaro Albuquerque, Jan 02 '20 at 02:51
I guess it's the best way to look at this question. I will edit an answer based on the twos answers. Thank you. — 2ndYearFreshman, Jan 02 '20 at 03:36
You need a sequence of functions whose sup norm is less than or equal to 1 but whose derivatives become unbounded. How about $[-1,1],\ x=0$ and $f_n(x) = \frac{2}{\pi}\arctan(nx)$? — Matematleta, Jan 02 '20 at 03:42

score 2 · Accepted Answer · answered Jan 02 '20 at 03:38

2

You need to consider $f$ with the following norm:

$$ ||f|| = \sup_{||x||=1}|f(x)| = \sup_{\max_{a \le t \le b}|x(t)|=1}|x'(c)| $$

where $x \in C^1[a, b]$.

Then you can take a non-negative $x_n' \in C[a, b]$ such that $x_n'(a)=x_n'(b)=0$, $x_n'(c) \rightarrow \infty$ and $x_n(b)=\int_a^b x'(s)ds=1$. For the existence of such functions check this.

You now have $|f(x_n)| \rightarrow \infty$.

answered Jan 02 '20 at 03:38

Lázaro Albuquerque

2,413

Thank you. I'll accept the answer. – 2ndYearFreshman Jan 02 '20 at 03:50

2ndYearFreshman · Answer 2 · 2020-01-02T03:49:50.507

2

Assuming the usual $sup$ norm on $C[a,b]$. It suffices to find a sequence of bounded functions $x_n(t)$ such that $|x'_n(c)|$ is unbounded.

for reference:this post

Another approach is to consider $f_n(x) = \frac{2}{\pi}\arctan(nx)$ on $[-1,1]$.

edited Jan 02 '20 at 03:49

answered Jan 02 '20 at 03:39

2ndYearFreshman

478

score 0 · Answer 3 · answered Apr 10 '22 at 00:50

You can also choose the sequence $x_{n}(t) = \sin(nt - nc).$

So $$ \| x_{n} \|_{C^{0}[a,b]} = \max_{a \leq t \leq b} | x_{n}(t) | = 1. $$

By the other side, you will have

$$ | f ( x_{n} )| = x^{'}_{n}(c) = n \cos(nc - nc) = n \cos(0) = n.$$

So $ | f ( x_{n} )| \to \infty $ and then, $f$ not could be bounded.

Kreyszig's FA Prob. $6$ section $2.8$

3 Answers3