2
$\begingroup$

I have the following question from a textbook.

Let $f:C[a,\,b]\rightarrow\mathbb{R}$ be defined by $f(x)=x(t_{0})$ for a fixed $t_{0}\in[a,\,b]$. Show that $f$ is a bounded linear function and $\|f\|=1$.

In order to show this I thought I should show that $f$ is continuous and use the following theorem

Theorem: $f$ is continuous if and only if it is bounded. How ever am stuck on how to do this. Is there a way to do this with this technique?

$\endgroup$
  • 1
    $\begingroup$ What norm are you using for f? And by $f(x) = x(t_0)$, is this just multiplication by $t_0$? $\endgroup$ – Matt Mar 9 '17 at 21:11
  • $\begingroup$ @Matt since the domain of $f$ is the set of continuous functions, $x$ is a function and $x(t_0)$ is the value of $x$ at the point $t_0$, i.e. $f$ could be called evaluation functional. $\endgroup$ – Roland Mar 9 '17 at 21:35
  • $\begingroup$ @Roland thanks for clarifying! $\endgroup$ – Matt Mar 9 '17 at 22:37
  • $\begingroup$ It would be good if you could cite the reference of the book! $\endgroup$ – BAYMAX Sep 14 '17 at 2:32
6
$\begingroup$

Lets look at $\|f(x)\|$ and try to express it (or bound) by using $\|x\| = \max_{t\in[a,b]}x(t)$.

$$\|f(x)\| = |x(t_0)| \le \|x\| \Rightarrow \frac{\|f(x)\|}{\|x\|} \le 1$$

But there always exists a constant continous function $x_c(t)$ for which we have $\|x_c\| = x_c(t_0)$ and

$$\|f(x_c)\| = |x_c(t_0)| = \|x_c\| \Rightarrow \frac{\|f(x_c)\|}{\|x_c\|} = 1$$

So, we have $$\|f\| = \sup_{x\in C[a,b]}\frac{\|f(x)\|}{\|x\|} = 1$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Won't we need to show that $$|f(x)|\leq M ||x||$$ for some $M\in\mathbb{R}$? so as to show boundedness? $\endgroup$ – Mafeni Alpha Mar 12 '17 at 13:10
  • $\begingroup$ @MafeniAlpha, the first inequality states exactly this $\endgroup$ – Andrei Kulunchakov Mar 12 '17 at 13:13
  • $\begingroup$ I have edited the post - the part saying "$\|f(x)\| = x(t_0)$" was missing absolute value - I assume it was just a typo. (Of course, if you prefer to have $\|x(t_0)\|$ rather than $|x(t_0)|$, as to have the notation consistent with what is mostly used for normed spaces, certainly go ahead and edit the post to the form you are satisfied with. $\endgroup$ – Martin Sleziak Sep 15 '17 at 2:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.