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?

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    $\begingroup$ 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

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

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  • $\begingroup$ 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

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