I'm unsure of how to solve this problem (it's number 5 in section 18 of Kolmogorov's Introductory Real Analysis).
Prove that f(x) = ax(0) + bx(1) is a bounded linear functional on C[0,1]. What is its norm?
The book doesn't clarify what a and b are, but my attempt to solve the problem is as follows:
Let x,y $\in$ C[0,1]. Then f($\alpha x+\beta y$) = a$\alpha x(0) + a\beta y(0) +b\alpha x(1) + b\beta y(1)$ = $\alpha (ax(0) + bx(1)) + \beta (ay(0) + by(1)) $ (after re-arranging) = $\alpha f(x) + \beta f(y)$, thus f is linear.
Boundedness is what I'm unsure of. I've used the fact that continuous functions are bounded on closed intervals:
Let x$\in$C[0,1]. Then x(t) is bounded since it is continuous on a closed interval. Hence, $\exists M \in \mathbb{R}$ such that x(t) $\lt$ M and so f(x) = ax(0) + bx(1) $\lt$ (a+b)M, thus f is bounded.
I'm unsure of what a and b are, can I assume they are just real numbers and so this would work for showing boundedness? Or did I screw up? As for the norm of f, I believe it just is:
$\|f||$ = $sup_{||x||=1} {|f(x)|}/{||x||}$ = sup |f(x)| = a+b.
You have $\|f(x)\|=\|ax(0)+bx(1)\|\leq |a||x(0)|+|b||x(1)|\leq (|a|+|b|)\|x\|$ so the norm of $f$ is inferior to $|a|+|b|$.
On the other hand, construct $x$ such that $x(0)=1$ if $a>0$, $x(0)=-1$ if $a\leq 0$, $x(1)=1$ if $b>0$, $x(1)=-1$ if $b\leq 0$, $|f(x)|\leq 1$, you have $\|x\|=1$ and $\|f(x)\|=|a|+|b|$. You can suppose for example that $x$ is linear between in $[0,1/2]$ with $x(1/2)=0$ and $x$ is linear in $[1/2,1]$.