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Suppose $f$ is a linear functional $(1)$ defined on $C[a,b]\quad(3)$, and

$\forall x\in C[a,b],x(t)\geq 0\Rightarrow f(x)\geq 0 \quad(2)$.

Prove that f is continuous. Furthermore, prove that there exists a monotonically increasing function $v(t)$ defined on $[a,b]$ such that,$f(x)=\int_a^bx(t)dv(t)$

I have known that $(C[a,b])^* \cong V_0[a,b]$.

To prove $f$ is bounded. I have three ideas.

Idea 1：$\|x_n-x\|_\infty\rightarrow0 \Rightarrow |f(x_n)-f(x)|\rightarrow 0$

Idea 2:$\forall x_n$,$\|x_n\|=1\Rightarrow f(x_n)\leq M $ for some M.

Idea 3:$\|x_n\|_\infty\rightarrow0 \Rightarrow |f(x_n)|\rightarrow 0$

For idea 2:if not,with (1) and (2) there will be a series $x_n,x_n\geq 0,\|x_n\|_\infty=1$, $f(x_n)\geq n$

And I know that if $f$ is bounded, then the "furthermore" part is obvious.

I have tried linear combination between $x_n$, but in vain.

My question is how to prove $f$ is bounded? Which condition really matters ?

Any idea will be appreciated :)

Yuan
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    I recommend the $|f(x)|\leq C ||x||_\infty$ method. If you take an arbitrary function $x(t)$, what inequality can you set up regarding $x$ and $||x||$? What happens when you plug that into $f$ and use the fact that $f$ is linear? – ProfOak Mar 24 '20 at 02:42

1 Answers1

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To prove $f$ is bounded or continuous.

Use idea 3.

Given a series $x_n$, s.t.$\lim_{n\to\infty}x_n=0$. Let $y_n=\sup_{k\geq n,t\in [a,b]}|x_k(t)|$,then $\lim_{n\to\infty}y_n=0$ and $y_n$ decreases as $n$ increases.

$\forall m,\exists N, n>N\Rightarrow y_n<1/m$. With condition(3), $f(y_n)<f(1/m)=f(1)/m$. So $\lim_{n\to\infty}y_n=0$. And then $\lim_{n\to\infty}f(|x_n|)\leq f(y_n)\rightarrow 0$.By $-f(|x_n|\geq f(x_n)\leq f(x_n))$(condition (3)),$\lim f(x_n)=0$. So $f$ is continous.

Yuan
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