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$F$ is a continuous linear functional on $C_0(0,1)$, I want to prove that there exists a unique $g\in V_0[0,1]$, $g(0+)=g(0)$, and $g(1-)=g(1)$, $F(x)=\int_0^1x(t)dg(t), \forall x \in C_0(0,1)$, and $\|g\|=\|F\|$.

$C_0(0,1):=\{f: f $ is continuous on $[0,1]$,and $f(0)=f(1)=0 \}$.

$V_0[0,1]:=\{f:f\in V[0,1] $ is the bounded variation on $[0,1]$, $f(0)=0$, and $f$ is right-continuous in $(0,1)\}$.

I have been thinking using the theorem that $F$ is a continuous functional on $C[0,1]$, there exist a unique $g\in V_0[0,1]$, $F(x)=\int_0^1x(t)dg(t), \forall x \in C[0,1]$, and $\|g\|=\|F\|$. I can extend $F$ on $C_0(0,1)$ to a functional $\tilde{F}$ on $C[0,1]$, but I don't know how to prove the $g$ is right-continuous at $0$ and left-continuous at $1$.

shwsq
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Let $\varphi$ be a bounded linear functional on $C_0[0,1].$ Let $$(Pf)(x)=f(x)-[f(1)-f(0)]x-f(0))$$ Then $Pf$ vanishes at $0$ and $1.$ The formula $\varPhi(f)=\varphi(Pf) $ gives a bounded linear functional on $C[0,1].$ Since $P^2f=Pf$ we get $\varPhi(f)=\varPhi(Pf).$ There exists a unique (up to the additive constant) function $g$ with bounded variation on $[0,1],$ such that $$\varPhi(f)=\int\limits_0^1f(x)\,dg(x)=\int\limits_0^1(Pf)(x)\,dg(x)\quad (*)$$

Let $$ h(x)=\begin{cases}g(x)& 0<x<1\\ g(0+) & x=0\\ g(1-) & x=1 \end{cases}$$ Then $h$ is left continuous at $0$ and right continuous at $1.$ Since $Pf$ vanishes at $0$ and $1$ we get $$ \int\limits_0^1(Pf)(x)\,dh(x) = \int\limits_0^1(Pf)(x)\,dg(x)$$ As the function $g$ in $(*)$ is unique we get $g(x)=h(x)+c$ for a constant $c.$ In particular $g$ is right continuous at $0$ and left continuous at $1.$ Moreover $(*)$ implies $$\varphi(f)=\varPhi(f)=\int\limits_0^1f(x)\,dg(x),\quad f\in C_0[0,1]$$

Remark The functional $\varPhi$ may have the norm larger than the norm of $\varphi,$ but this is not essential for the above reasoning.

Ryszard Szwarc
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