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We are given $T$ is an operator on $C[0,1]$ as follows

$T(g(x))=\sum\limits_{k=1}^{m}p_kg(f_k(x)), p_k\in [0,1], f_k\in C[0,1]$, could anyone tell me how to show adjoint of this operator is as follows?

Myshkin
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The dual of $C[0,1]$ is the space of all signed/complex measures on $[0,1]$ with total variation norm. we have $T^{*}(\mu)(g)=\mu (Tg)=\int Tg d\mu=\int \sum p_kg(f_k)d\mu=\sum p_k \int gd(\mu\circ f_k^{-1})$ and hence $T^{*}=\sum p_k \mu\circ f_k^{-1}$

Myshkin
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Kavi Rama Murthy
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  • So, on which object $T^*$ will act? I am not understanding that. will it act on some subset of $[0,1]$ or the space I am considering and give me a measure of that subset or it will give me a new measure? – Myshkin Mar 31 '19 at 10:45
  • $T:C([0,1]) \to C([0,1])$ so $T^{}:(C([0,1])^{} \to C([0,1])^{}$. $C([0,1])^{}$ ois the space of measures on Borel subsets of $[0,1]$. – Kavi Rama Murthy Mar 31 '19 at 11:28
  • Then why did you write $T^(\mu)(g)$? Would it be $T^(\mu)(A)$? where $A$ is some Borel subset of $[0,1]$? – Myshkin Mar 31 '19 at 12:19
  • When you think of the dual of $C[0,1]$ as the space of measures you make an identification: a measure $\mu$ is also regarded as a continuous linear functional on $C[0,1]$ defined by $\mu (f)=\int_0^{1}f d\mu$. – Kavi Rama Murthy Mar 31 '19 at 12:21
  • Okay! so $T^: (C[0,1])^\to (C[0,1])^, T^(\mu)(g)= \mu(Tg)$ Thanks!!! I Love you! – Myshkin Mar 31 '19 at 12:23
  • Another obscurity: why or how this $f_k^{-1}$ thing is appearing in the definition of $T^*$? – Myshkin Mar 31 '19 at 12:28
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    @DingDong $\int g(f) d\mu=\int g d(\mu\circ f^{-1})$. This is the change of variables formula. – Kavi Rama Murthy Mar 31 '19 at 23:13