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Let $\varphi : \left [ 0, 1 \right ] \to \left [ 0,1 \right ]$ be the continuous function. Let $A : C \left [ 0,1 \right ] \to C \left [ 0,1 \right ]$ be an operator defined by $$\left ( Af\right ) \left ( t \right ) = f \left ( \varphi \left ( t \right ) \right ) $$ for $f \in C \left [ 0,1 \right ]$. Prove that $A$ is a continuous linear operator. Determine norm of $A$ and find the adjoint operator $A^*$.


I proved that $A$ is continuous linear and $\left \| A \right \|=1$. I don't know how to find the adjoint operator $A^*$ of $A$. Denote $X = C \left [ 0,1 \right ]$, we define $A^*$ by for $y^* \in X^*$ we have $A^*y^* = y^* \circ A$. Hence, for $f^* \in X^*$ we get $$\left ( A^*f^*\right ) \left ( f \right ) = \left ( f^* \circ A\right ) \left ( f \right ) = f^* \left ( Af\right ) = f^* \left ( f \circ \varphi\right ).$$ How can I keep on?

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Recall that $C([0,1])^*$ is the space $M([0,1])$ of Radon measures on $[0,1]$ with the total variation norm. For $\mu \in M([0,1])$ and $f \in C([0,1])$, we have \begin{align*} A^*\mu(f) &= \mu(Af)\\ &= \int_{[0,1]} f \circ \phi\, d\mu\\ &= \int_{[0,1]} f \,d(\phi_*\mu)\\ &= (\phi_*\mu)(f) \end{align*} where $\phi_*\mu \in M([0,1])$ denotes the pushforward measure $(\phi_*\mu)(B) = \mu\bigl(\phi^{-1}[B]\bigr)$ for Borel sets $B \subseteq [0,1]$.

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