How to find the Hilbert Adjoint operator of $T(f)(t)=\int_{0}^tf(s)\ ds$? 
Possible Duplicate:
Finding the adjoint of an operator 

Consider the vector space $C[0, 1]$ with inner product,
\begin{align*}
\langle f, g\rangle=\int_{0}^1f(t)g(t)\ dt.
\end{align*}
Let $T:C[0, 1]\rightarrow C[0, 1]$ the bounded linear operador given by,
\begin{align*}
T(f)(t)=\int_{0}^tf(s)\ ds.
\end{align*}
How can I find the Hilbert adjoint operator $T^*$ of $T$?
 A: $\def\sp#1{\left\langle#1\right\rangle}$For $f,g \in C[0,1]$ we have
\begin{align*}
  \sp{Tf, g} &= \int_0^1 Tf(t)g(t)\,dt\\
             &= \int_0^1 \int_0^t f(s)\,ds \cdot g(t)\,dt\\
             &= \int_0^1 \int_s^1 f(s)g(t)\,dt\,ds\\
             &= \int_0^1 f(s)\int_s^1 g(t)\,dt\,ds\\
             &=: \sp{f, T^*g}
\end{align*}
with 
$$ (T^*g)(s) = \int_s^1 g(t)\, dt, \qquad s \in [0,1] $$
A: We have
$$
(Tf)'=f \quad \forall\ f \in X:=C([0,1]).
$$ 
Using integration by parts we have, for every $f, g \in X$:
\begin{eqnarray}
\langle Tf,g\rangle &=&\int_0^1(Tf)(t)g(t)\,dt=\int_0^1(Tf)(t)(Tg)'(t)\,dt\\
&=&\left[(Tf)(t)(Tg)(t)\right]_0^1-\int_0^1(Tf)'(t)(Tg)(t)\,dt\\
&=&(Tf)(1)(Tg)(1)-\int_0^1(Tf)'(t)(Tg)(t)\,dt\\
&=&\int_0^1f(t)(Tg)(1)dt-\int_0^1f(t)(Tg)(t)\,dt\\
&=&\int_0^1f(t)[(Tg)(1)-(Tg)(t)]\,dt\\
&=&\langle f,T^*g\rangle,
\end{eqnarray}
where
$$
(T^*g)(t)=(Tg)(1)-(Tg)(t)=\int_0^1g(s)\,ds-\int_0^tg(s)\,ds=\int_t^1g(s)\,ds.
$$
Hence
$$
(T^*f)(t)=\int_t^1f(s)\, ds \quad \forall\ f \in X.
$$
