# How to find the Hilbert Adjoint operator of $T(f)(t)=\int_{0}^tf(s)\ ds$? [duplicate]

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 $T$?

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## marked as duplicate by Mhenni Benghorbal, Per Manne, Norbert, Marvis, Hagen von EitzenNov 22 '12 at 10:46

$\def\sp#1{\left\langle#1\right\rangle}$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 $$(T^*g)(s) = \int_s^1 g(t)\, dt, \qquad s \in [0,1]$$

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Ah, I see, integral over the triangle $(0,0)$, $(1,0)$, $(1,1)$. –  Berci Nov 22 '12 at 0:45
Very interesting @martini, but I still have a doubt, how did you get the equality \begin{align*} \int_{0}^1\int_{0}^tf(s)ds\cdot g(t)\ dt=\int_{0}^1\int_{s}^1f(s)g(t)\ dt\ ds? \end{align*} –  PtF Nov 22 '12 at 0:49
It's Fubini. Both are the integral of $(s,t)\mapsto f(s)g(t)$ over $\{(s,t) \mid 0 \le s \le t \le 1\}$ (@Berci's triangle). –  martini Nov 22 '12 at 0:55
I see =D Thanks for the help @martini.. –  PtF Nov 22 '12 at 1:00
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 $$(T^*f)(t)=\int_t^1f(s)\, ds \quad \forall\ f \in X.$$