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I want to find the transpose of the Volterra operator $Vf(x) = \int_0^x f(t)dt, \;\; x\in(0,1)$ acting in $V:L^2(0,1) \rightarrow V:L^2(0,1) $. The transpose is defined as $\textbf{M}':U'\rightarrow V'$. For Hilbert spaces the transpose is replaced by the adjoint. I would guess that the transpose is also a map $\textbf{M}':L^2(0,1)\rightarrow L^2(0,1)$ Since $L^2$ is self dual. But how do I find the map?

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    Do you consider $L_2((0,1))$ over $\mathbb{R}$ or over $\mathbb{C}$. Are you looking for Banach adjoint or Hilbert adjoint operator $M$?2012-12-04

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Following the hint from Zach $\langle Vf,g\rangle = \int_0^1\int_0^1 K(t,s)f(t)g(s)dtds$ Changing the order of integration $\langle Vf,g\rangle = \int_0^1 \left(\int_0^1 K(t,s)g(s)ds \right)f(t) dt$ Hence $\textbf{K}'f = \int_0^1 K(t,s) f(t)dt,$ where $K(t,s) =1$ if $t\leq s$ hence $\textbf{K}'f = \int_x^1 f(t)dt$

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    It is perfectly reasonable to accept your own answers provided they are a demonstration of your knowledge gained from Math.SE! :-) See the FAQ for more.2012-12-05