Transpose of Volterra operator

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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Hint: The Volterra operator is actually an integral operator. If we set $K(x,y) = 1$ if $y \leq x$ and $0$ if $y > x$, then $Vf(x) = \int_0^1 K(x,y) f(y) dy$. Now think about the defining relationship the adjoint satisfies: $\langle Vf, g\rangle = \langle f, V^\ast g\rangle$. –  Zach L. Dec 4 '12 at 16:23
Do you consider $L_2((0,1))$ over $\mathbb{R}$ or over $\mathbb{C}$. Are you looking for Banach adjoint or Hilbert adjoint operator $M$? –  no identity Dec 4 '12 at 16:23

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 $$\textbf{K}'f = \int_x^1 f(t)dt$$
No idea about answering your own question. However, be very careful with the variables. Note that $Kf(x) = \int_0^1 K(x,y)f(y)dy.$ Take note of where the $x$ appears and where the $y$ appears. Your answer needs a small change to be correct. And if you're working over $\mathbb{C}$, make sure to use conjugates. –  Zach L. Dec 4 '12 at 16:49
It's correct if you drop the conjugation on $f$. I just mentioned using the conjugate so you would not lose points on a proof, if this is for homework. In your proof, your identities involving the inner product need to use the complex conjugate. –  Zach L. Dec 4 '12 at 19:32