We are given $Ax(t) = t\int_0^{t^2} \frac{x(s)}{s}ds$ so I know that I need to find $A^*$ such that $(Ax, y) = (x, A^*y)$.
$(Ax, y) = \int_0^1 t \int_0^{t^2} \frac{x(s)}{s}ds y(t) dt = \int_0^1 \int_0^{t^2} ty(t)\frac{x(s)}{s}ds dt$
Now, I need to get this to equal $(x, A^*y)$, but I don't know how to figure out how to do this. $(x, A^*y) = \int_0^1 x(s)A^*yds$. I'm guessing $A^*$ is some integral but I don't know how to get the integral from above to the form $\int_0^1 x(t) [\int$something $ds] dt$ where the thing in brackets would be my adjoint operator.

On another note, I don't really know what $D(A)$ would be. My guess is its {$x \in L^2(0,1)$}, but I'm not sure so could someone please let me know if this is correct

  • 1
    $\begingroup$ You wrote $(Ax,y)$ wrong which is probably why you are confused. $\endgroup$
    – Ian
    May 10 '18 at 2:59
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    $\begingroup$ You have no y there at all. $\endgroup$
    – Ian
    May 10 '18 at 3:02
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    $\begingroup$ Change the order of integration and then group everything except $x(s)$ into one aggregate, which will be $A^*y$. $\endgroup$
    – Ian
    May 10 '18 at 3:04
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    $\begingroup$ so thats what I had except I meant $x(s)$ in mine, not $x(t)$. Is that correct? $A^*y(s) = \frac{1}{s} \int _\sqrt{s} ^1 ty(t) dt$? $\endgroup$ May 10 '18 at 3:47
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    $\begingroup$ The domain of $A$ indeed requires some decay condition on $x(t)$ near $t=0$. $\endgroup$
    – Ian
    May 10 '18 at 9:56

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