I have solved other problems like this using integration by parts. In this case, I can't figure out what to make each part for the integration. The question is true/false. Ultimately to show this you have to find the adjoint and see if it is what they claim it is, i.e., $\displaystyle uv - \int vdu$.

What do you make $u$ and $v$?

If $\langle f \vert g \rangle = \displaystyle \int_0^{\infty} f(x) g(x) e^{-x} dx$ for functions $f,g \in L_2([0,\infty))$ and $L = \left(x + \dfrac{d}{dx} \right)$ (assume that all elements of $L_2([0,\infty))$ are differentiable), its adjoint is $L^* = \left(x - \dfrac{d}{dx} \right)$.


We want $L^*$ such that $$\langle f \vert Lg \rangle = \langle L^*f \vert g \rangle$$ We have $$S=\int_0^{\infty} e^{-x}f(x)\left(x+\dfrac{d}{dx}\right)g(x) dx = \int_0^{\infty}x e^{-x}f(x) g(x) dx + \int_0^{\infty} e^{-x} f(x) dg(x)$$ We have $$\int_0^{\infty} e^{-x} f(x) dg(x) = \int_0^{\infty} d\left(e^{-x} f(x)g(x)\right) - \int_0^{\infty} g(x) d\left(e^{-x} f(x)\right)$$ Hence, $$\int_0^{\infty} g(x) d\left(e^{-x} f(x)\right) = \int_0^{\infty} g(x)\left(e^{-x} \dfrac{df}{dx} - e^{-x} f(x)\right) dx$$ Hence, $$S = \int_0^{\infty}e^{-x}\overbrace{\left(xf(x)-\dfrac{df}{dx}+f(x)\right)}^{L^*f}g(x)dx + \int_0^{\infty} d\left(e^{-x} f(x)g(x)\right)$$ Hence, the adjoint is $$L^* = \left(x+1 - \dfrac{d}{dx}\right)$$

  • $\begingroup$ I'm not seeing how the last term reduces to zero. $\endgroup$ – Josephine Moeller May 11 '13 at 3:32
  • $\begingroup$ @JohnMoeller The last term integrates to $\left. e^{-x} f(x) g(x) \right\vert_0^{\infty}$. For $L^2$ functions, with $f(0) = g(0) = 0$, the term goes off. Hence, $L^*$ is the adjoint of the operator on the subspace of function $S^* = \{f(x) \in L^2([0,\infty)): f(x) = 0\}$. $\endgroup$ – user17762 May 11 '13 at 3:37
  • $\begingroup$ But what if $f(x) = e^{-x}$? Then $f$ is square-integrable on $[0,\infty]$. I don't see why $f(0)$ has to be $0$. $\endgroup$ – Josephine Moeller May 11 '13 at 3:43
  • $\begingroup$ @JohnMoeller The adjoint is usually only defined on a subspace of the original space. $\endgroup$ – user17762 May 11 '13 at 3:45
  • $\begingroup$ But the OP's question doesn't have the constraint that $f(0) = g(0) = 0$. $\endgroup$ – Josephine Moeller May 11 '13 at 3:46

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