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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 $L^* = \left(x - \dfrac{d}{dx} \right)$.

olliepower
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1 Answers1

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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 $$L^* = \left(x+1 - \dfrac{d}{dx}\right)$$

  • I'm not seeing how the last term reduces to zero. – Josephine Moeller May 11 '13 at 03:32
  • @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 $S^ = {f(x) \in L^2([0,\infty)): f(x) = 0}$. –  May 11 '13 at 03:37
  • 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 $0$. – Josephine Moeller May 11 '13 at 03:43
  • @JohnMoeller The adjoint is usually only defined on a subspace of the original space. –  May 11 '13 at 03:45
  • But the OP's question doesn't have the constraint that $f(0) = g(0) = 0$. – Josephine Moeller May 11 '13 at 03:46
  • @JohnMoeller What I meant is that the adjoint need not always exist on the entire space. If it exists on a sufficiently large subspace, which in this case is the subspace $S^* \subset L^2$, we typically call $L^$ as the adjoint without specifying the subspace $S^$. –  May 11 '13 at 03:48
  • @JohnMoeller I googled about this and found a relevant pdf with some nice examples. You might also be interested in this.(http://www.math.vt.edu/people/russell/m2k_fsp_diffop.pdf). The author also seems to agree that the adjoint need not always exist on the entire space. –  May 11 '13 at 03:51
  • Ok, fair enough. I'm still struggling to figure out how to drop that extra "1" in the adjoint. – Josephine Moeller May 11 '13 at 03:53
  • @JohnMoeller Since the question is a true or false question, I would assume the answer is false. :) $L^*$ would have been $\left(x- \dfrac{d}{dx}\right)$ if the inner product was just $\displaystyle \int_0^{\infty} f(x) g(x) dx$. Probably the motivation of the question was to highlight the fact that the adjoint depends on the inner product. –  May 11 '13 at 03:54
  • Ah, good point. My brain put "show that" in the consequent and I didn't notice the specification in the OP's question. – Josephine Moeller May 11 '13 at 03:56
  • how did you derive the this term xf(x) in the third line. I get 1− ddx but i cannot derive the x – olliepower May 11 '13 at 05:49
  • @olliepower Which $xf(x)$? Note that $\left(x+ \dfrac{d}{dx}\right)g(x) = x g(x) + \dfrac{d g(x)}{dx}$. –  May 11 '13 at 06:25