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Given two functions $f,g:(0,1)\to\mathbb{R}$ such that $f''+\lambda f = 0$ and $g''+\gamma g = 0$, $f(0) = f(1)$ and $g(0) = g(1)$.$\lambda,\gamma \ge 2\pi$

Show that (with minimal fuss any) $$\lvert|f+g|\rvert_{L^2}^2 = \lvert|f|\rvert_{L^2}^2 + \lvert|g|\rvert_{L^2}^2$$

Appreciate if you don't do any integration or use of any trigonometry.

Robert Lewis
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Rajesh D
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  • I don't see how you can discuss $\Vert \cdot \Vert_{L^2}$ without some reference to integration . . . – Robert Lewis Jul 31 '17 at 15:22
  • @RobertLewis : I mean avoiding any integration based on fomulae or integration of closed form expressions. – Rajesh D Jul 31 '17 at 15:24
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    Look up "Sturm-Liouville self-adjoint", you should find something rather general. (It is a bit odd because you selected the eigenvalues rather than selecting the boundary conditions, which makes it not obvious how to use general considerations about self-adjoint operators.) – Ian Jul 31 '17 at 15:27
  • Yeah, I think something about boundary conditions might be needed . . – Robert Lewis Jul 31 '17 at 15:47
  • @Ian updated. I think they make sense. – Rajesh D Jul 31 '17 at 16:12
  • Then yes, in this case you have a self-adjoint Sturm-Liouville eigenvalue problem and general theory applies to give the orthogonality required. – Ian Jul 31 '17 at 16:19
  • @Ian : Is there any S-L like theory for functions in higher dimensional Euclidean spaces. – Rajesh D Jul 31 '17 at 16:21
  • Are you sure you don't want $\Vert f + g \Vert_{L^2}^2 = \Vert f \Vert_{L^2}^2 + \Vert g \Vert_{L^2}^2$? I'm pretty sure your conjectured conclusion is false as stated. – Robert Lewis Jul 31 '17 at 17:44
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    @RobertLewis Ah forgot about squareroot. Need to square. Thanks for pointing out. – Rajesh D Jul 31 '17 at 17:54
  • I'm trying to solve this with "minimal fuss" but the shifting conditions in the question create more stuff to fuss over! – Robert Lewis Jul 31 '17 at 17:55
  • OK, don't mean to be obnoxious. But you also need to consider saying something about $\lambda$ and $\gamma$, I'll warrant. Why did you take that stuff out? Just trying to help. No offense meant. Cheers! – Robert Lewis Jul 31 '17 at 17:56
  • I added the tag, "differential equations". – Robert Lewis Aug 01 '17 at 05:35

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