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So we are given the following:

  • Operator $A$ with $Au=-u''$;
  • $u \in D_A = \{u\colon[a,b]\rightarrow R,u\in C^2([a,b]),u(a)=u(b)=0\}$;
  • $D_A$ is dense in $L^2((a,b))$.

Find the minimum value that is possible for the maximum eigenvalue of $A$.

So I have proven that:

  • operator $A$ is self-adjoint
  • Operator $A$ is positively defined that is $\langle Ax ,x\rangle >0$. But afterwards I'm really not sure how to handle the problem.
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Isn't $\sqrt{\lambda}=\frac{k\pi}{b-a}$ an eigenvalue for $k=0,1,2,...$? In this case there would be no maximum eigenvalue. – Jose27 Apr 14 '12 at 20:45
Ok I found the same eigenvalues as you did. I thought due to the "operator" concept things were handled differently. So probably the minimum value would for k=0 – bubbly Apr 15 '12 at 18:09
Actually, $k=0$ is not an eigenvalue (by the boundary conditions), so the minimum eigenvalue would be for $k=1$. – Jose27 Apr 16 '12 at 3:36

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