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I saw this theorem somewhere and I have problem finding proof for the third part. Is it even true? If yes I'd be glad if someone could suggest me a book for reference or give me some hints for proving it myself: Consider the eigenvalue problem: $$\mathbf Ty_n=\lambda_nw(x)y_n(x),n=0,1,...,x\in\Omega.$$ Where $\mathbf T$ is linear and dense in $L^2([a,b]).$ If $\mathbf T$ is a self-adjoint operator then:
1. All the eigenvalues $\lambda_n$ are real.
2. Any two eigenfunctions of $\mathbf T$ belonging to different eigenvalues are orthogonal to each other.
3.The set of eigenfunctions is complete(dense) in the corresponding space.

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  • $\begingroup$ What sort of operator is $T$? If you're considering an arbitrary Hilbert space, the operator $f \to xf$ on $L^2([0, 1])$ is self-adjoint but has no (nonzero) eigenfunctions. (Its spectrum is nonempty, of course.) $\endgroup$
    – anomaly
    Commented Dec 8, 2017 at 20:40
  • $\begingroup$ @anomaly $\mathbf T$ is linear and it's range is dense in $L^2([a, b])$ $\endgroup$
    – shvd1991
    Commented Dec 8, 2017 at 21:00
  • $\begingroup$ Ah, that's a very different problem. $\endgroup$
    – anomaly
    Commented Dec 8, 2017 at 21:28

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You don't need to require dense range. Because $T$ is selfadjoint, you have $$ \ker T=(\text{ran}\,T^*)^\perp=(\text{ran}\,T)^\perp. $$ So you can take an orthonormal basis for the kernel and one for the closure of the range. So, whether $T$ has dense range or not is irrelevant.

For selfadjoint $T$, one can get an orthonormal basis of eigenvectors if $T$ is compact. Or, more generally, if every point in the spectrum is isolated. Otherwise, $T$ may has as little as zero eigenvectors. Since you are working in $L^2[0,1]$, the most natural examples of selfadjoint operators are the operators of multiplication by a function. One can easily show that if $M_f$ is the operator of multiplication by a function $f$ its spectrum is the essential range of $f$. And it is selfadjoint when $f$ is real. In the particular case where $f$ is monotone, it cannot have eigenvectors: if $$ \tag{!}M_f g=\lambda g, $$ you the equality $f(t)g(t)=\lambda g(t)$ a.e. So, wherever $g$ is nonzero, $f(t)=\lambda$. As we assumed that $f$ is monotone, the equality $(1)$ is impossible. So $M_f$ has no eigenvectors whatsoever.

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    $\begingroup$ thanks but I don't see how it helps me to prove the third part $\endgroup$
    – shvd1991
    Commented Dec 9, 2017 at 12:28

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