Consider in L2 (-∞, ∞) the operator Kv = ∫-∞^∞ exp (-(x²+y²)/2) v (y) dy. (9.5.79) (a) Prove that K is self-adjoint. (b) Prove that K is completely continuous. (c) Find the eigenvectors of K.

Question

Consider in [latex]\mathcal{L}_{2} (-\infty , \infty )[/latex] the operator

[latex]Kv = \int_{-\infty}^{\infty } exp \left(\frac{-(x^{2} + y^{2} )}{2} \right) v (y) dy.[/latex] (9.5.79)

(a) Prove that K is self-adjoint.
(b) Prove that K is completely continuous.
(c) Find the eigenvectors of K.

Accepted Answer

Since [latex]k (x, y) = exp (-(x^{2} + y^{2} / 2) = k^{*} (y, x),[/latex] the operator is self-adjoint.

Also,

[latex]\int_{-\infty}^{\infty }\int_{-\infty}^{\infty } [k(x, y)]^{2} dx dy = \int_{-\infty}^{\infty } exp (-x^{2} ) dx \int_{-\infty}^{\infty } exp (-y^{2} ) dy = \pi \lt \infty ,[/latex] (9.5.80)

which is a sufficient condition that K is completely continuous.
The eigenproblem is

[latex]exp \left( \frac{- x^{2}}{2} \right) \int_{-\infty}^{\infty } exp \left( \frac{- y^{2}}{2} \right) v (y) dy = \lambda v (x).[/latex] (9.5.81)

One eigenvector is [latex]v_{0} (x) = \alpha exp (-x^{2}/2 ),[/latex] for which Eq. (9.5.81) yields

[latex]exp \left(\frac{- x^{2}}{2} \right) \alpha \int_{-\infty}^{\infty } exp (-y^{2} ) dy = \lambda _{0} \alpha exp \left(\frac{-x^{2} }{2} \right)[/latex] (9.5.82)

or [latex]\lambda _{0} = \sqrt{\pi } .[/latex] Normalizing [latex]v_{1}[/latex] we obtain

[latex]1 = \left\|v_{1} \right\|^{2} = \alpha^{2} \int_{-\infty }^{\infty } exp \left(\frac{-x^{2} }{2} \right) dx = \alpha^{2} \sqrt{\pi }[/latex] (9.5.83)

or

[latex]v_{0} (x) = \frac{1}{\pi ^{1/4} } exp \left(\frac{-x^{2} }{2} \right), \lambda_{0} = \pi^{1/2}[/latex] (9.5.84)

This is the only eigenfunction of K with a nonzero eigenvalue. The eigenfunctions corresponding to λ = 0 obey the equation

[latex]\int_{-\infty}^{\infty } exp (\frac{-y^{2} }{2} ) v (y) dy = 0.[/latex] (9.5.85)

The first two of these are

[latex]v_{1} (x) = \frac{1}{\sqrt{ 2\sqrt{\pi } } } 2x exp (\frac{-x^{2} }{2} )[/latex] (9.5.86)

and

[latex]v_{2} (x) = \frac{1}{2\sqrt{ \sqrt{\pi } } } (4x^{2} -2 ) exp (\frac{-x^{2} }{2} )[/latex] (9.5.87)

In general,

[latex]v_{n} (x) = \frac{1}{\sqrt{2^{n} n! \sqrt{\pi } } } H_{n} (x) exp (\frac{-x^{2} }{2} ),[/latex] (9.5.88)

where [latex]H_{n} (x)[/latex] is a Hermite polynomial generated by the formula

[latex]H_{n} (x) = (-1)^{n} exp (x^{2} ) \frac{d^{n} }{dx^{n} } (exp(-x^{2} )) .[/latex] (9.5.89)

From the theorem for completely continuous self-adjoint operators, it follows that [latex]v_{0} (x), v_{1} (x),v_{2} (x), [/latex]. . . form an orthonormal basis set in [latex]\mathcal{L}_{2} (-\infty , \infty )[/latex]. The functions

[latex]\frac{H_{n} (x) }{\sqrt{2^{n} n! \sqrt{\pi }} } , n = 0, 1, 2,[/latex]. . . (9.5.90)

therefore, form an orthonormal basis set in [latex]\mathcal{L}_{2} (-\infty , \infty ; exp (-x^{2} )).[/latex]

Question 9.5.4: Consider in L2 (-∞, ∞) the operator Kv = ∫-∞^∞ exp (-(x²+y²)...

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