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Set $v=x^n{e}^{-x}$. Direct differentiation gives $x{v}^{\prime }=(n-x)v$. Apply the Leibniz rule to differentiate $n+1$ times:Hence $x{v}^{(n+2)}+(x+1)v^{(n+1)}+(n+1)v^{(n)}=0$. By Rodrigues' formula, $v^{(n)}=n!e^{-x}{L}_n$, so the next two derivatives are $n!e^{-x}(L_n^{\prime }-L_n)$ and $n!e^{-x}(L_n^{\prime \prime }-2L_n^{\prime }+L_n)$. Substitution gives the displayed Laguerre differential operator on polynomials eigenvalue equation. This derivation also covers $n=0$.

This equation has a regular singular point at zero and an irregular singular point at infinity. For a nonnegative integer $\nu =n$, the solution analytic at zero with value one is the Laguerre polynomial $L_n$. Its weighted self-adjoint form is $(x{e}^{-x}{y}^{\prime }{)}^{\prime }+\nu e^{-x}y=0$, connecting it to the Laguerre differential operator on polynomials.

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

Differentiating $v=x^n{e}^{-x}$ gives $x{v}^{\prime }=(n-x)v$. Applying the Leibniz rule $n+1$ times to this identity givesOnly two terms survive on each side because the multiplying factors are linear. Rearranging yields $x{v}^{(n+2)}+(x+1)v^{(n+1)}+(n+1)v^{(n)}=0$. Since $v^{(n)}=n!e^{-x}{L}_n$, the next two derivatives are $n!e^{-x}(L_n^{\prime }-L_n)$ and $n!e^{-x}(L_n^{\prime \prime }-2L_n^{\prime }+L_n)$. Substituting and cancelling $n!e^{-x}$ givesThis is deriving the Laguerre equation from Rodrigues' formula, exhibiting $L_n$ as an eigenfunction of the Laguerre differential operator on polynomials.

Multiply the two displayed equations by $e^{-x}$ and subtract the one with indices reversed. The product terms combine into the Lagrange identity for a Sturm-Liouville operator:Integrating from zero to infinity givesAll factors $L_j,L_j^{\prime }$ are polynomials. The endpoint term at zero vanishes because of the factor $x$, and the one at infinity vanishes because the exponential function dominates every polynomial. Therefore, if $m\ne n$,The integral is absolutely convergent. This is orthogonality in the weighted inner product with weight $e^{-x}$; the boundary cancellation is the self-adjoint mechanism behind the Laguerre differential operator on polynomials.

For real positive weight $w$, the weighted inner product isThe functions must have finite weighted norm; the conjugate is omitted for real functions. Multiplying the differential equation by $x{e}^{-x}$ givesThus the Sturm-Liouville form has $p=x{e}^{-x}$, $q=0$ and $w=e^{-x}$. With the positive-sign spectral convention defineThe existing Laguerre differential operator on polynomials uses the opposite operator $L=-\mathcal{A}$, with eigenvalues $-n$; keeping this sign convention explicit avoids confusing the two descriptions.

For admissible $f,g$, integration by parts givesThe boundary form must vanish for every pair in the self-adjoint domain. In the usual regular-at-zero Laguerre realization one imposes ${\lim }_{x\downarrow 0}x{e}^{-x}{y}^{\prime }=0$, excluding the logarithmic branch; at infinity one uses the weighted square-integrable domain with vanishing boundary form. A condition $y(\infty )=0$ would be inappropriate for polynomial eigenfunctions. For any pair of polynomials, the factor $x$ makes the boundary form zero at zero, while the exponential beats all polynomial growth at infinity. These functions also have finite weighted norms.

Self-adjointness yields $(n-m)⟨y_m,y_n{⟩}_w=0$, giving orthogonality of Sturm-Liouville eigenfunctions:For the specified polynomials, direct differentiation gives $\mathcal{A}u=u$ and $\mathcal{A}v=3v$. Hence their eigenvalues in the displayed convention are 1 and 3, respectively, or −1 and −3 for $L=-\mathcal{A}$.

To find the quadratic, put $h=a{x}^2+bx+c$. The equation $x{h}^{\prime \prime }+(1-x)h^{\prime }+2h=0$ reduces to $(4a+b)x+(b+2c)=0$. Thus $b=-4a$, $c=2a$. Since $h(1)=-a=1$,These are scalar multiples of the Laguerre polynomials of degrees one, three and two.

For $n=0$, the first integral is $x{e}^{-x}{y}^{\prime }=C$, so all solutions with the prescribed interior normalization areTwo independent choices areThey both equal one at $x_0$, and their Wronskian is $e^x/x\ne 0$. For sufficiently large $x$, the last unit interval contributes at least $e^{x-1}/x$, proving that $y_2\to \infty$. These are the zero-eigenvalue solutions of the Laguerre differential equation, with the second expressible through the exponential integral. It also has a logarithmic singularity at zero and is not an admissible polynomial or weighted square-integrable eigenfunction. An interior value alone does not enforce the endpoint conditions of the spectral problem.

At eigenvalue zero the Laguerre differential operator on polynomials gives $x{y}^{\prime \prime }+(1-x)y^{\prime }=0$, equivalently $(x{e}^{-x}{y}^{\prime }{)}^{\prime }=0$. Thus $y^{\prime }=D{e}^x/x$ and the displayed solutions follow from the exponential integral. For any $x_0>0$, all solutions with $y(x_0)=1$ can be written $y(x)=1+D{\int }_{x_0}^x{e}^{t}dt/t$. The choices $D=0$ and $D=1$ are linearly independent despite sharing that one prescribed value.

The second branch grows as $e^x/x$ at infinity and has a logarithmic singularity at zero. It is consequently excluded from the regular weighted square-integrable eigenproblem with weight $e^{-x}$. Only the constant zero mode remains in that self-adjoint realization. A normalization at one interior point alone is not enough to select a unique second-order solution.