I am currently studying Sturm-Liouville theory and I have a doubt about a proof I found in the book Sturm-Liouville Theory and Its Applications by M. A. Al-Gwaiz, about the eigenvalues of the SL operator.
For context, Al-Gwaiz defines the self-adjoint differential operator
$$L=p\frac{d^2}{dx^2}+p'\frac{d}{dx}+r,$$ under the boundary conditions \begin{align} \alpha_1u(a) & +\alpha_2u'(a)=0, & \lvert\alpha_1\rvert & +\lvert\alpha_2\rvert>0,\\ \beta_1u(b) & +\beta_2u'(b)=0, & \lvert\beta_1\rvert & +\lvert\beta_2\rvert>0. \end{align}
Edit
In the following lemma (p.70-71) he's studying the regular SL eigenvalue problem, $$Lu+\lambda\rho(x)u=0,\quad x\in(a,b).$$ where $p>0$, and for simplicity he takes the weight function as $\rho(x)=1$, so the SL eigenvalue problem takes the form $$Lu+\lambda u=0,\quad x\in(a,b).$$
Lemma 2.22. The eigenvalues of $-L$ are bounded below by a real constant.
Proof.
[...]
Seeking a contradiction, suppose $-L$ has three linearly independent eigenfunctions $u_1$, $u_2$, and $u_3$ with their corresponding eigenvalues $\lambda_1$, $\lambda_2$, and $\lambda_3$ all less than $\ell$. We can assume, without loss of generality, that the eigenfunctions are orthonormal. Since \begin{align} \alpha_1u_i(a) & +\alpha_2u_i'(a)=0, \\ \beta_1u_i(b) & +\beta_2u_i'(b) =0, & i=1,2,3, \end{align} we see that each of the six vectors $\bigl(u_i(a),u_i'(a)\bigr)$ and $\bigl(u_i(b),u_i'(b)\bigr)$ lies in a one-dimensional subspace of $\mathbb{R}^2$. Therefore the three vectors $\mathbf{u}_i=\bigl(u_i(a),u_i'(a),u_i(b),u_i'(b)\bigr)$ lie in a two-dimensional subspace of $\mathbb{R}^4$. We can therefore form a linear combination $c_1\mathbf{u}_1+c_2\mathbf{u}_2+c_3\mathbf{u}_3$, where not all the coeffcients are zeros, such that $$c_1\mathbf{u}_1+c_2\mathbf{u}_2+c_3\mathbf{u}_3=\mathbf{0}.$$ But this implies \begin{align} c_1u_1(a)+c_2u_2(a)+c_3u_3(a) & =0,\\ c_1u_1(b)+c_2u_2(b)+c_3u_3(b) & =0. \end{align} The function $$v(x)=c_1u_1(x)+c_2u_2(x)+c_3u_3(x)$$ is therefore an eigenfunction of $-L$ which satisfies $v(a)=v(b)=0$, and consequently its eigenvalue is bounded below by $\ell$. But this is contradicted by the inequality $$\langle-Lv,v\rangle=\lambda_1\lvert c_1\rvert^2+\lambda_2\lvert c_2\rvert^2+\lambda_3\lvert c_3\rvert^2<\ell(\lvert c_1\rvert^2+\lvert c_2\rvert^2+\lvert c_3\rvert^2)=\ell\lVert v\rVert^2.\quad\quad\square$$
I fail to see why he claims in the last paragraph that $v(x)$ is an eigenfunction of $-L$. Of course, I know that $L$ is a linear operator and $$-Lv=-c_1Lu_1-c_2Lu_2-c_3L_3=-c_1\lambda_1u_1-c_2\lambda_2u_2-c_3\lambda_3u_3,$$ but I can't follow from there.
P.S.: I am not interested in another proof, I have found a few that use other methods and I have no problem with them. I only want to know what I am not understanding in this case.
P.P.S.: Sorry for the long post. I'm sure the answer is a silly thing I'm no seeing and thought I should cite the relevant part of the proof fort those who don't have access to the book.
As far as I can tell, that’s an error in the proof, but the proof nevertheless works if you just delete “is an eigenfunction of $-L$ which” (leaving “The function $v(x)$ therefore satisfies $v(a)=v(b)=0$”) because the lower bound $\ell$ for the norm of $v$, though stated only for eigenfunctions, was in fact derived (in Equation $(2.38)$) “for any $u\in C^2([a,b])$” without assuming that $u$ is an eigenfunction.