I'm trying to check/prove this variational formulation of the first eigenvector of a compact, self-adjoint integral operator, and I'm having a bit of difficulty (explained in the paragraph after the next one).


Let $\Omega \subset \mathbb{R}^d$ be a bounded open subset (domain). We alwas take integrals w.r. to the Lebesgue measure, in what follows. Let $k$ be a symmetric kernel i.e. (1) $k(x,y)=k(y,x)$, and it's a Hilbert-Schimidt kernel, i.e. (2)$k \in L^{2}(\Omega \times \Omega)$. So we've the compact, self-adjoint operator $K: L^{2}(\Omega)\to L^{2}(\Omega)$ defined by $K[u](x):= \int_{\Omega}k(x,y)u(y)dy$, and hence, by Hilbert-Schmidt theorem, can be expanded using its eigencalues and egenfunctions as: $K[u](x)=\Sigma_{n=1}^{\infty}\lambda_n \phi_n(x)\int_{\Omega}u(y)\phi_n(y)dy$ for some orthonormal basis (ONB for short) $\{\phi_n\}$ of $L^{2}(\Omega)$. This also implies and is implied by: $k(x,y)=\Sigma_{n=1}^{\infty}\lambda_n\phi_n(x)\phi_n(y)$

Variational formulation of the first/principal eigenfunction:

Consider the functional $J: L^{2}(\Omega) \to \mathbb{R}$ defined by $J[f, \lambda]:= \int_{\Omega \times \Omega} (k(x,y)-\lambda f(x) f(y))^{2}dxdy$. I'd like to prove (if possible!) that: the minimizer of $J$, say $f_{min}$, is the first/principal eigenfunction $f_1$ of $K$ with eigenvalue $\lambda_1$. To do so, we attempt to minimize $J$ w.r.t. $(f, \lambda)$ subject to the constraint $||f||_{L^2(\Omega)}=1$. To do so, I took the partial derivative of $J$ w.r.t. $f$, evaluate it at the $L^2$ function $g$ and set it equal to zero.

This gives, after some straightforward calculation: $\forall g \in L^2(\Omega)$

$$0 = \frac{\partial{J}}{\partial{f}}(g)= \int_{\Omega \times \Omega}[f(x)g(y)+g(x)f(y)][k(x,y)-\lambda f(x)f(y)]dxdy..........(1)$$

Now, in order to obtain that, $f_{min}$ defined by the solution of $(1)$ above, is an eigenfunction of $K$ with eigenvalue $\lambda_{min} = arg \hspace{1.5mm}min_{f, \lambda} J[f, \lambda] $, we must first have $K[f]=\lambda f$ from $(1)$ algebraically, which'll lead to $K[f_{1}]=\lambda_{1} f_{1}$.

And if we've $K[f]=\lambda f$, then it's equivalent to as having

$$ \int_{\Omega} k(x,y)f(y)dy = \lambda f(x) \forall g \in \mathcal{C}_c(\Omega) $$

The above is equivalent to having, using $||f||_{L^2(\Omega)}=1$,

$$ \int_{\Omega} [k(x,y)- \lambda f(x)\lambda f(y)]f(y)g(x)dxdy = 0 \forall g \in \mathcal{C}_c(\Omega) ..........(2) $$

My problem is/questions are:

1) I'm unable to show, if possible, how to obtain $(2)$ from $(1)$? Note that, by symmetry of the kernel $k, (2)\implies (1)$, but how to go the other way?

2) Is there a literature or reference could you point me to that I can look up to see this treatment in detail?

3) Can this variational characterization be related to the min-max variational characterization using the Rayleigh quotient?

Thank you so much!!!


1 Answer 1


If $K : \mathcal{H}\rightarrow\mathcal{H}$ is a self-adjoint linear operator on a Hilber space $\mathcal{H}$ such that $$ \lambda \le \frac{\langle Kf,f\rangle}{\langle f,f\rangle},\;\;\; f\in\mathcal{H}\setminus\{0\}, $$ then $\langle (K-\lambda I)f,f\rangle \ge 0$ for all $f\in\mathcal{H}\setminus\{0\}$, which means that $[f,g]=\langle (K-\lambda I)f,g\rangle$ satisfies all of the properties of an inner product except possibly positive definiteness. So the Cauchy-Schwarz inequality holds: $$ |[f,g]|^2 \le [f,f][g,g] $$ Setting $g=(K-\lambda I)f$ gives $$ \|(K-\lambda I)f\|^4 \le \langle (K-\lambda I)f,f\rangle[g,g] $$ If $\langle Kf,f\rangle =\lambda\langle f,f\rangle$, then the right side is $0$, which forces $Kf=\lambda f$.

  • $\begingroup$ Hi @DisintegratingByParts, it’s not the first time that I see you here! Nice to meet you. I’m working in this theme but for the second eigenvalue (yes! its a different question). Can I caracterize the second eigenvalue of the problem by min <Kf,f> ($\|f\|=1$)as well Dirichlet Laplacian Problem, but in $L^2(\Omega)$ ? $\endgroup$ Jan 29 at 20:12
  • $\begingroup$ You can minimize over unit vectors that are orthogonal to the first eigenspace. $\endgroup$ Jan 29 at 22:32

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