# Variational formulation for first eigenfunction for compact self-adjoint integral operators, and request for references

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).

Background/theory:

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!!!

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 $$Kf=\lambda f$$.