Rayleigh-Ritz Theorem Let $U$ be an $n$-dimensional subspace of $L:=L_2([-1,1])$. Let $F$ be an acting on $L$, given at $f \in L$
$$
(Ff)(x):=\int_{-1}^1 \frac{\sin a(x-y)}{(x-y)}f(y) dy, \quad x \in [-1,1], \quad a>0.
$$
Let $\lambda_n(a), \, n=0,1,2\ldots,$ be the eigenvalues of $F$. Using Rayleigh-Ritz Theorem show that for all $n>a$
$$
\lambda_n(a)\geq \min_{U/\{0\}}\frac{\|Ff\|_L}{\|f\|_L}.
$$
Thank you.
 A: Since no one more knowledgeable has answered this, let me make an attempt. I will look at a closely related operator namely
$$
(Ff)(x) =  \int_{-1}^1 \frac{\sin (a (x-y))}{ (x-y)} 1_{[-2,2]}(x-y) f(y) dy
$$
which seems to be exactly what you have (check?) Here $1_{[-2,2]}$ is the indicator of the set $[-2,2]$. Introducing it does not change the integral over $x \in [-1,1]$.
I will use $\|\cdot\|_2$ for the norm of $L^2([-1,1]) =: L^2[-1,1]$.
Now, we can use the corollary 4.28 of Brezis's functional analysis (p. 114 on English version) which says that if $g$ is a fixed function in $L^1(\mathbb{R})$ and $\mathcal{B}$ is a bounded set in $L^p(\mathbb{R})$ for $p \in [1,\infty)$, and we let $\mathcal{F} = g * \mathcal{B}$, then $\mathcal{F}|_{\Omega}$ has compact closure in $L^p(\Omega)$ for any measurable set $\Omega$ with finite measure. 
Apply the above with $g(y) = \frac{\sin(ay)}{y} 1_{[-2,2]}(y)$ and $\mathcal{B}= \{ f: \| f\|_2 \le 1\}$, $p = 2$ and $\Omega = [-1,1]$ to conclude that $F:L^2[-1,1] \to L^2[-1,1]$ is a compact operator.
Now, here is a general statement of the Rayleigh-Ritz from Garling's Inequalities (p. 246) 
Suppose that $T = \sum_{n=1}^\infty s_n(T) \langle \cdot,x_n\rangle y_n \in K(H_1,H_2)$ (that is compact from $H_1$ to $H_2$) where $(x_n)$ and $(y_n)$ are orthonomral sequences in $H_1$ and $H_2$, respectively, and $(s_n(T))$ is a decreasing sequence of non-negative real numbers, then
$$
 s_n(T) = \inf \big\{ \| T_{|J^\perp} \|:\; \text{dim} J < n\big\}
$$
where $ \|T_{|J^\perp}\|$ is the norm of $T$ restricted to $J^\perp$, that is
$$
  \|T_{|J^\perp}\| = \sup\{ \|T(x)\|:\; \|x\| \le 1, x \in J^\perp\}
$$
Note that this is also the same as
$$
\|T_{|J^\perp}\| = \sup\Big\{ \frac{\|T(x)\|}{\|x\|}:\; x \in J^\perp \setminus\{0\}\Big\}
$$
Take $T$ to be our $F$ operator (and $H_1 = H_2 = L^2[-1,1]$). Since $F$ is compact it has such a representation and putting the pieces together should give you the result you want for the singular values, and it seems that you are missing a $\perp$ in your statement. You might also try to show that the operator is positive (self-adjoint) in which case the result holds for eigenvalues too.
