Proof: Inequality in Mercer's theorem

Question

In the outline of the Mercer's theorem proof there is an inequality assumed without any explanation:

$$\sum_{i=0}^{\infty} \lambda_i \vert e_i(t) e_i(s) \vert \le \sup_{x \in [a,b]} \vert K(x,x)\vert^2$$

Why does this need to hold?

Answer 1

The Parseval-Bessel Theorem leads to $$f= \sum_{j=1}^\infty (f,e_j)\, e_j~~ \text{for every }~~f\in L^2(a,b) \tag{1}$$ Which implies by linearity and continuity along with the fact that $Ke_j =\lambda_je_j$ that
$$ T_Kf =Kf =\sum_{j=1}^\infty \lambda_j\,(f,e_j)\, e_j~~ \text{for every }~~f\in L^2(a,b) \tag{2}$$ Hence, one can carefully check starting with finite summation and Bessel inequality $$ (Kf,f)= \sum_{j=1}^\infty \lambda_j\,|(f,e_j)|^2 \tag{3}$$ We also set the Kernel $$K_n(t,s) = \sum_{j=1}^{n} \lambda_j e_j(t) e_j(s)\tag{Kn}$$

then,

$$TK_nf(t) = \sum_{j=1}^{n} \lambda_j e_j(t) \int_{a}^{b}f(s)e_j(s)\,ds = \sum_{j=1}^{n} \lambda_j (f\, ,e_j)e_j(t) $$ hence, $$ (K_nf,f)= \sum_{j=1}^{n} \lambda_j |(f\, ,e_j)|^2.$$

Next we consider the truncated Kernel $$ R_n(t,s) =K(t,s)- \sum_{j=1}^n \lambda_j\,e_j(t)\, e_j(s)\tag{4}$$

It derives from the foregoing that

$$ (R_nf,f)= \sum_{j=n+1}^\infty \lambda_j\,|(f,e_j)|^2\ge 0~~\text{for every }~~f\in L^2(a,b) \tag{5}$$

i.e $(R_nf,f)\ge0$ then there is one result claiming that $R_n(t,t)\ge0$ for almost every $t\in(a,b)$ which leads to $$ R_n(t,t) =K(t,t)- \sum_{j=1}^n \lambda_j\,e_j(t)\, e_j(t)\ge 0 \tag{6}$$ ie $$ \sum_{j=1}^n \lambda_j\,e_j(t)\, e_j(t)\le K(t,t) \tag{7}$$ This holds true for abitratry $n\in\mathbb{N}$. whence,

$$ \sum_{j=1}^\infty \lambda_j\,e_j(t)\, e_j(t)\le \sup_{t\in [a,b]} K(t,t) $$ Applying Cauchy-Schwartz inequality we get, \begin{split} \Big|\sum_{j=1}^\infty \lambda_j\,e_j(s)\, e_j(t)\Big|^2 &\le& \Big(\sum_{j=1}^\infty \lambda_j\,e_j(s)\, e_j(s)\Big ) \Big(\sum_{j=1}^\infty \lambda_j\,e_j(t)\, e_j(t)\Big)\\ &\le& \sup_{t\in [a,b]} K^2(t,t) \end{split}

Prove of the Claim in the complex case suppose there is $x_0$ such that $K(x_0,x_0)<0$ then there are $c,d $ such that $a\le c