I have been studying the solutions $f(x)$, $x \in [0,1]$ to integral equations of the form \begin{equation} f(x) = \int_{0}^{1}K(x,y)\frac{f(y)}{\sqrt{h^{2}+(f(y))^{2}}}dy \end{equation} where the kernel $K(x,y)$ is a real-valued function over the domain $[0,1] \times [0,1]$ and it is bounded as $0 \leq K(x,y) \leq 1$. Here $h$ is a non-zero real-valued constant. Naturally $f(x) = 0 \ \forall x$ is always a solution but typically there is another, less trivial solution for some $-h_{c} \leq h \leq h_{c}$ where I call $h_{c}$ the critical value of $h$. In certain cases (such as when the kernel is degenerate) I know how to construct this solution.
I am wondering more generally if, given a continuous kernel $K(x,y)$ it is possible to analytically find the critical point $h_{c}$ above which only the trivial solution exists. Is it possible to use Banach's fixed point theorem to do this? E.g. identifying the values of $h$ at which there is only a single unique solution $f(x)$ (which presumably must be $f(x) = 0$) this would then tell us the points where the non-trivial solution can exist?
Let $A\colon \Bbb R\to \Bbb R$, $A(x)=\frac x{\sqrt{h^2+x^2}}$ and $$F\colon C([0,1])\to C([0,1]),\quad F(f)(x)=\int_0^1 K(x,y)A(f(y))dy.$$ Observe that $A$ is Lipschitz function satisfying with a Lipschitz constant $1/h$ (w.l.o.g. we assume $h>0$). Therefore, for any $f,g\in C([0,1])$ we have $$ |F(f)(x)-F(g)(x)| \leq \int_0^1K(x,y)|A(f(y))-A(g(y))|dy \leq \frac 1h\int_0^1K(x,y)|f(y)-g(y)|dy \\ \leq \frac 1h\|f-g\|_\infty\int_0^1K(x,y)dy \leq \frac{k}{h}\|f-g\|_\infty $$ for any $x\in [0,1]$, where $k:=\sup_x\int_0^1K(x,y)dy$. Therefore $$ \|F(f)-F(g)\|_\infty \leq \frac{k}{h}\|f-g\|_\infty. $$ This shows that $F$ is a contraction if $h>k$ and therefore the equation $F(f)=f$ has a unique solution. In this case the only solution to the given equation is a trivial one. Observe that the value $k$ may not be the optimal value and I don't know how to find the threshold (I suppose it's very difficult to find it analytically).