@BalarkaSen okay according to the question f is a continuous function on the interval [0,1]
Let's define the function
g(x)=f(x)-x let's take g(x)=0 then f(x)=x and x=0
similarly if we take x=1 and g(1)=0 then
f(x)=x otherwise f(x) is smaller than or equal to 1 and similarly if g(0) is not equal to 0 then f(0)-0 is greater then or equal to 0.in that case by the indeterminate value theorem there will be a value c $$c \to [0,1]$$ such that g(c) =0 and f(c)=c and therefore c is the fixed point
The Eigen value $\lambda(t)$ which is characterised by the Rayleigh quotient (where $t$ is a scalar variable):
$$R(u,\Omega_t)= \frac{\int_{\Omega_t} |\nabla u|^2 dy }{\int_{\Omega_t} u^2 dy}$$
$\lambda(\Omega_t) = \mathbf{min} \{R(v, \Omega_t) : v \in H^{1,2}(\Omega_t)\}$
The minimiser $u*$ o...
