Consider the following functional equation

begin {equation *}

phi (x) = x phi (a + (1-a) x) + (1-x) phi ((1-b) x), phi (0) = 0, phi (1) = 1,

end {equation *}

when $ a> frac {1} {2} $ It can be studied using shrinkage mappings.

Consider the Banach space of all limited lipschitz functions in $ (0.1) $ such that for each $ f $ in the space,

begin {equation *}

m (f) = sup_ {x neq y} frac { left vert

f (x) -f (y) right vert} { left vert x-y right vert}.

end {equation *}

We denote by $ CL 0 the subspace of all functions $ f $ for which $ f (0) = 0 $. Clearly

begin {equation *}

left Vert f right Vert _ {CL ^ {0}} = m (f)

end {equation *}

it's a norm but $ CL 0 that is, with the $ left Vert cdot right Vert _ {CL ^ {0}} $, a Banach space.

Theorem: for $ a in ( frac {1} {2}, 1) $mapping

begin {equation *}

(T phi) (x) = x phi (a + (1-a) x) + (1-x) phi ((1-b) x)

end {equation *}

is a contraction mapping in $ CL ^ {0} cap lbrace f, f in CL ^ {0}, f (1) leq 1 rbrace. $

I have two questions:

1) Can I replace the limited Lipschitz functions with the continuous linear operators? Does the space of continuous linear operators (under the given norm) form a Banach space?

2) How can I show that $ CL ^ {0} cap lbrace f, f in CL ^ {0}, f (1) leq 1 rbrace $ is closed?

Please guide me about it. Thank you