Prove that the mapping F:C[0,1]->C[0,1] dfnd by F(f)=q(x)f(x)+(0 to 1(defnt int)∫ p ((x)) f (x)) dx,((where

Question

Prove that the mapping F:C[0,1]->C[0,1] defined by F(f)=q(x)f(x)+(0 to 1(defnt int)∫ p ((x)) f (x)) dx,((where
q, p ∈ C[0, 1] are given fixed functions )), is contunious

Answer 1

We need to prove that for any f ∈ C[0,1] and any ε > 0 there exists a δ > 0 such that if ||f - g|| < δ for any g ∈ C[0, 1] then ||F(f) - F(g)|| < ε. So let f ∈ C[0,1] and ε > 0.

Note: in the following, all integrals are from 0 to 1, but I won't write this explicitly because it's awkward in text.

Now for g ∈ C[0,1] we have
F(g) - F(f) = (q(x)g(x) + ∫ (p(x) g(x)) dx) - (q(x)f(x) + ∫ (p(x) f(x)) dx)
= q(x) (g(x) - f(x)) + ∫ (p(x) (g(x) - f(x))) dx

By the triangle inequality, ||F(g) - F(f)|| <= ||q(x) (g(x) - f(x))|| + ||∫ (p(x) (g(x) - f(x))) dx||
<= ||q(x)|| ||g(x) - f(x)|| + ||p(x)|| ||g(x) - f(x)|| . 1
(note that all norms here are in C[0, 1] and not in the complex numbers; that is, they are function norms, not pointwise)
= ||g - f|| (||q|| + ||p||)
Hence, choose δ = ε / (||q|| + ||p||). Then if ||g - f|| < δ, ||F(g) - F(f)|| < δ (||q|| + ||p||) = ε as required.