Anonymous asked in Science & MathematicsMathematics · 1 decade ago

To mathematicians(real analysis)?

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)

2 Answers

  • Favorite Answer

    I'll assume that the topology on the space C[0,1] is induced by the sup-norm (that is, the distance between two functions f,g is given by sup |f(x)-g(x)| as x ranges over [0,1]; a famous theorem says that the set of polynomials is a dense subset of this space).

    Now it is a matter of playing games with the triangle inequality and the fact that the absolute value of the integral is smaller or equal to the integral of the absolute value. You'll need to use the fact that p and q are bounded functions in absolute value.

  • The sum or product of two continuous functions is continuous.

    The integral of a continuous function is continuous.

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