Question

Assume that f:[a,b]→R has continuous derivative on [a,b] and ϵ>0. Show that there exist a polynomial p:[a,b]→R such that supₓ∈[ₐ,ᵦ]

∣f(x)−p(x)∣<ϵ,supₓ∈[ₐ,ᵦ] . ∣f f(x)−p (x)∣<ϵ , supₓ∈[ₐ,ᵦ] f(x)−p (x)∣<ϵ

Answer 1

The Weierstrass Approximation Theorem states that for a continuous function with a continuous derivative on a closed interval, there exists a polynomial that can approximate the function within a given tolerance. Therefore, there exists a polynomial p(x) that can approximate f(x) within the given tolerance ϵ. This polynomial satisfies the specified suprema inequalities.

Step-by-step explanation:

To show that there exists a polynomial p(x) such that supₓ∈[ₐ,ᵦ] ∣f(x)−p(x)∣<ϵ, we can use the Weierstrass Approximation Theorem. This theorem states that if a function f(x) is continuous on a closed interval [a,b], then for any positive value of ϵ, there exists a polynomial p(x) such that |f(x)-p(x)|<ϵ for all values of x in the interval [a,b].

Since f(x) has a continuous derivative on [a,b], it is also continuous on [a,b]. Therefore, we can apply the Weierstrass Approximation Theorem to find a polynomial p(x) that approximates f(x) within a given tolerance ϵ.

To prove the suprema inequalities, we can use the fact that since p(x) approximates f(x) within a tolerance ϵ, the difference between f(x) and p(x) at any point x in the interval [a,b] will be less than ϵ. Therefore, the suprema of the absolute difference between f(x) and p(x) will also be less than ϵ.