Polynomial interpolation

[math8]

Let $$x_{0}, x_{1}, \cdots , x_{n}$$ be distinct points in the interval [a,b] and $$f \in C^{1}[a,b]$$.

We show that for any given $$\epsilon >0$$ there exists a polynomial p such that

$$\left\| f-p \right\|_{\infty} < \epsilon$$ and $$p(x_{i}) = f(x_{i})$$ for all $$i=1,2, \cdots , n$$

I know $$\left\| f\right\| _{\infty}= max_{x \in [a,b]}|f(x)|$$ and I wonder if the polynomial they are asking for is the Lagrangian polynomial interpolating f at the nodes $$x_{0}, x_{1}, \cdots , x_{n}$$. If yes, I am not sure how to prove the problem.

 

[mighty2000]

Thank you for your post. You are correct that the polynomial they are asking for is the Lagrangian polynomial interpolating f at the given points. To prove this, we can use the Weierstrass Approximation Theorem which states that for any continuous function f on a closed interval [a,b], there exists a sequence of polynomials that converge uniformly to f on that interval.

Since f \in C^1[a,b], it is also continuous on [a,b]. Thus, by the Weierstrass Approximation Theorem, there exists a sequence of polynomials p_n that converge uniformly to f on [a,b]. This means that for any given \epsilon >0, there exists an N such that for all n > N, we have \left\| f-p_n \right\|_{\infty} < \epsilon.

Now, let's consider the Lagrangian polynomial p(x) that interpolates f at the given points x_0, x_1, \cdots , x_n. By construction, we know that p(x_i) = f(x_i) for all i=1,2, \cdots , n. We can also show that p(x) converges uniformly to f on [a,b]. This is because p(x) is a polynomial, and as n \rightarrow \infty, the degree of p_n increases, making the approximation of f by p_n better and better. Therefore, for any given \epsilon >0, there exists an N such that for all n > N, we have \left\| f-p \right\|_{\infty} < \epsilon.

Therefore, we have shown that for any given \epsilon >0, there exists a polynomial p that satisfies the conditions stated in the forum post. Thus, the problem is proved.

I hope this helps. Let me know if you have any further questions or if you need any clarification.