1. ## Polynomial interpolation

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.

2. Lol you might wanna check out your post again... Whatever you copied of physics forum has been saved as an image so won't show up I guess.

3. Its probably cached by their browser so they can't tell. I'll try to repair it...

Originally Posted by 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 $||f-p||_\infty < \epsilon$ and $p(x_i)=f(x_i)$ for all $i=1,\cdots, n$.

I know $||f||_\infty = \max_{x\in[a,b]}|f(x)|$ and I wonder if the polynomial they are asking for is the Lagrangian polynomial interpolating f at the nodes $x_1,\cdots,x_n$. If yes, I am not sure how to prove the problem.

4. I'll give it a shot I guess...

Assume WLOG that both a and b are in the $x_i$, and replace f with f-g, where g is the polynomial which interpolates the nodes $x_i$. Clearly if f-g can be approximated by a polynomial, f can be approximated by that polynomial plus g. So we have to find a polynomial which is within $\epsilon$ of f and vanishes at $x_i$.

Chuse as g a polynomial which is everywhere within $\epsilon\over 2n$ of f by the Stone–Weierstrass theorem. Let $d_i = f(x_i) - g(x_i)$ which is no more than $\epsilon\over 2n$ in magnitude. We will chuse polynomials $h_i$ to add to g to correct these errors so that the final approximating polynomial is $g+\sum h_i$. Then $||f-g-\sum h_i||_\infty \le ||f-g||_\infty + \sum ||h_i|| \le {\epsilon\over 2n} + n{\epsilon \over 2n} \le \epsilon$ if all the $||h_i||_\infty \le {\epsilon\over 2n}$.

Therefore we have only to prove the following claim: for every $\epsilon>0$ and $x,y\in[a,b]$ ( $y<\epsilon$), there is a polynomial h which is everywhere less than $\epsilon$ in magnitude on [a,b] and h(x)=y.

To do this, chuse a cosine wave centered at x and scale it down to amplitude y. We can take its Taylor polynomial to enough terms that inside [a,b] it is closer to zero at every point than the cosine wave [perhaps someone should check this, as I only have a rough sketch], so that this Taylor polynomial is y at x and it is $\le \epsilon$ on [a,b], as desired.

...god, what a mess. Its probably wrong too. Be warned!