Let's consider function $f: E \rightarrow K$, where $E \subset \mathbb C$. Let's consider Lagrange interpolation of $f$ for $a_1, a_2,...,a_n \in E$:
$$L_nf(x) = \sum_{j = 0}^n f(a_j)l_j(x)$$
where $l_j(x) = \prod_{k=0, k \neq j}^n\frac{x - a_k}{a_j - a_k}$
I want to prove that $\|L_n\| = \max_{x \in E}\sum_{j = 0}^n |l_j(x)|$.
My work so far
Let's write what is $\|L_nf\|$:
$$\|L_nf\| = \|\sum_{j = 0}^n f(a_j) l_j(x)\| = \max|\sum_{j = 0}^n f(a_j)l_j(x)| \le \max \sum_{j = 0}^n |f(a_j)l_j(x)| \le$$
$$\le \max\sum_{j =0 }^n \|f\| |l_j(x)| = \|f\|\max\sum_{ j =0}^n|l_j(x)|$$
So we have fact, that:
$$\frac{\|L_n(f)\|}{\|f\|} \le \max\sum_{j=0}^n|l_j(x)|$$
So now to obtain equality, we need to find such function $f$, for which $\frac{\|L_n(f)\|}{\|f\|} = \max \sum_{j = 0}^n|l_j(x)|$. Could you please give me a hand in finding such example?
My solution
Let's denote $x_0$ as: $\max_{x \in E}\sum_{j = 0}^n |l_j(x)| = \sum_{j = 0}^n |l_j(x_0)|$
Let's choose $f(x_j) = \textrm{sgn}l_j(x_0)$. The problem with $\textrm{sgn}$ is its definition on set of complex numbers. I am going to understand it as: $\textrm{sgn}(x) = \frac{|x|}{x}$. Then: $$\|L_nf\| \ge |L_n(f)(x_0)| = |\sum_{j = 0}^n \textrm{sgn}l_j(x_0) l_j(x_0)| = |\sum_{j = 0}^n |l_j(x_0)||$$
Now, since we only sum positive numbers (modules of complex number), we'll have that:
$$|\sum_{j = 0}^n |l_j(x_0)|| = \sum_{j = 0}^n |l_j(x_0)|$$
And that ends the proof