Norm of a linear form in a function space

Question

I would appreciate some help with the following problem. Let $u, u_n:C([0,1])\to\mathbb{R}$ the linear forms defined by $$u(f)=\int_0^1f(x)dx\quad\&\quad u_n(f)=\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right).$$ I'm trying to prove that $$||u_n-u||=\sup_{||f||_\infty=1}|u_n(f)-u(f)|=2.$$

I already have that $||u_n-u||\leq2$ and I've tried to show that the $\sup$ is attained (I'm not sure that it is true) by using trigonometric functions but I haven't been able to get anything. Could anyone give me a hint?

Thanks in advance.

Answer 1

You indeed can do it using trigonometric functions. Following the suggestion of @Berci, define $$f_k(x) = -1+2|\sin(n\pi x)|^{1/k}$$ so that $u_n(f_k) = -1$ for all $k \in \Bbb{N}$. On the other hand, $f_k \to 1$ almost everywhere so by the Lebesgue dominated convergence theorem we have $$\lim_{k\to\infty} u(f_k) = \lim_{k\to\infty}\int_0^1 f_k(x)\,dx = \int_0^1 \lim_{k\to\infty} f_k(x) = 1.$$ Hence $$\|u_n-u\| \ge \lim_{k\to\infty} \frac{|u_n(f_k)-u(f_k)|}{\|f_k\|_\infty} = 2.$$