Norm convergence of functionals to the Riemann integral

Question

Consider $\ell_{n} \in C[0,1]^{'}$ s.t $\ell_{n}(f)=\frac{1}{n} \sum_{k=0}^{n-1}f(\frac{k}{n})$, does $\ell_{n}$ converge in norm to Riemann integation?

I think yes; since every contionous function is Riemann integable, thus $\lim_{n} \sup_{\mid x\mid=1} | \ell_{n} - R | = 0$

@DanielFischer aahh, since you can always "inductivly" define a "worse" function? which make a uniform bound impossible? — user123124, Dec 29 '15 at 11:54
Basically. For every given $n \in \mathbb{N}$ and $\varepsilon > 0$, you can easily construct a function $f \in C[0,1]$ with $\lVert f\rVert_{\infty} = 1$, $\ell_k(f) = 0$ for all $k \leqslant n$ and $\int_0^1 f(x),dx > 1 - \varepsilon$. — Daniel Fischer, Dec 29 '15 at 11:57

score 1 · Accepted Answer · answered Dec 29 '15 at 12:19

1

The answer is NO.

Let $\,f_n(x)=\lvert\sin(2\pi n x)\rvert$. Then $\max_{x\in[0,1]}\lvert\,f(x)\rvert=1$, $\,\ell(\,f)=\int_0^1 f_n(x)\,dx=\frac{2}{\pi}$, while $\ell_n(\,f_n)=0$.

answered Dec 29 '15 at 12:19

Yiorgos S. Smyrlis

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Norm convergence of functionals to the Riemann integral

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