Riemann-Lebesgue Lemma With Infinite Orthonormal Base In an inner product space of infinite dimension, $V$, having an orthonormal base $\{e_k\}_{k=1}^\infty$,
There's the Riemann-Lebesgue lemma: $\forall u \in V$,  $\lim_{k\to\infty} \langle u, e_k\rangle = 0$.
Now looking at $\mathbb{R}^n$, with the series of $(1,1,1,1,1,1,\ldots)$,
How come the Riemann-Lebesgue Lemma doesn't comply with it?
Thanks.
 A: The most frequently seen infinite-dimensional inner product space is isomorphic to the space of all sequences $(a_n)_{n=1}^\infty$ of scalars satisfying $\sum_{n=1}^\infty |a_n|^2<\infty$. (If "scalars" means real numbers then one can write $a_n^2$ rather than $|a_n|^2$, but if it means complex numbers then one must write $|a_n|.)$ The sequence $(1,1,1,\ldots)$ is not a member of that space, since $1^2+1^2+1^2+\cdots\not<\infty.$
The Riemann–Lebesgue lemma applies to integrable functions, and "integrable" means the integral of the absolute value of the function is finite. Applied to functions on a bounded interval $[a,b]$, that means
$$
\int_a^b |f(x)|\,dx<\infty, \tag 1
$$
and on bounded intervals, that is a weaker statement than
$$
\int_a^b |f(x)|^2\,dx<\infty.
$$
Functions on a bounded interval that satisfy the condition $(1)$ stated above have Fourier series
$$
\sum_{n=-\infty}^\infty c_n e^{2\pi in(x-a)/(b-a)}
$$
and the Riemann–Lebesgue lemma says $c_{\pm n}\to0$ as $n\to\infty.$ But the Riemann–Lebesgue lemma is not applicable when the condition $(1)$ given above does not hold.
BTW you'll notice I wrote Reimann–Lebesgue and not Riemann-Lebesgue. Some of the more fastidious publishers and Wikipedia insist on the former punctuation.
