I recently came across the following lemma.

Lemma. Suppose $\Omega_n^N = [-\pi n, \pi n]^N$ and $f \in L^1(\Omega_n^N)$ is $2\pi n$-periodic. Let $A$ be a subset of $[-\pi n, \pi n]$. There exists some $\alpha_N \in [-\pi n, \pi n] \setminus A$ such that \begin{equation} \left\vert \int_{[-\pi n, \pi n]^{N-1} \times \lbrace \alpha_n \rbrace} f(x) dx \right\vert \leq \frac{1}{2 \pi n -\vert A \vert } \int_{\Omega_n^N} \vert f(x) \vert dx \tag{1} \end{equation}

Proof. Integrate the l.h.s. of (1) for $\alpha_N \in [-\pi n, \pi n] \setminus A$ and use the mean-value theorem.

Now there are a couple of things that seem strange.

  • First of all the lemma seems to be plain wrong if you choose $A=[-\pi n, \pi n]$. There is certainly no $\alpha_N \in \emptyset$. Let us therefore suppose that the author meant "proper subset".

  • Now let's do as the author tells us: We call the l.h.s of (1) \begin{equation} g(\alpha) := \left\vert \int_{[-\pi n, \pi n]^{N-1} \times \lbrace \alpha \rbrace} f(x) dx \right\vert \end{equation} and integrate \begin{equation} I:= \int_{[-\pi n, \pi n] \setminus A} g(\alpha) d \alpha \end{equation} How can one apply the mean value theorem here? $[-\pi n, \pi n] \setminus A$ is not an interval, it need not even be connected.

Can anyone tell me what the author actually meant, when he gave the idea of his proof? Any help would be greatly appreciated!


I think what the authors call the "mean-value theorem" is something like this: if $B$ is a measurable subset of $\mathbb R$ (with $\vert B\vert >0$) and $g:B\to \mathbb R$ is in $L^1(B)$, then one can find $\alpha\in B$ such that $g(\alpha)\leq \frac{1}{\vert B\vert} \int_B g(t)dt$. This is easy to check by contradiction.

Concerning your lemma, you may assume that $f\geq 0$ because in the left-hand side $\left\vert \int\right\vert$ is $\leq \int\vert\;\; \vert$. In other words, you may omit all absolute values.

Instead of $\int_{[-n\pi,n\pi]^{N-1}\times\{ \alpha\}} f(x) dx$, I will write $\int_{[-n\pi,n\pi]^{N-1}} f(x,\alpha)dx$, which seems more correct.

Applying the "mean-value theorem" to $B=[-n\pi,n\pi]\setminus A$ (which requires that $\vert A\vert\neq 2n\pi$) and $g(t)=\int_{[-n\pi,n\pi]^{N-1}} f(x,t)dx$, this gives the result because $$\int_B g(t)dt=\int_{[-n\pi,n\pi]^{N-1}\times B }f(x,t)dxdt\leq \int_{\Omega_n^N} f(u)du\, .$$

(Actually, I should be a little bit more precise: $g$ may take infinite values, but it is finite almost everywhere. Alternatively, the "mean-value theorem" still holds with functions with values in $[0,\infty]$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.