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$ \newcommand{\bN}{\mathbb{N}} \newcommand{\eps}{\varepsilon} \newcommand{\diff}{\mathop{}\!\mathrm{d}} $ We fix $T \in (0, \infty)$ and $f \in L^\infty (0, T)$. Let $n \in \bN^*$ and $\eps_n := T/n$. Let $t_k^n := k \eps_n$ for $k \in \{0, \ldots,n\}$. Let $\tau^n_t := t_{k+1}^n$ if $t \in [t_k^n, t_{k+1}^n)$ for some $k \in \{ 0, \ldots,n-1\}$.

It is mentioned (without proof) in my lecture notes that

$$ \lim_{n \to \infty} n \int_0^T (\tau^n_t - t) f(t) \diff t = \frac{T}{2} \int_0^T f(t) \diff t . $$

Below is my approach. Is there simpler alternative approach without using properties of $L^p$ spaces?

We have $L^\infty (0, T) \subset L^1 (0, T)$. On the other hand, the space of Lipschitz functions on $(0, T)$ is dense in $L^1 (0, T)$. Hence we can safely assume $f$ is Lipschitz and thus Riemann integrable. It suffices to prove \begin{align} \lim_{n \to \infty} \sum_{k=0}^{n-1} \int_{t^n_k}^{t^n_{k+1}} n (t^n_{k+1} - t) f(t) \diff t = \lim_{n \to \infty} \frac{T}{2} \sum_{k=0}^{n-1} (t^n_{k+1} - t^n_k) f(t_k) . \end{align}

Let $L$ be the Lipschitz constant of $f$. Notice that $|n (t^n_{k+1} - t)| \le T$ and $|f(t) - f(t^n_k)| \le \frac{LT}{n}$ for $t \in [t_k^n, t_{k+1}^n)$. Then \begin{align} & \Big | \sum_{k=0}^{n-1} \int_{t^n_k}^{t^n_{k+1}} n (t^n_{k+1} - t) f(t) \diff t - \sum_{k=0}^{n-1} \int_{t^n_k}^{t^n_{k+1}} n (t^n_{k+1} - t) f(t_k) \diff t \Big | \\ \le & \frac{LT^2}{n} \sum_{k=0}^{n-1} (t^n_{k+1} - t^n_{k}) = \frac{LT^3}{n}. \end{align}

Notice that $t^n_{k+1} - t^n_k = \frac{T}{n}$. Then \begin{align} & \sum_{k=0}^{n-1} \int_{t^n_k}^{t^n_{k+1}} n (t^n_{k+1} - t) f(t_k) \diff t - \frac{T}{2} \sum_{k=0}^{n-1} (t^n_{k+1} - t^n_k) f(t_k) \\ = & \sum_{k=0}^{n-1} \int_{t^n_k}^{t^n_{k+1}} n (t^n_{k+1} - t) f(t_k) \diff t - \frac{1}{2} \sum_{k=0}^{n-1} \int_{t^n_k}^{t^n_{k+1}} n (t^n_{k+1} - t^n_k) f(t_k) \diff t \\ = & n \sum_{k=0}^{n-1} f(t_k) \int_{t^n_k}^{t^n_{k+1}} \frac{t^n_{k+1} + t^n_k - 2t}{2} \diff t = 0 . \end{align}

This completes the proof.

Mittens
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Akira
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1 Answers1

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The integral in the OP can be expressed as an integral over oscillating periodic functions. Define the function $$\phi(t)=\lceil t\rceil -t=1+\lfloor t\rfloor -t=1-\{t\}$$ The function $\phi$ is $1$-periodic.

The change of variables $u=t/T$ yields $$I_n=n\int_{(0,T]}(t^n_{k+1}-t)f(t)\,dt=T^2\int_{(0,1]}\phi(nu) f(Tu)\,du$$ By Fejer's theorem (a cousin of the Riemann-Lebesgue theorem)

$$I_n\xrightarrow{n\rightarrow\infty}T^2\Big(\int_{(0,1]}\phi(u)\,du\Big)\,\int_{(0,1]}f(Tu)\,du=\frac{T}{2}\int_{(0,T]}f(s)\,ds$$

Mittens
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