We fix two real numbers $a$ and $b$ such that $a < b$. For all integers $n \geqslant 1$ such that $a + \sqrt{n\lambda} > 0$, we define $$I_{n} = \{k \in \mathbb{N} \mid n\lambda + a\sqrt{n\lambda} \leqslant k \leqslant n\lambda + b\sqrt{n\lambda}\}$$ For $k \in \mathbb{Z}$, we define $x_{k,n} = \frac{k - n\lambda}{\sqrt{n\lambda}}$. We consider the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \mathrm{e}^{-\frac{1}{2}x^{2}}$ for all $x \in \mathbb{R}$, which is $M$-Lipschitz for some $M > 0$. a) Show that, if $x, h \in \mathbb{R}$ and $h > 0$, then $\left| hf(x) - \int_{x}^{x+h} f(t) \mathrm{d}t \right| \leqslant M \frac{h^{2}}{2}$. b) Deduce from this, when $I_{n}$ is non-empty, an upper bound for $$\left| \frac{1}{\sqrt{n\lambda}} \sum_{k \in I_{n}} f\left(x_{k,n}\right) - \int_{x_{p,n}}^{x_{q+1,n}} f(t) \mathrm{d}t \right|$$ where $p$ is the smallest element of $I_{n}$ and $q$ is the largest. c) Show that $$\lim_{n \rightarrow +\infty} \frac{1}{\sqrt{n\lambda}} \sum_{k \in I_{n}} f\left(x_{k,n}\right) = \int_{a}^{b} f(x) \mathrm{d}x$$
We fix two real numbers $a$ and $b$ such that $a < b$. For all integers $n \geqslant 1$ such that $a + \sqrt{n\lambda} > 0$, we define
$$I_{n} = \{k \in \mathbb{N} \mid n\lambda + a\sqrt{n\lambda} \leqslant k \leqslant n\lambda + b\sqrt{n\lambda}\}$$
For $k \in \mathbb{Z}$, we define $x_{k,n} = \frac{k - n\lambda}{\sqrt{n\lambda}}$. We consider the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \mathrm{e}^{-\frac{1}{2}x^{2}}$ for all $x \in \mathbb{R}$, which is $M$-Lipschitz for some $M > 0$.
a) Show that, if $x, h \in \mathbb{R}$ and $h > 0$, then $\left| hf(x) - \int_{x}^{x+h} f(t) \mathrm{d}t \right| \leqslant M \frac{h^{2}}{2}$.
b) Deduce from this, when $I_{n}$ is non-empty, an upper bound for
$$\left| \frac{1}{\sqrt{n\lambda}} \sum_{k \in I_{n}} f\left(x_{k,n}\right) - \int_{x_{p,n}}^{x_{q+1,n}} f(t) \mathrm{d}t \right|$$
where $p$ is the smallest element of $I_{n}$ and $q$ is the largest.
c) Show that
$$\lim_{n \rightarrow +\infty} \frac{1}{\sqrt{n\lambda}} \sum_{k \in I_{n}} f\left(x_{k,n}\right) = \int_{a}^{b} f(x) \mathrm{d}x$$