Show that Riemann sum equals integral

Suppose we have some continuous function $\textbf{F}\colon \mathbb{R}^{n} \to \mathbb{R}^n$ (i.e. a force field) and a curve $\gamma\colon [a, b]\subseteq \mathbb{R} \to \mathbb{R}^{n}$ in $C^1$. Considering the partition $P$ of the interval $[a, b]$ \begin{align*} P= \left\{t_{0}=a, t_{1}=a + \frac{b-a}{n}, a + 2 \frac{b-a}{n} , \ldots,t_{n}= b\right\} \implies \Delta t = \frac{b-a}{n} .\end{align*} I want to show that \begin{align*} \lim_{n \to\infty} \sum_{k=1}^{n}\left\langle \textbf{F}\left(\gamma(t_{k}^{*})\right), \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)\right\rangle \Delta t &= \int_{a}^{b}\left\langle \textbf{F}\left(\gamma(t)\right),\gamma'(t)\right\rangle \mathrm{d}t \\ &= \lim_{n \to\infty} \sum_{k=1}^{n}\left\langle \textbf{F}\left(\gamma(t_{k}^{*})\right),\gamma'(t_{k}^{*}) \right\rangle\Delta t .\end{align*} So let $\epsilon>0$ be arbitrary. We can apply the mean value theorem componentwise, namely for all $1\le i\le m$ we have \begin{align*} \forall x,y \in [a, b] \colon \gamma_{i}(y)-\gamma_{i}(x)=\gamma_{i}'(\xi_i) (y-x), \quad \xi_i \in [x, y] .\end{align*}

By uniform continuity of $\gamma'$ there exists some $\delta$ s.t. for all $x$ and for all $y \in (x-\delta, x+\delta)$ \begin{align*} \left\|\gamma'(y)-\gamma'(x) \right\|_{2} <\epsilon .\end{align*} Now, we choose $N>0$ s.t. \begin{align*} \frac{b-a}{N}< \delta .\end{align*} Then, for all $n\ge N$ we find \begin{align*} &\left| \sum_{k=1}^{n}\left(\left\langle \textbf{F}\left(\gamma(t_{k}^{*})\right), \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)\right\rangle \Delta t -\left\langle \textbf{F}\left(\gamma(t_{k}^{*})\right),\gamma'(t_{k}^{*}) \right\rangle\Delta t \right) \right | \\ = &\left| \sum_{k=1}^{n}\Delta t\left\langle \textbf{F} \left(\gamma(t_{k}^{*})\right), \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right) -\gamma'(t_{k-1})+\gamma'(t_{k-1}) -\gamma'(t_{k}^{*})\right\rangle \right | \\ \le & \Delta t\sum_{k=1}^{n}\left\| \textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2}\left\| \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)-\gamma'(t_{k-1})+\gamma'(t_{k-1}) -\gamma'(t_{k}^{*})\right\|_{2} \\ \le &\Delta t\sum_{k=1}^{n} \left\| \textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2}\left(\left\| \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)-\gamma'(t_{k-1})\right\|_2+\left\|\gamma'(t_{k-1}) -\gamma'(t_{k}^{*})\right\|_{2}\right) \\ < &\Delta t \sum_{k=1}^{n} \left\|\textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2} \cdot \epsilon(\sqrt{m} +1) .\end{align*} The last inequality follows from the fact that \begin{align*} \frac{\gamma_{i}(t_{k})-\gamma_{i}(t_{k-1})}{t_{k}-t_{k-1}} = \gamma_{i}'(\xi_{i}), \quad \xi_{i} \in (t_{k-1}, t_{k}) \end{align*} and since $\xi_i-t_{k-1}<\delta$ we have \begin{align*} |\gamma_{i}'(\xi_{i})-\gamma_{i}'(t_{k-1})| < \epsilon \end{align*} for all $1\le i\le m$.

Since $\left\|\textbf{F}\left(\gamma(t)\right)\right\|_{2} $ is a continuous function on a compact set it takes a maximal value, let this be $M$. It follows that \begin{align*} \Delta t \sum_{k=1}^{n} \left\|\textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2} \cdot \epsilon(\sqrt{m} +1) \le (b-a)^{2} m\cdot M\cdot \epsilon(\sqrt{m} +1) .\end{align*} Choosing $\epsilon$ in the beginning accordingly concludes the proof.

Solution 1:

The overall structure of your proof is good but there is a problem with applying the MVT since $\gamma$ is a vector valued function.

So I just write what you need to fix avoiding some detail also.

Then $$\left |\sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma(t_k)-\gamma(t_{k-1})\right> - \sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma' (t_k^*) \right> \Delta t\right|$$$$=\left |\sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma(t_k)-\gamma(t_{k-1})-\gamma' (t_k^*) \Delta t\right> \right|$$$$\le\sum_{k=1}^n \|\mathbf F(\gamma(t_k^*))\|_2\cdot \|\gamma(t_k)-\gamma(t_{k-1})-\gamma' (t_k^*) \Delta t\|_2$$$$=\sum_{k=1}^n \|\mathbf F(\gamma(t_k^*))\|_2\cdot \|\theta_k(t_k)-\theta_k(t_{k-1})\|_2$$where $\,\theta_k (t)=\gamma (t)-\gamma'(t_k^*)\;t$ .

Apply now to $\,\theta_k\,$ a weakened version of the MVT valid for a vector valued function.
See Baby Rudin, third ed., p. 113 .

Then for every $k$ there exists $\xi_k \in [t_{k-1},t_k]$ such that$$\|\theta_k(t_k)-\theta_k(t_{k-1})\|_2\le \|\theta_k'(\xi_k)\|_2\,\Delta t$$

Since $$\theta_k'(\xi_k)=\gamma' (\xi_k)-\gamma' (t_k^*)$$you get finally$$\left |\sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma(t_k)-\gamma(t_{k-1})\right> - \sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma' (t_k^*) \right>\Delta t \right|$$$$\le \sum_{k=1}^n \|\mathbf F(\gamma(t_k^*))\|_2\cdot \|\gamma' (\xi_k)-\gamma' (t_k^*)\|_2\,\Delta t$$Anyway know that you can succeed even using just the classic monodimensional MVT: this means you must work componentwise of course

Edit

By the uniform continuity of $\gamma'$, for every $\varepsilon>0$ there exists $\delta_i>0$ such that $$|\gamma_i'(x)-\gamma_i'(y)|<\varepsilon\quad\text{if}\quad |x-y|<\delta_i$$Let $\,\delta=\min\, {\delta_i}$ .

Then, by the monodimensional MVT, there exists $\,\xi_{ki}\in[t_{k-1},t_k]\,$ such that $$\|\gamma(t_k)-\gamma(t_{k-1})-\gamma' (t_k^*) \Delta t\|_2$$$$= \sqrt {\sum_{i=1}^m (\gamma_i' (\xi_{ki})-\gamma_i' (t_k^*))^2}\,\Delta t < \sqrt m\, \varepsilon\, \Delta t\quad \text{if}\quad \Delta t < \delta$$