Upper bound of multi-dimensional integral

Question

It seems clear to me that for a function $f: [a,b] \rightarrow \mathbb{R}$, the following upper bound holds, $$ \left| \int_u^v f(x) dx \right| \leq ||f||_{\infty} h $$ where $h = b - a$ and $u,v \in [a,b]$. This seems clear just using the Riemann sum definition of an integral. A proof I am studying in a paper by Ciarlet, Schultz and Varga (1967), they use a higher dimensional version of this result (I am modifying their notation slightly for the purposes of asking this question). The result they use (last step of their Theorem 9) is: $$ \left| \int_{u_k}^{\xi} d v_k \int_{u_{k-1}}^{v_k} d v_{k-1} \cdots \int_{u_1}^{v_2} f(v_1) d v_1 \right| \leq ||f||_{\infty} \frac{h^k}{k!} $$ Here, $u_i,\xi$ are distinct points in $[a,b]$. My question is how does one prove this result, and in particular where does the $k!$ come from in the above result? Without the $k!$ this appears to be a trivial extension of the 1D case. Including the $k!$ makes a tighter upper bound, but I don't understand where it comes from.

Answer 1

Although the link from the comment answers it, you can also just prove it straightforwardly by induction. Putting $h=\lvert\xi-u_k\rvert$, you can estimate the left-hand side for $k+1$ by induction hypothesis after the substitution $v=\lvert v_{k+1}-u_{k+1}\rvert$ by $$\int_0^{\lvert\xi-u_{k+1}\rvert}\lVert f\lVert_\infty\frac{v^k}{k!}dv= \lVert f\lVert_\infty\frac{\lvert\xi-u_{k+1}\rvert^{k+1}}{(k+1)!}$$