[proofplan] We first express each forward difference as an integral of the derivative of the previous difference. Iterating this identity gives an $r$-fold integral formula for $\Delta_h^r f(x)$ over the cube $[0,h]^r$. Since every point sampled by the integral lies in $[a,b]$, the integrand is bounded by $\|f^{(r)}\|_\infty$, and the cube has $\mathcal{L}^r$-measure $h^r$. Taking the supremum over all admissible $h$ and $x$ gives the estimate. [/proofplan]

[step:Represent one forward difference as an integral of the derivative]For each $m \in \mathbb N$, let $\mathcal{L}^m$ denote $m$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb R^m$. Let $h>0$ and let $u \in [a,b-h]$. Since $f \in C^1([a,b])$, the restriction of $f$ to $[u,u+h]$ is continuously differentiable, and the [Fundamental Theorem of Calculus](/theorems/632) gives \begin{align*} \Delta_h^1 f(u)=f(u+h)-f(u)=\int_0^h f'(u+s)\,d\mathcal{L}^1(s). \end{align*} More generally, if $g \in C^1([a,b-(k-1)h])$ and $u \in [a,b-kh]$, then \begin{align*} \Delta_h^1 g(u)=\int_0^h g'(u+s)\,d\mathcal{L}^1(s). \end{align*} We will apply this identity successively to the functions $g=\Delta_h^{k-1}f$.[/step]

[guided]For each $m \in \mathbb N$, let $\mathcal{L}^m$ denote $m$-dimensional Lebesgue measure on $\mathbb R^m$. The basic identity is the integral form of a first forward difference. Fix $h>0$ and $u \in [a,b-h]$. The interval $[u,u+h]$ is contained in $[a,b]$, so $f'$ is continuous on this interval. Therefore the [Fundamental Theorem of Calculus](/theorems/632), applied to the continuously differentiable restriction $f|_{[u,u+h]}$, gives \begin{align*} f(u+h)-f(u)=\int_0^h f'(u+s)\,d\mathcal{L}^1(s). \end{align*} By the definition of the first forward difference, the left-hand side is $\Delta_h^1 f(u)$, so \begin{align*} \Delta_h^1 f(u)=\int_0^h f'(u+s)\,d\mathcal{L}^1(s). \end{align*} This is the only idea that will be iterated. At the next stage, the role of $f$ is played by $\Delta_h^1 f$; then by $\Delta_h^2 f$; and so on. The domain condition $u \in [a,b-kh]$ ensures that all points appearing after $k$ forward steps remain inside $[a,b]$.[/guided]

[step:Iterate the integral identity to obtain an $r$-fold formula] We prove by induction on $k \in \{1,\dots,r\}$ that, for every $h>0$ and every $x \in [a,b-kh]$, \begin{align*} \Delta_h^k f(x)=\int_{[0,h]^k} f^{(k)}(x+s_1+\cdots+s_k)\,d\mathcal{L}^k(s). \end{align*} Here $s=(s_1,\dots,s_k) \in [0,h]^k$. The case $k=1$ is the identity from the previous step. Assume the formula holds for some $k<r$. Then $\Delta_h^k f \in C^1([a,b-kh])$, and for $x \in [a,b-(k+1)h]$, \begin{align*} \Delta_h^{k+1}f(x)=\int_0^h \frac{d}{d\tau}\bigl(\Delta_h^k f\bigr)(x+\tau)\,d\mathcal{L}^1(\tau). \end{align*} Differentiating the inductive formula under the integral is justified because $f^{(k+1)}$ is continuous on $[a,b]$ and the integration domain $[0,h]^k$ is compact. Thus \begin{align*} \frac{d}{d\tau}\bigl(\Delta_h^k f\bigr)(x+\tau)=\int_{[0,h]^k} f^{(k+1)}(x+\tau+s_1+\cdots+s_k)\,d\mathcal{L}^k(s). \end{align*} For $(\tau,s) \in [0,h]\times [0,h]^k$, the map \begin{align*} (\tau,s) \mapsto f^{(k+1)}(x+\tau+s_1+\cdots+s_k) \end{align*} is continuous on the compact cube $[0,h]\times [0,h]^k$, hence integrable with respect to $\mathcal{L}^1 \otimes \mathcal{L}^k$. Substituting the differentiated formula and identifying $[0,h]\times[0,h]^k$ with $[0,h]^{k+1}$ by the map $(\tau,s_1,\dots,s_k)\mapsto (s_1,\dots,s_{k+1})$ with $s_{k+1}=\tau$, the product measure identity $\mathcal{L}^1 \otimes \mathcal{L}^k=\mathcal{L}^{k+1}$ gives \begin{align*} \Delta_h^{k+1}f(x)=\int_{[0,h]^{k+1}} f^{(k+1)}(x+s_1+\cdots+s_{k+1})\,d\mathcal{L}^{k+1}(s). \end{align*} This completes the induction. Taking $k=r$ gives the desired representation of $\Delta_h^r f(x)$. [/step]

[step:Bound the $r$-fold integral by the supremum norm of $f^{(r)}$] Let $0<h\le t$ and $x \in [a,b-rh]$. For every $s=(s_1,\dots,s_r) \in [0,h]^r$, \begin{align*} a \le x+s_1+\cdots+s_r \le x+rh \le b. \end{align*} Hence \begin{align*} |f^{(r)}(x+s_1+\cdots+s_r)|\le \|f^{(r)}\|_\infty. \end{align*} Using the integral formula from the previous step and the triangle inequality for integrals, \begin{align*} |\Delta_h^r f(x)|\le \int_{[0,h]^r} |f^{(r)}(x+s_1+\cdots+s_r)|\,d\mathcal{L}^r(s). \end{align*} Therefore \begin{align*} |\Delta_h^r f(x)|\le \int_{[0,h]^r} \|f^{(r)}\|_\infty\,d\mathcal{L}^r(s). \end{align*} Since $\mathcal{L}^r([0,h]^r)=h^r$, we obtain \begin{align*} |\Delta_h^r f(x)|\le h^r\|f^{(r)}\|_\infty. \end{align*} [/step]

[step:Take the supremum over admissible differences] Because $0<t\le (b-a)/r$, the interval $[a,b-rh]$ is nonempty for every $0<h\le t$. From the previous step, for every admissible $h$ and $x$, \begin{align*} |\Delta_h^r f(x)|\le h^r\|f^{(r)}\|_\infty\le t^r\|f^{(r)}\|_\infty. \end{align*} Taking first the supremum over $x \in [a,b-rh]$ and then the supremum over $0<h\le t$ gives \begin{align*} \omega_r(f,t)\le t^r\|f^{(r)}\|_\infty. \end{align*} This is the asserted derivative bound for the $r$-th modulus of smoothness. [/step]