I want to write a function in terms of its mollification and higher order forward differences. Given a function $u:\mathbb{R}\rightarrow\mathbb{R}$ and $h>0$, we set $u_{h}(x):=\frac{1}{h}u\left( \frac{x}{h}\right) $, $x\in\mathbb{R}$. We also define inductively $\Delta_{h}^{1}u(x)=u(x+h)-u(x)$ and $\Delta_{h}^{m}u(x)=\Delta_{h}^{1}(\Delta_{h}^{m-1}u(x))$.
Ilin proved the following result (the proof is in the first volume of O. V. Besov, V. P. Ilin and S. M. Nikolski, “Integral Representation of Functions and Embedding Theorem,”) Let $u:\mathbb{R}\rightarrow\mathbb{R}$ be a function of class $C^{m}$, $m\in\mathbb{N}$. Then for every $h>0$ and $x\in\mathbb{R}$, \begin{align*} u(x) & =\int_{\mathbb{R}}u\left( x+y\right) \omega_{h}\left( y\right) \,dy\\ & \quad+\int_{0}^{h}\frac{1}{\xi}\int_{\mathbb{R}}\phi_{\xi}(t)\int% _{\mathbb{R}}\Delta_{\delta z}^{(m)}u(x+t+z)\psi_{\xi}(z)\,dzdtd\xi, \end{align*} where $\omega,\phi,\psi\in C_{c}^{\infty}\left( \mathbb{R}\right) $, with support contained in $[0,1]$ and $\int_{\mathbb{R}}\omega(x)\,dx=1$, and $0<\delta<1$.
The proof is rather long. When $m=1$ there is a much simpler representation formula, which I think is due to Grisvard.
Let $u\in C^{1}(\mathbb{R})$. Then for all $\ell>0$ and $x\in\mathbb{R}$, $$ u(x)=\frac{1}{\ell}\int_{0}^{\ell}u(x+t)\,dt-\int_{0}^{\ell}\frac{1}{y^{2}% }\int_{0}^{y}(u(x+y)-u(x+t))\,dtdy. $$ The proof goes as follows. By translation it is enough to prove the previous identity at $x=0$. Let $0<\varepsilon<\ell$. Using integration by parts we have \begin{align*} \int_{\varepsilon}^{\ell}\frac{1}{y^{2}} & \int_{0}^{y}(u(y)-u(t))\,dtdy=\int% _{\varepsilon}^{\ell}\frac{u(y)}{y}\,dy-\int_{\varepsilon}^{\ell}\frac {1}{y^{2}}\int_{0}^{y}u(t)\,dtdy\\ & =\int_{\varepsilon}^{\ell}\frac{u(y)}{y}\,dy-\int_{\varepsilon}^{\ell}% \frac{u(y)}{y}\,dy+\left[ \frac{1}{y}\int_{0}^{y}u(t)\,dt\right] _{y=\varepsilon}^{y=\ell}\\ & =\frac{1}{\ell}\int_{0}^{\ell}u(t)\,dt-\frac{1}{\varepsilon}\int% _{0}^{\varepsilon}u(t)\,dt\rightarrow\frac{1}{\ell}\int_{0}^{\ell }u(t)\,dt-u(0) \end{align*} as $\varepsilon\rightarrow0^{+}$ since $u$ is continuous at $0$. On the other hand, by the fundamental theorem of calculus $u(y)-u(t)=\int_{t}^{y}u^{\prime }(\tau)\,d\tau$. Hence, \begin{align*} \left\vert \frac{1}{y^{2}}\int_{0}^{y}(u(y)-u(t))\,dt\right\vert & \leq \frac{1}{y^{2}}\int_{0}^{y}\int_{t}^{y}|u^{\prime}(\tau)|\,d\tau dt\leq \frac{1}{y}\int_{0}^{y}|u^{\prime}(\tau)|\,d\tau.\\ & \leq\Vert u^{\prime}\Vert_{L^{\infty}((0,\ell))}. \end{align*} It follows by the Lebesgue dominated convergence theorem that $$ \int_{\varepsilon}^{\ell}\frac{1}{y^{2}}\int_{0}^{y}% (u(y)-u(t))\,dtdy\rightarrow\int_{0}^{\ell}\frac{1}{y^{2}}\int_{0}% ^{y}(u(y)-u(t))\,dtdy, $$ which concludes the proof.
I am interested in a representation of the type established by Grisvard for $m\geq2$ and which is simpler than the one proved by Ilin.
