formula for repeated finite differences

Question

I am looking for a proof of a well-known fact, whose proof must be very easy, though I've been struggling to find it. Let $\Delta$ be the map from real-valued functions of a real variable, given by $(\Delta g)(x) = g(x+1/2)-g(x-1/2)$. Let $u(x)=0$ if $x\le 0$ and $u(x)=x$ otherwise. We set $0^0 =0$. Then, for smooth functions $h$, and $n>0$, $$(\Delta^n h)(0) = \frac{1}{(n-1)!}\int_{-\infty}^\infty h^{(n)}(x). (\Delta^n(u^{n-1}))(x)\,dx.$$ Here $h^{(n)}$ denotes the $n$th derivative of $h$.

I tried proving this by induction and also by using the exact integral remainder in Taylor's Theorem, but couldn't get either of these two obvious approaches to work. I seem to be missing some trick in manipulation.

Answer 1

Taylor $$g(y)=\sum_{k=0}^{n-1} g^{(k)}(x)\frac{(y-x)^k}{k!}+\int_x^y \frac{(y-z)^{n-1}}{(n-1)!} g^{(n)}(z)dz.$$ By induction on can prove $$\Delta^n g(x)=\sum_{k=0}^n \binom{n}{k}(-1)^k g(x+n/2-k).$$ Note that $\Delta^n p=0$ for polynomials $p$ of degree smaller than $n$. Now \begin{eqnarray} \Delta^n(u^{n-1})(x)&=&\sum_{k<n/2+x} \binom{n}{k}(-1)^k (x+n/2-k)^{n-1}\\ &=&-\sum_{k\geq n/2+x} \binom{n}{k}(-1)^k (x+n/2-k)^{n-1}\\ &=&\sum_{j<n/2-x} \binom{n}{j}(-1)^j (-x+n/2-j)^{n-1}\\ &=&\Delta^n(u^{n-1})(-x) \end{eqnarray} and \begin{eqnarray} \Delta^nh(0)&=&\sum_{k=0}^n \binom{n}{k}(-1)^k h(n/2-k)\\ &=&\sum_{k=0}^n \binom{n}{k}(-1)^k \left(\sum_{l=0}^{n-1} h^{(l)}(0)\frac{(n/2-k)^l}{l!}+\int_0^{n/2-k} \frac{(n/2-k-z)^{n-1}}{(n-1)!} h^{(n)}(z)dz\right)\\ &=&\sum_{k=0}^n \binom{n}{k}(-1)^k \int_0^{n/2-k} \frac{(n/2-k-z)^{n-1}}{(n-1)!} h^{(n)}(z)dz\\ &=&\frac{1}{(n-1)!}\int_{0}^{n/2} h^{(n)}(z) \sum_{k<n/2-z} \binom{n}{k}(-1)^k (n/2-k-z)^{n-1}dz\\ &&+\frac{1}{(n-1)!}\int_{-n/2}^{0} h^{(n)}(z) \sum_{k>n/2-z} \binom{n}{k}(-1)^k (n/2-k-z)^{n-1}dz\\ &=&\int_{-\infty}^{\infty} h^{(n)}(z) \Delta^n(u^{n-1})(z)dz. \end{eqnarray}

Answer 2

In fact, the proof by induction works perfectly well, starting with the obvious case $n=1$. To go from $n-1$ to $n$, note:
1. If $j\ge 1$, $\Delta^i(u^j)$ is a $C^{j-1}$-function.
2. If $i>j\ge1$, then $\Delta^i(u^j)$ has compact support. (By induction on $i$, $\Delta^i(u^j)$ is zero on $[-\infty,0]$ and a polynomial of degree $j-i$ near $\infty$.)
3. For any integrable function $g$, $\int_{-\infty}^\infty (\Delta g)(x)\,dx=0$.
4. $(\Delta(\alpha.\beta))(x)=\alpha(x+1/2)(\Delta\beta)(x) +(\Delta\alpha)(x)\beta(x-1/2).$

Taking $\alpha(x)=h^{(n-1)}(x)$ and $\beta(x)=\Delta^{n-1}u^{n-2}(x+1/2)$, we find $$(\Delta^n h)(0) = (\Delta^{n-1}(\Delta h))(0)= \frac1{(n-2)!}\int\Delta h^{(n-1)}(x).\Delta^{n-1}u^{n-2}(x)\,dx\text{ induction hypothesis}\\ =\frac{-1}{(n-2)!}\int h^{(n-1)}(x+1/2).\Delta^n u^{n-2}(x+1/2)\,dx \text{ by 3 and 4 above}\\ =\frac{-1}{(n-2)!}\int h^{(n-1)}(x).\Delta^n u^{n-2}(x)\,dx\\ = \frac{1}{(n-1)!}\int h^{(n)}(x).\Delta^n u^{n-1}(x)\,dx\text{ by integration by parts}.$$