For \(f\) a real-valued function on the real line, define the function \(\Delta f\) by \(\Delta f(x)=f(x+1)-f(x)\). For \(n\ge2\), define \(\Delta^n f\) recursively by \(\Delta^n f=\Delta(\Delta^{n-1}f)\). Prove that \(\Delta^n f=0\) if and only if \(f\) has the form \[ f(x)=a_0(x)+a_1(x)x+\cdots+a_{n-1}(x)x^{n-1}, \] where \(a_0,a_1,\ldots,a_{n-1}\) are periodic functions of period \(1\).
Solution
For \(k\geq0\), write \[ B_k(x)=\binom{x}{k}, \] where \(B_0(x)=1\) and \[ B_k(x)=\frac{x(x-1)\cdots(x-k+1)}{k!} \] for \(k\geq1\).
1. If \(a\) is periodic of period \(1\), then\[ \Delta\bigl(aB_0\bigr)=0 \]and, for every \(k\geq1\),\[ \Delta\bigl(aB_k\bigr)=aB_{k-1}. \]
Since \(a(x+1)=a(x)\), \[ \begin{aligned} \Delta\bigl(aB_k\bigr)(x) &= a(x+1)B_k(x+1)-a(x)B_k(x)\\ &= a(x)\bigl(B_k(x+1)-B_k(x)\bigr). \end{aligned} \] For \(k=0\), the last difference is \(0\). For \(k\geq1\), Pascal’s polynomial identity gives \[ B_k(x+1)-B_k(x)=B_{k-1}(x). \] This proves both formulas.
2. If\[ f(x)=\sum_{k=0}^{n-1}a_k(x)B_k(x) \]with each \(a_k\) periodic of period \(1\), then \(\Delta^n f=0\).
By step 1, each application of \(\Delta\) lowers the index of every nonzero binomial term by one and annihilates the \(B_0\) term. Hence \[ \Delta^{k+1}\bigl(a_kB_k\bigr)=0 \] for every \(0\leq k\leq n-1\). Since \(k+1\leq n\), one has \[ \Delta^n\bigl(a_kB_k\bigr)=0 \] for every term. Linearity of \(\Delta^n\) therefore gives \(\Delta^n f=0\).
3. Conversely, if \(\Delta^n f=0\), then there are period-one functions \(a_0,\ldots,a_{n-1}\) such that\[ f(x)=\sum_{k=0}^{n-1}a_k(x)B_k(x). \]
Proceed by induction on \(n\), beginning with the auxiliary case \(n=1\).
If \(\Delta f=0\), then \[ f(x+1)=f(x) \] for every \(x\), so \(f\) itself is period one and \[ f(x)=f(x)B_0(x). \]
Now suppose the assertion holds for \(n-1\), and assume \(\Delta^n f=0\). Put \[ g=\Delta f. \] Then \[ \Delta^{n-1}g=0. \] By the induction hypothesis, there are period-one functions \(b_0,\ldots,b_{n-2}\) such that \[ g(x)=\sum_{k=0}^{n-2}b_k(x)B_k(x). \] Define \[ h(x)=\sum_{k=0}^{n-2}b_k(x)B_{k+1}(x). \] Step 1 gives \[ \Delta h=g=\Delta f. \] Therefore \[ \Delta(f-h)=0, \] so \[ a_0=f-h \] is period one. Taking \[ a_{k+1}=b_k \] for \(0\leq k\leq n-2\) yields \[ f(x)=a_0(x)+\sum_{k=0}^{n-2}a_{k+1}(x)B_{k+1}(x), \] which is the required representation.
4. A representation\[ f(x)=\sum_{k=0}^{n-1}a_k(x)B_k(x) \]with period-one coefficients exists if and only if a representation\[ f(x)=a_0'(x)+a_1'(x)x+\cdots+a_{n-1}'(x)x^{n-1} \]with period-one coefficients exists.
The polynomials \[ B_0(x),B_1(x),\ldots,B_{n-1}(x) \] form a basis of the real polynomials of degree less than \(n\): each \(B_k\) has degree \(k\) and leading coefficient \(1/k!\), so the change-of-basis matrix from \((B_0,\ldots,B_{n-1})\) to \((1,x,\ldots,x^{n-1})\) is triangular with nonzero diagonal.
Consequently each \(B_k\) is a constant linear combination of \(1,x,\ldots,x^k\), and each \(x^k\) is a constant linear combination of \(B_0,\ldots,B_k\). Passing from either representation to the other therefore replaces the coefficient functions only by constant linear combinations of them. Constant linear combinations of period-one functions are again period one.
5. The stated characterization holds:\[ \Delta^n f=0 \quad\Longleftrightarrow\quad f(x)=a_0(x)+a_1(x)x+\cdots+a_{n-1}(x)x^{n-1} \]for period-one functions \(a_0,\ldots,a_{n-1}\).
If \(\Delta^n f=0\), step 3 gives a binomial-basis representation and step 4 converts it to the power-basis representation in the statement. Conversely, a power-basis representation converts by step 4 to a binomial-basis representation, and step 2 then gives \(\Delta^n f=0\).
6. Q.E.D.
Step 5 proves both implications.