Let \(f:\mathbb R\to\mathbb R\) be continuous and satisfy \[ |f(x)|\le\frac{C}{1+x^2} \] for some \(C>0\). Define \[ F(x)=\sum_{n=-\infty}^{\infty}f(x+n). \]
- Prove that \(F\) is continuous and \(1\)-periodic.
- If \(G\) is continuous and \(1\)-periodic, prove that \[ \int_0^1F(x)G(x)\,dx =\int_{-\infty}^{\infty}f(x)G(x)\,dx. \]
Solution
1. The series\[ \sum_{n\in\ZZ}f(x+n) \]converges absolutely and uniformly on every compact interval.
Fix a compact interval \(K=[a,b]\) and set \[ R\coloneqq\max\{\abs{a},\abs{b}\}. \] If \(x\in K\) and \[ \abs{n}\geq2R+1, \] then \[ \abs{x+n} \geq \abs{n}-\abs{x} \geq \abs{n}-R \geq \frac{\abs{n}}{2}. \] Hence \[ \abs{f(x+n)} \leq \frac{C}{1+(x+n)^2} \leq \frac{4C}{n^2}. \] The series \[ \sum_{n\in\ZZ\setminus\{0\}}\frac1{n^2} \] converges. For the finitely many \(n\) with \(\abs{n}<2R+1\), the term \(\abs{f(x+n)}\) is bounded by \(C\) on \(K\), so the Weierstrass M-test gives uniform absolute convergence on \(K\).
2. The function\[ F(x)=\sum_{n\in\ZZ}f(x+n) \]is continuous on \(\RR\).
Each function \[ x\longmapsto f(x+n) \] is continuous. By step 1, their series converges uniformly on every compact interval. Therefore its sum is continuous on every compact interval, hence on all of \(\RR\).
3. The function \(F\) is \(1\)-periodic.
Absolute convergence from step 1 permits reindexing: \[ \begin{aligned} F(x+1) &= \sum_{n\in\ZZ}f(x+1+n)\\ &= \sum_{m\in\ZZ}f(x+m)\\ &= F(x), \end{aligned} \] where \(m=n+1\).
4. If \(G\) is continuous and \(1\)-periodic, then \(G\) is bounded on\(\RR\).
Continuity on the compact interval \([0,1]\) gives a number \(M_G\) such that \[ \abs{G(x)}\leq M_G \] for \(x\in[0,1]\). For arbitrary \(y\in\RR\), choose \(k\in\ZZ\) with \(y-k\in[0,1]\). Periodicity gives \[ G(y)=G(y-k), \] so the same bound holds on all of \(\RR\).
5. One may integrate the defining series for \(F\) term by term against\(G\) on \([0,1]\):\[ \int_0^1F(x)G(x)\,dx = \sum_{n\in\ZZ} \int_0^1 f(x+n)G(x)\,dx. \]
By step 1, the series defining \(F\) converges uniformly on \([0,1]\). By step 4, \(G\) is bounded there. Hence \[ \sum_{n\in\ZZ}f(x+n)G(x) \] also converges uniformly on \([0,1]\). Termwise integration of a uniformly convergent series of continuous functions is therefore valid.
6. For each \(n\in\ZZ\),\[ \int_0^1 f(x+n)G(x)\,dx = \int_n^{n+1}f(y)G(y)\,dy. \]
Use the substitution \[ y=x+n. \] Then \[ \begin{aligned} \int_0^1 f(x+n)G(x)\,dx &= \int_n^{n+1}f(y)G(y-n)\,dy\\ &= \int_n^{n+1}f(y)G(y)\,dy, \end{aligned} \] because \(G\) has period \(1\) and \(n\in\ZZ\).
7. The function \(fG\) is absolutely integrable on \(\RR\).
By step 4, \[ \abs{f(x)G(x)} \leq \frac{CM_G}{1+x^2}. \] The majorant is integrable on \(\RR\), so \[ \int_{-\infty}^{\infty}\abs{f(x)G(x)}\,dx<\infty. \]
8. Therefore\[ \boxed{ \int_0^1F(x)G(x)\,dx = \int_{-\infty}^{\infty}f(x)G(x)\,dx }. \]
By steps 5 and 6, \[ \int_0^1F(x)G(x)\,dx = \sum_{n\in\ZZ} \int_n^{n+1}f(y)G(y)\,dy. \] Absolute integrability from step 7 allows the unit intervals \([n,n+1]\) to be summed over all \(n\in\ZZ\), giving exactly the improper integral over \(\RR\).
9. Q.E.D.
Steps 2 and 3 prove part 1, and step 8 proves part 2.