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Suppose that $f:[0,\infty)\rightarrow \mathbb C$ is a $C^1$ function satisfying $f(0)=0$ and $\int_0 ^\infty (|f(y)| + |f^{'}(y)|)dy<\infty$. Show that $$\left\vert \sum_{0\leq n< \infty} f(n) - \int_0 ^\infty f(y)dy \right|\leq \int_0 ^\infty |f^{'}(y)|dy.$$

This problem is from Barry Simon's Basic Complex Analysis, A Comprehensive Course in Analysis, Part 2A, problem 2.3.6.

Thank you.

This is a deleted question. People with good reps decided to close it down. I am reposting it here with a solution below because I think the idea behind this could save some time for the others. Also, the additional condition $f(0)=0$ is sufficient for the exercises that depend on this problem in the book.

Anindya Biswas
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1 Answers1

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Thanks to @ArcticChar, whose idea in the original post led me to this solution. However, I could not quite grasp his idea but it helped me significantly to find my solution.

First note that for any $C^1$ function $g:[0,1]\rightarrow \mathbb C$ satisfies $\int_0 ^1 g(t)dt=g(1)-\int_0 ^1 tg^{\prime}(t)dt$. So $|g(1)|\leq \int_0 ^1 (|g(t)|+|g^{\prime}(t)|)dt$ and $|g(1)-\int_0 ^1 g(t)dt|\leq \int_0 ^1 |g^{\prime}(t)|dt$ (note that $t\in [0,1]$). Now for each $n\geq 1$ we define $g_n (t)=f(n-1 +t)$. Then we have

$g_n (1)=f(n)$,

$\int_0 ^1 g_n(t)dt=\int_{n-1} ^n f(t)dt$,

$\int_0 ^1 |g_n^\prime(t)|dt=\int_{n-1} ^n |f^\prime(t)|dt$,

$|f(n)-\int_{n-1} ^n f(t)dt|\leq \int_{n-1} ^n |f^{\prime}(t)|dt$, and

$\Sigma_{n=1} ^\infty |f(n)|\leq \int_0 ^\infty (|f(t)|+|f^\prime (t)|)dt<\infty$ (hence the series is absolutely convergent).

Since $f(0)=0$ we see that

$|\Sigma^\infty _{n=0} f(n) -\int_0 ^\infty f(t)dt|\leq \Sigma^\infty _{n=1}|f(n)-\int_{n-1} ^n f(t)dt|\leq \Sigma^\infty _{n=1}\int_{n-1} ^n |f^{\prime}(t)|dt=\int_0 ^\infty |f^{\prime}(t)|dt$.

This completes the proof.

Anindya Biswas
  • 810