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Proving Uniform Continuity of Periodic Functions on the Real Line

March 26, 2025Technology3290
Proving Uniform Continuity of Periodic Functions on the Real Line In t

Proving Uniform Continuity of Periodic Functions on the Real Line

In this article, we will delve into the proof that a continuous and periodic function $f: mathbb{R} to mathbb{R}$ is uniformly continuous. This involves utilizing the properties of periodic functions and the definition of uniform continuity. The proof will be structured to ensure clarity and comprehensibility for both students and professionals alike. Let's begin by defining the necessary terms and then proceed with the proof.

Definitions

Periodic Function

A function $f$ is periodic with period $T geq 0$ if for all $x in mathbb{R}$, $f(x T) f(x)$.

Uniform Continuity

A function $f$ is uniformly continuous on a set $A$ if for every $epsilon > 0$, there exists a $delta > 0$ such that for all $x, y in A$ with $|x - y| , we have $|f(x) - f(y)| .

Proof

Compactness of the Period

Since a function $f$ is periodic with period $T$, it suffices to consider $f$ on one period, say the interval $[0, T]$. The function $f$ is continuous on this closed interval, which is a compact set.

Application of Heine-Cantor Theorem

The Heine-Cantor theorem states that any continuous function on a compact set is uniformly continuous. Therefore, since $f$ is continuous on the compact interval $[0, T]$, it follows that $f$ is uniformly continuous on $[0, T]$.

To show that $f$ is uniformly continuous on $mathbb{R}$, we need to demonstrate that the uniform continuity on $[0, T]$ extends to the entire real line.

- Let $epsilon > 0$ be given. By the uniform continuity of $f$ on $[0, T]$, there exists $delta > 0$ such that for any $x, y in [0, T]$ with $|x - y| , we have $|f(x) - f(y)| .

- Consider points outside the interval: For any $x, y in mathbb{R}$ such that $|x - y| , we can find integers $n, m in mathbb{Z}$ such that $x$ and $y$ can be expressed in the form $x a nT$ and $y b mT$ for $a, b in [0, T]$.

- Without loss of generality, assume $n m$. If not, we can shift $x$ or $y$ by $T$ to bring them into the same period. Then, since $f$ is periodic, we have $f(x) f(a nT) f(a)$ and $f(y) f(b nT) f(b)$.

-Bounding the Difference:

- If $|x - y| and both $x$ and $y$ belong to the same period, then $|f(x) - f(y)| directly follows from the uniform continuity on $[0, T]$.

Conclusion

Since any two points in $mathbb{R}$ can be brought into the same period using the periodicity of $f$, we conclude that $f$ is uniformly continuous on $mathbb{R}$.

Thus, we have shown that if $f$ is continuous and periodic, then $f$ is uniformly continuous on $mathbb{R}$.

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