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So we want to prove that if $f(x+P) = f(x)$ for some $P > 0$, and $f$ is continuous, then $f$ is uniform continuous.

I just started on this topic, so I'm really uneasy with these methods, but here's my attempt:

Since $f$ is continuous, it's continuous on [0,P+1], and here it must be uniform, so there's a $\delta '$ that solves our problem.

Now set |x-y| < 1 and assume that $x \le y$. Then pick an integer $k$ such that $$kP \le x \le kP + P$$ Since $|x-y| < 1$, we get $$kP \le x \le y \le kP + P + 1 $$ or $$0 \le x - kP \le y - kP \le P + 1$$ Then set $x' = x - kP$ and $y' = y - kP$. These are members of $[0,P+1]$, and thus we use that $sin(x') = sin(x)$ and same for $y$ and $y'$.

Then I can say all the stuff about $\delta$ and $\epsilon$ and how it works about, but I am unsure about above.... does it even make sense? Am I taking care of all $x,y \in \mathbb{R}$?

Chan Heni
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1 Answers1

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Here are a few points:

  • It seems that you mistook the function $f$ for the particular sine function.

  • To formalize your proof, you should say that $\delta=1$, for the particular $\epsilon$ used in the uniform continuity in the interval $[0,P+1]$.

Other than that, I think that's a great proof.

user1337
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