Let $\gamma:\mathbb{R}\to\mathbb{R}^n$. Suppose $\gamma$ is periodic function with period 1.

Question

Let $\gamma:\mathbb{R}\to\mathbb{R}^n$. Suppose $\gamma$ is periodic function with period 1. Let $a\in\mathbb{R}$ and

$$\eta=\gamma\big |_{[a;a+1]}:[a;a+1]\to\mathbb{R}^n$$ of $\gamma$. Prove:

• $\gamma$ is continuous iff $\eta$ is continuous

• $\gamma$ is smooth iff $\eta$ is smooth and $\eta^{(k)}(a)=\eta^{(k)}(a+1)$ for all $k$.

I know $\gamma$ continuous (smooth) then $\eta$ continuous (smooth) but i have no ideal to solve the converse.

Help me prove the converse by definition.

Answer 1

We have learnt in calculus 101 that $\lim_{x\to0} f(x)$ exists iff both one-sided limits $\lim_{x\to0-} f(x)$, $\lim_{x\to0+} f(x)$ exist and are equal. As a consequence $f'(0)$ exists iff both one-sided derivatives $f^{'+}(0)=\lim_{x\to0+}{f(x)-f(0)\over x}$ and $f^{'-}(0)$ exist, and are equal. Similarly for higher derivatives.

There is not more at stake in your problem. Note that $\eta^{(k)}(a+1)$ is meant to be a one-sided limit, which coincides with the left hand derivative $\gamma^{(k)-}(a)$ due to the assumed periodicity of $\gamma$.