Let f and g be perodic functions, not necessary same period, and $\lim_{x \rightarrow \infty}f(x)-g(x)=0$. ST f(x)=g(x) for all x.x

Question

Can anyone suggest me some hint?

I define a function $\phi(x)=f(x)-g(x)$. we want to show that $\phi(x)=0$.

I think the functions $f(x)$ and $g(x)$ may not be continuous, so we cannot use IVT for this question.

Can anyone suggest me some hint?

Answer 1

First, notice that if $f$ and $g$ have a common period $T > 0$ then the result is easily obtained by noticing that if $x \in \mathbb{R}$ and $k \in \mathbb{N}$ then we have $$ f(x) - g(x) = f(x + k T) - g(x + k T) $$ and then letting $k$ tends to $+ \infty$.

Hence, we only need to prove that $f$ and $g$ have a common period. Let then $T,T' > 0$ be periods respectively for $f$ and for $g$. Then for $x \in \mathbb{R}$ and $k \in \mathbb{N}$ write $$ \begin{split} g(x+T) - g(x) & = g(x + T + k T') - g(x + kT') \\ & = \left[g(x+ T + k T') - f(x + T + k T') \right] - \left[g(x+kT') - f(x + kT')\right]. \end{split} $$ In the second line, we introduced artificially the term $f(x + kT') - f(x +T+ kT') = 0$. Then, $x$ being fixed, we may let $k$ tends to $+ \infty$. We see that both terms between brackets tend to $0$, so that $g(x) = g(x+T)$. That is, $T$ is a period for $g$, which ends the proof since $T$ is by definition a period for $f$.