How do we conclude that $f(x)=0, \forall x\in \mathbb{R}$ ?

Question

Suppose that $f:\mathbb{R}\rightarrow \mathbb{R}$ is a periodic function such that $\displaystyle{\lim_{x\rightarrow +\infty}f(x)=0}$.

I want to show that $f(x)=0$ for all $x\in \mathbb{R}$.

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Let $T$ be the period of $f$, then $f(x)=f(x+T)$.

Therefore, we have that $$0=\lim_{x\rightarrow +\infty}f(x)=\lim_{x\rightarrow +\infty}f(x+T)=\lim_{x\rightarrow +\infty}f(x+2T)=\dots =\lim_{x\rightarrow +\infty}f(x+nT), \ \forall n\in \mathbb{Z}$$

So, $$|f(x)|=|f(x+T)|=|f(x+2T)|=\dots =|f(x+nT)|\leq \epsilon$$

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But how exactly do we conclude that $f(x)=0, \forall x\in \mathbb{R}$ ?

Answer 1

Let $x\in \mathbb{R}$. Then $x_n:=x+nT$ tends to $+\infty$ ($T$ is the period of $f$) and $f(x_n)=f(x)$ for all $n$, since $f$ is $T-$periodic. Since $\displaystyle \lim_{x\rightarrow +\infty}f(x)=0$, we get that $\displaystyle \lim_{n\rightarrow +\infty}f(x_n)=0$ and by the uniqueness of the limit, we get $f(x)=0$, as desired.

Answer 2

$\lim_{x\to\infty}f(x)=0$ if and only if $\lim_{n\to\infty}f(x_n)=0$ for all sequences with $\lim x_n=\infty$. If there exist some $x\in\mathbb R$ for which $f(x)\neq0$, then let $x_n=x+nT$ and get a contradiction.

Answer 3

Hint :From your argument - if there is some point $a$ such that $f(a) \neq 0$ then there is a sequence of such points (In specific, a sequence that us not bounded)

Answer 4

Notice that $\lim_{x \rightarrow \infty} f(x) =0$ is equivalent to the condition that for all $\epsilon >0$ that there exists $N$ large enough such that we have $\left|f(x)\right| < \epsilon$ when $x>N$. There must also exist an integer $m$ such that $mT > N$. So then $\left|f(x)\right| = \left|f(x + mT)\right| < \epsilon$ and this was regardless of our choice of $\epsilon$.

Answer 5

You definitely have the right idea by showing that $\lim_{x \to +\infty} f(x)=\lim_{x \to +\infty} f(x+nT)$ for $n \in \Bbb{Z}$. However, we need to use that with the definition of limits to prove this theorem. Now, I am going to say that we are proving $f(x_0)=0$ for some arbitrary $x_0 \in \Bbb{R}$. That way, when I use $x$ below, I'm talking about a general input into $f$ while $x_0$ is the value we are trying to prove that $f(x_0)=0$ for the theorem.

Let's say we pick an arbitrary $\epsilon > 0$, as you did. Then, by the definition of limits as $x \to +\infty$, there exists an $M$ such that $|f(x)-0|=|f(x)|<\epsilon$ for all $x > M$. Now, let us choose $x > M$ such that $x=x_0+nT$ for some $n \in \Bbb{Z}$. If we choose $x=M+1=x_0+M-x_0+1=x_0+\frac{M-x_0+1}{T}T$, then we have $x > M$ and $x=x_0+nT$ for some $n \in \Bbb{R}$, but we can't be sure that $n$ is an integer. To mak $n$ an integer, we take the ceiling function of it, so we have $x=x_0+\lceil\frac{M-x_0+1}{T}\rceil T$. Now, $x > M$ still holds true, but we are sure that $n \in \Bbb{Z}$.

Now, since $x=x_0+nT$ for $n \in \Bbb{Z}$, we know that $f(x)=f(x_0)$. Thus, $|f(x_0)|<\epsilon$. We can do this for any $\epsilon > 0$, meaning we can bring $f(x_0)$ arbitrarily close to $0$. Now, we will prove $f(x_0)=0$ by contradiction.

Let's say $f(x_0) \neq 0$. Then, choose $\epsilon=\frac{|f(x_0)|}{2}$. Since $\epsilon > 0$, $|f(x_0)|<\frac{|f(x_0)|}{2}$. By dividing both sides by $|f(x)|$, we get $1<\frac 1 2$, which is a contradiction. Thus, our original assumption that $f(x_0) \neq 0$ is false and $f(x_0)=0$ for all $x_0 \in \Bbb{R}$.