Periodic Functions, Why does T have to be greater than 0

Question

If a periodic function can be describe has:

$$\forall x\in\mathbb{R},\exists t>0, st. f(x +t) = f(x)$$

Why does 't' have to be greater than 0?

Answer 1

Why does $t$ have to be greater than 0?

The definition does not work for $t=0$ obviously because then any function would be called periodic. Thus you have to exclude 0 at least.

And due to the symmetry of $=$, it makes no difference whether you use $t\neq 0$, $t>0$ or $t<0$:

$$f(x+t)=f(x) \iff f(x) =f(x+t) \iff f(x-t) =f(x) $$

where the rightmost equivalence follows if you replace $x$ by $x-t$ which can be done as $x\in\Bbb R$.

Answer 2

Actually, the idea is just to rule out $T=0.$ Now since if $T$ is a period then $kT$ is also a period for all nonzero integers $k,$ hence it suffices to define using $T>0.$

So why don't we want $T=0$? Because that would imply that there's no period.

Answer 3

If $t= 0$ then $f(x+0) = f(x)$ is trivially always true and not useful.

And the idea of a function being "periodic" means intuitively that the function well repeat on a regular pattern indefinitely because the will be an infinite number of $x_0=a$ for any real initial value; $x_1 = x_0 + t; x_2=x_0 + 2t; x_3 = x_0 + 3t$, etc where all the $f(x_i)$ are equal. Bot if $t = 0$ and $x_i = x_0 + i*0 = x_0=a$ and all the $x_i=a$ are equal then $f(a) = f(a)$ and.... that's pointless.

......

However there's an interesting counters example in finding what the value of $t$ is and what the least value of $t$ is.

$\sin x$ is periodic because for any $x$ we have $\sin (x+2\pi) = \sin x$ so $\sin x$ is periodic with $t =2\pi$.

But for a constant function $f(x) = c$. Then it is true that $f(x+2\pi) = c=f(x)$. Does that mean a constant function is periodic with period $t= 2\pi$? Well, no not really, as there's nothing special about $t= 2\pi$. We could have $t=$ anything and that would be true.