Constant function

[Papapetros Vaggelis]

Let \(\displaystyle{f:\mathbb{R}\to \mathbb{R}}\) be a continuous function such that

\(\displaystyle{f(x)=f(x+1)=f(x+2\,\pi)\,,\forall\,x\in\mathbb{R}}\).

Prove that \(\displaystyle{f}\) is constant on \(\displaystyle{\mathbb{R}}\).

[Tolaso J Kos]

Let \(\displaystyle{f:\mathbb{R}\to \mathbb{R}}\) be a continuous function such that

\(\displaystyle{f(x)=f(x+1)=f(x+2\,\pi)\,,\forall\,x\in\mathbb{R}}\).

Prove that \(\displaystyle{f}\) is constant on \(\displaystyle{\mathbb{R}}\).

Suppose ,for the sake of contradiction, that $f$ is non constant. As non constant, continuous and periodic $f$ has a least period. Let us denote that period with $T>0$. It is well known that every other period is a multiple of it. The assumptions tell us that both $1$ and $\pi$ are periods of our function. Those periods, however, cannot be both multiple of $T$. This implies that $f$ is constant.

Vaggelis, do we have a different arguement? Perhaps that $\mathbb{Z}+\pi \mathbb{Z}$ is dense on $\mathbb{R}$?

[Papapetros Vaggelis]

Hi Tolaso.

I have a different solution. Here it goes.

The function \(\displaystyle{f}\) is continuous and \(\displaystyle{2\,\pi}\) - periodical.

Let \(\displaystyle{k\in\mathbb{Z}-\left\{0\right\}}\). Then,

\(\displaystyle{\begin{aligned} f\hat(k)&=\dfrac{1}{2\,\pi}\,\int_{-\pi}^{\pi}f(x)\,e^{-i\,k\,x}\,\mathrm{d}x\\&=\dfrac{1}{2\,\pi}\,\int_{-\pi-1}^{\pi-1}f(x+1)\,e^{-i\,k\,(x+1)}\,\mathrm{d}x\\&=e^{-i\,k}\,\dfrac{1}{2\,\pi}\,\int_{-\pi}^{\pi}f(x)\,e^{-i\,k\,x}\,\mathrm{d}x\\&=e^{-i\,k}\,f\hat (k)\end{aligned}}\).

Since \(\displaystyle{k\in\mathbb{Z}-\left\{0\right\}}\), we have that \(\displaystyle{e^{-i\,k}\neq 1}\)

so, \(\displaystyle{f\hat(k)=0}\).

Considering the continuous and \(\displaystyle{2\,\pi}\) - periodical function

\(\displaystyle{g=f-f\hat(0):\mathbb{R}\to \mathbb{R}}\),

we get \(\displaystyle{g\hat (k)=0\,,\forall\,k\in\mathbb{Z}\,\,(\ast)}\) and then

\(\displaystyle{g=\mathbb{O}\implies f=f\hat(0)}\).

\(\displaystyle{(\ast)}\) :

\(\displaystyle{\begin{aligned} g\hat(0)&=\dfrac{1}{2\,\pi}\,\int_{-\pi}^{\pi}g(x)\,\mathrm{d}x\\&=f\hat(0)-f\hat(0)\\&=0\end{aligned}}\)

and for \(\displaystyle{k\neq 0}\),

\(\displaystyle{\begin{aligned} g\hat(k)&=\dfrac{1}{2\,\pi}\,\int_{-\pi}^{\pi}g(x)\,e^{-i\,k\,x}\,\mathrm{d}x\\&=\dfrac{1}{2\,\pi}\,\int_{-\pi}^{\pi}f(x)\,e^{-i\,k\,x}\,\mathrm{d}x-\dfrac{f\hat(0)}{2\,\pi}\,\int_{-\pi}^{\pi}e^{-i\,k\,x}\,\mathrm{d}x\\&=0\end{aligned}}\)