How to show $\|e^{ikx}-e^{ix}\|_{L^{1}}$ does not converge to $0$ when $k\rightarrow\infty$?

Question

I am in the middle of a proof, and I need to show that

$$\lim_{k\rightarrow\infty}\int_{-\pi}^{\pi}|e^{ikx}-e^{ix}|dx\ \text{does not converge to}\ 0.$$

For now, I got this:

\begin{align*} \int_{-\pi}^{\pi}|e^{ikx}-e^{ix}|dx=\int_{-\pi}^{\pi}|e^{ix}||e^{ix(k-1)}-1|dx&=\int_{-\pi}^{\pi}|e^{ix(k-1)}-1|dx\\ &=\int_{-\pi(k-1)}^{\pi(k-1)}\dfrac{1}{k-1}|e^{iy}-1|dy,\ \text{replacing}\ y:=x(k-1)\\ &=\int_{-\pi}^{\pi}|e^{iy}-1|dy,\ \text{since}\ e^{iy}\ \text{is}\ 2\pi-\text{periodic}. \end{align*}

Now we get rid of $k$, but what should I do to show the RHS is not zero?

Thank you!

Answer 1

Continuoing your last equation:

$\int_{-\pi}^{\pi} |e^{iy}-1| dy = \int_{-\pi}^{\pi} |cos(y) +isin(y) -1|dy =\int_{-\pi}^{\pi}\sqrt{(cos(y)-1)^2 +sin(y)^2)}dy =\int_{-\pi}^{\pi}\sqrt{2-2cos(y)}dy=2\sqrt{2}\int_0^\pi \sqrt{1-cos(y)}dy $.

Now, $\int_0^{\pi} \sqrt{1-cos(y)}dy =\int_0^{\pi} \sqrt{1-cos(y)} \dfrac{\sqrt{1+cos(y)}}{\sqrt{1+cos(y)}}dy=\int_0^{\pi} \dfrac{\sqrt{1-cos^2(y)}}{\sqrt{1+cos(y)}} dy$

Using $sin^2(y)+cos^2(y)=1$ in the last equation we get

$= \int_0^{\pi} \dfrac{\sqrt{sin^2(y)}}{\sqrt{1+cos(y)}}dy = \int_0^{\pi} \dfrac{sin(y)}{\sqrt{1+cos(y)}}dy$

Now using the substitution $u = 1+cos(y)$ we get

$= -\int_2^0\dfrac{1}{\sqrt{u}}du =\int_0^2\dfrac{1}{\sqrt{u}}du= 2\sqrt{u}|_0^2 = 2\sqrt{2}$.

Returnning to our equation we get :

$$ 2\sqrt{2}\int_0^\pi \sqrt{1-cos(y)}dy= 8$$

Answer 2

The primitive of $e^{ikx}-e^{ix} $ tends to $ \frac{1-e^{ix}}{i}$ uniformly, this is not the zero function