2
$\begingroup$

If we consider $0<r<1$ and $j\in\mathbb{N}$, is this inequality true?

$$\left|\sum_{k=0}^jr^ke^{ikt}\right|\leq \left|\sum_{k=0}^je^{ikt}\right|$$

EDIT: A counterexample is in the comments. I'm wondering, however, if there is actually an inequality when you take integrals:

$$\int_0^{2\pi}\left|\sum_{k=0}^jr^ke^{ikt}\right|dt\leq \int_0^{2\pi}\left|\sum_{k=0}^je^{ikt}\right|dt$$

$\endgroup$
3
  • 3
    $\begingroup$ $j = 1, t = \pi, r = .5$ $\endgroup$
    – mathworker21
    Oct 3, 2018 at 17:11
  • $\begingroup$ Thanks. Could be possible that although the inequality is not true for all $t$, the integrals of those modulus from $0$ to $2\pi$ actually held the inequality ? $\endgroup$
    – Mark_Hoffman
    Oct 3, 2018 at 17:19
  • 2
    $\begingroup$ not an answer, but just wanted to let you know that $$\int_0^{2\pi} |\sum_{k=0}^j r^ke^{ikt}|^2 dt \le \int_0^{2\pi} |\sum_{k=0}^j e^{ikt}|^2dt.$$ Indeed, the LHS is $\sum_{k=0}^j r^{2k}$ while the RHS is $j$. $\endgroup$
    – mathworker21
    Oct 3, 2018 at 17:29

1 Answer 1

Reset to default
0
$\begingroup$

Going to answer myself. The inequality with the integrals is true. If you call $g(t):=\sum_{k=0}^je^{ikt}\in P(\mathbb{T})\subset L^1(\mathbb{T})$ and denote the Poisson kernel by $P_r(t)=\sum_{k\in\mathbb{Z}}r^{|k|}e^{ikt}$, using Young's inequality for $p=1$ you have:

$$\int_0^{2\pi}\left|\sum_{k=0}^jr^ke^{ikt}\right|dt=\|P_r\ast g \|_1\leq \|P_r\|_1\cdot\|g\|_1= \|g\|_1=\int_0^{2\pi}\left|\sum_{k=0}^je^{ikt}\right|dt$$

$\endgroup$
2
  • 1
    $\begingroup$ So this shows the inequality for any $L^p$ norm, right? $\endgroup$
    – mathworker21
    Oct 4, 2018 at 17:45
  • $\begingroup$ Yes, that is correct, for $1\leq p \leq\infty$. $\endgroup$
    – Mark_Hoffman
    Oct 8, 2018 at 8:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .