# Inequality regarding complex numbers

If we consider $$0 and $$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:

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

• $j = 1, t = \pi, r = .5$ – mathworker21 Oct 3 '18 at 17:11
• 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 ? – Mark_Hoffman Oct 3 '18 at 17:19
• 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 $j$. – mathworker21 Oct 3 '18 at 17:29

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:
$$\int_0^{2\pi}\left|\sum_{k=0}^jr^ke^{ikt}\right|dt=\|P_r\ast g \|_1\leq \|P_r\|_1\cdot\|g\|_1= \|g\|_1=\int_0^{2\pi}\left|\sum_{k=0}^je^{ikt}\right|dt$$
• So this shows the inequality for any $L^p$ norm, right? – mathworker21 Oct 4 '18 at 17:45
• Yes, that is correct, for $1\leq p \leq\infty$. – Mark_Hoffman Oct 8 '18 at 8:47