# $L^1$ inequality between ordered real numbers implies $L^2$ norm inequality

Let $$x_1,\ldots,x_n$$ be real numbers such that $$|x_1|\geq \ldots\geq |x_n|$$ and $$\displaystyle \sum_{i=k+1}^n |x_i| \leq \alpha \sum_{i=1}^k |x_i|$$ where $$1\leq k \leq n-1$$ and $$\alpha >0$$.

Prove that $$\sum_{i=k+1}^n x_i^2 \leq \alpha \sum_{i=1}^k x_i^2$$

I have managed to prove a looser inequality: \begin{align} \alpha \sum_{i=1}^k x_i^2 &\geq \frac{\alpha}{k}\left(\sum_{i=1}^k |x_i| \right)^2 \quad \quad \text{Cauchy-Schwarz}\\ &\geq \frac 1k \sum_{i=k+1}^n |x_i| \sum_{i=1}^k |x_i|\\ &\geq \frac 1k \left(\sum_{i=k+1}^n |x_i| \right)^2\\ &\geq \frac 1k \sum_{i=k+1}^n x_i^2 \quad \quad \text{since } \|z\|_1 \geq \|z\|_2 \end{align}

Since $$\frac 1k$$ is not $$\geq 1$$, I'm stuck...

## 1 Answer

It is quite simple: Note that $$|x_i| \le |x_k|$$ for any $$i \in \{k+1,\ldots,n\}$$. Thus $$\tag{1}\sum_{i=k+1}^n x_i^2 \le |x_k| \sum_{i=k+1}^n |x_i| \le \alpha |x_k| \sum_{i=1}^k |x_i|,$$ where the given condition is used in the last step. Now $$|x_k x_i| \le x_i^2$$ for an $$i=1,\ldots,k$$, because $$|x_k| \le |x_i|$$ for all $$i=1,\ldots,k$$. So proceeding in (1) gives the claimed estimate $$\sum_{i=k+1}^n x_i^2\le\alpha \sum_{i=1}^k x_i^2.$$

• Nice, thank you. This line of thought did not occur to me. – Issou Chankla Nov 21 '18 at 11:17
• In general you lose information, if you apply Cauchy-Schwarz. The constants between $\|\cdot\|_1$ and $\|\cdot\|_2$ are also strict. In fact, the key point here is the condition $|x_1| \ge \ldots \ge |x_n|$. (The statement is wrong without this condition!) – p4sch Nov 21 '18 at 12:04