Exercise: Prove that $\|x\|_\infty \leq \|x\|_2\leq \|x\|_1$ for any $x\in\mathbb{R}^n$.
What I've tried: I think I successfully proved the first inequality. For each $k\in\{1,\ldots,n\}$, one has $\left|x_k\right|\leq \bigg(\sum\limits_{j=1}^n\left|x_j\right|^2\bigg)^{1/2} = \|x\|_2$. If we combine this with $\|x\|_\infty = \sup_n\left|x_n\right|$ we know that $\|x\|_\infty \leq \|x\|_2.$ I wasn't able to prove the second inequality, but I have the following solution:
Let $\lambda = \|x\|_1$. Consider the vector $y = \lambda - 1x$. Then by the properties of a norm we obtain $\|y\|_1 = \lambda^{-1}\|x\|_1 = 1.$ Therefore, $\left|y_k\right|\leq 1$ for all $k\in\{1,\ldots,n\}$. It follows that $\left|y_k\right|^2 \leq \left|y_k\right|$, and thus $$\|y\|_1^2\sum_{k=1}^n\left|y_k\right|^2\leq \sum_{k=1}^n\left|y_k\right| =\|y\|_1= 1$$ We can conclude that $\|y\|_2 \leq 1$. Since $x = \lambda y$, we obtain $$\|x\|_2 = \left|\lambda\right|\|y\|_2 \leq \left|\lambda\right| = \|x\|_1$$
Question: Why do we have that $\|y\|_1 = \lambda^{-1}\|x\|_1$ if $y = \lambda - 1x$? (I don't understand why there is a one before the $x$). I understand the rest of the solution except for the part "$x =\lambda y$".
Thanks!
It should read $y= \lambda^{-1}x$ and not $y = \lambda - 1x$. Then we have $\|y\|_1 = \lambda^{-1}\|x\|_1$ and $x= \lambda y$.