An inequality including L1 and L2 norms

Question

When I'm studying a proof an expression confused me. It is stated in the article that $$ \frac{1}{\left\Vert \mathbf{y}\right\Vert _{1}}\left|\left(\mathbf{x}_{i}^{\top}-\mathbf{x}_{j}^{\top}\right)\hat{\mathbf{r}}\right|\le\frac{1}{\left\Vert \mathbf{y}\right\Vert _{2}}\left\Vert \mathbf{x}_{i}^{\top}-\mathbf{x}_{j}^{\top}\right\Vert _{2}\left\Vert \hat{\mathbf{r}}\right\Vert _{2}. $$ This expression would be correct by Cauchy Schwarz inequality if both sides included $\left\Vert \mathbf{y}\right\Vert _{2}$ term instead of $\left\Vert \mathbf{y}\right\Vert _{1}$ term. Is it still correct if there is $\left\Vert \mathbf{y}\right\Vert _{1}$ on the left and $\left\Vert \mathbf{y}\right\Vert _{2}$ on the right?

Answer 1

yes, as you can prove $$ \left\lVert \bf y \right\rVert _{2}^{2} \leq \left\lVert \bf y \right\rVert _{1}^{2} . $$ I assume that you are in $\mathbb{R}^{n}$ then let $\textbf{y} = \left( y_{1} , \ldots , y_{n} \right)$, we have \begin{align*} \left\lVert \bf y \right\rVert _{2}^{2} & = \sum\limits_{i=1}^{n} \left\lvert y_{i} \right\rvert ^{2} , \\ \left\lVert \bf y \right\rVert _{1}^{2} & = \left( \sum\limits_{i=1}^{n} \left\lvert y_{i} \right\rvert \right) ^{2} = \sum\limits_{i=1}^{n} \left\lvert y_{i} \right\rvert ^{2} + 2 \sum\limits_{i < j} \left\lvert y_{i} \right\rvert \left\lvert y_{j} \right\rvert \geq \sum\limits_{i=1}^{n} \left\lvert y_{i} \right\rvert ^{2} . \\ \end{align*}