Prove that $\textrm{lip} L \leq \left(\sum_{j=1}^n || L(e_j) ||^2\right)^{1/2}$ where $e_j = (\delta_{jk})_{1\leq k \leq n}$ is standar basis, $L$ is linear transformation $L: \mathbb{R}^n \to Y$, $Y$ is normed vector space and norm on $\mathbb{R}^n$ is defined as standard Euclidean norm.
What I have tired is following: $\textrm{lip} L = \sup_{a\ne b} \frac{||L(a) - L(b)||}{||a-b||}$, then I tried writting vectors $a$ and $b$ using basis vectors and then using linearity of $L$ have got something but not what I need. Can someone please help me.
Let $\|x\|_2$ denote the Euclidean norm of any $x\in \Bbb R^n,$ i.e. $$\|x\|_2=\left(\sum_{j=1}^nx_j^2\right)^{1/2}.$$ Then, $$\begin{align}\textrm{lip} L&= \sup_{\|x\|_2=1} \|L(x) \| \\&\le\sup_{\|x\|_2=1}\sum_{j=1}^n|x_j|\|L(e_j)\| \\&=\sup_{\|x\|_2=1}\sum_{j=1}^nx_j\|L(e_j)\| \\&= \left(\sum_{j=1}^n\| L(e_j)\|^2\right)^{1/2}, \end{align} $$ the last equality being the Cauchy-Schwarz theorem $$\sup_{\|x\|_2=1}\sum_{j=1}^nx_jy_j=\|y\|_2$$ applied to $y_j=\| L(e_j)\|.$