I'm just starting out on a course on approximation theory based almost solely on the textbook by Powell, and must say that certain concepts still evade me. This is the solution to a question asking to prove that the function $\phi$ is continuous. It seems like a very basic thing, but I'm stuck on figuring out where the last line of the inequality in the following comes from:
For a function $\phi: \mathscr{R^n}\to \mathscr{A}$ given by
$$\phi(\lambda)=\lambda_1b_1+...+\lambda_n b_n, \quad (\lambda_1,...,\lambda_n)\in\ \mathscr{R^n},$$ If $\lambda=(\lambda_1,...,\lambda_n),\mu=(\mu_1,...,\mu_n) \in\ \mathscr{R^n}$, then $$\phi(\lambda)-\phi(\mu)=(\lambda_1-\mu_1)b_1+...+(\lambda_n-\mu_n)b_n.$$ Hence $$\begin{align}\|\phi(\lambda)-\phi(\mu)\| & \leq \lvert \lambda_1-\mu_1\rvert \|b_1\|+...+\lvert \lambda_n-\mu_n\rvert \|b_n\| \\ & \leq n \| \lambda-\mu\|_2 \max_{1\leq i\leq n} \|b_i\|.\end{align} $$
$\mathscr{A}$ being an finite-dimensional subspace of a normed linear space.
For n=2, visualising it makes sense, but I feel like I'm missing something with regards to the actual meaning of the $\mathscr {L_2}$-norm generally. Any help would be greatly appreciated.
First, as all the $\|b_k\|$ are less than $\max_{1\leq i\leq n} \|b_i\|$, we have
$$\lvert \lambda_1-\mu_1\rvert \|b_1\|+...+\lvert \lambda_n-\mu_n\rvert \|b_n\| \leq \left(\lvert \lambda_1-\mu_1\rvert + \dots + \lvert \lambda_n-\mu_n\rvert\right) \cdot \max_{1\leq i\leq n} \|b_i\|.$$
Now, note that $\|\lambda - \mu\|_2 = \sqrt{\displaystyle\sum\limits_{k=1}^n |\lambda_k-\mu_k|^2}$. Let $i \in \{1,..,n\}$ : as this big sum contains only positive terms and as $|\lambda_i-\mu_i|^2$ is one of them, we get : $$|\lambda_i-\mu_i|^2 \leq \displaystyle\sum\limits_{k=1}^n |\lambda_k-\mu_k|^2 $$ and therefore $$\sqrt{|\lambda_i-\mu_i|^2} \leq \sqrt{\displaystyle\sum\limits_{k=1}^n |\lambda_k-\mu_k|^2} $$ hence $$|\lambda_i-\mu_i| \leq \|\lambda - \mu\|_2.$$
Therefore
$$\begin{array}{rll}\lvert \lambda_1-\mu_1\rvert \|b_1\|+...+\lvert \lambda_n-\mu_n\rvert \|b_n\| &\leq& \left(\lvert \lambda_1-\mu_1\rvert + \dots + \lvert \lambda_n-\mu_n\rvert\right) \cdot \max_{1\leq i\leq n} \|b_i\| \\ &\leq & \left(\|\lambda - \mu\|_2+ \dots + \|\lambda - \mu\|_2\right) \cdot \max_{1\leq i\leq n} \|b_i\|\\ &\leq & n \cdot \|\lambda - \mu\|_2 \cdot \max_{1\leq i\leq n} \|b_i\|.\end{array}$$
$\rhd$ Note that we can get a better inegality with the Cauchy-Schwarz inequality? The Cauchy-Schwarz inequality tells us that $\displaystyle\sum\limits_{i=1}^n a_ib_i \leq \sqrt{\displaystyle\sum\limits_{i=1}^n a_i^2}\sqrt{\displaystyle\sum\limits_{i=1}^n b_i^2}$. Applying this inequality with $a_i = 1$ and $b_i = |\lambda_i-\mu_i|$, we get $$\displaystyle\sum\limits_{i=1}^n |\lambda_i-\mu_i|\leq \sqrt{\displaystyle\sum\limits_{i=1}^n 1}\sqrt{\displaystyle\sum\limits_{i=1}^n |\lambda_i-\mu_i|^2} = \sqrt{n} \|\lambda - \mu \|_2.$$
This then gives : $$\left(\lvert \lambda_1-\mu_1\rvert + \dots + \lvert \lambda_n-\mu_n\rvert\right) \cdot \max_{1\leq i\leq n} \|b_i\| \leq \sqrt{n} \cdot \|\lambda - \mu\|_2 \cdot \max_{1\leq i\leq n} \|b_i\|.$$