Characterizing linear functionals of norm less than $1$

Question

To my embarassement, I am struggling with the following elementary statement: if $V$ is a finite dimensional real vector space, endowed with an inner product and an orthonormal basis $\{ e_1, \dots, e_n \}$, then

$$\{ \phi \in V^* \mid | \phi (x) | \le \| x \| \} = \left\{ \phi \in V^* \mid \sum _{i=1} ^n \phi (e_i) ^2 \le 1 \right\} .$$

The $\supseteq$ inclusion is easy, using the Cauchy-Schwarz inequality:

$$ | \phi (x) | = \left| \sum _{i=1} ^n x_i \phi (e_i)\right| \le \| x \| \cdot \sqrt {\sum _{i=1} ^n \phi (e_i) ^2} \le \| x \| .$$

The $\subseteq$ inclusion, though, makes me feel miserable, since all the approaches that I attempt lead to the much coarser $\sum \limits _{i=1} ^n \phi (e_i) ^2 \le n $. What is then the smart inequality to use here, please?

Answer 1

Maybe this is silly, but if you know $|\phi(x)| \leqslant \|x\|$ for all $x \in V$, can't you take $x = \sum_{i=1}^n \phi(e_i)e_i$, so $\phi(x) = \sum_{i=1}^n \phi(e_i)^2$, and $\|x\|^2 = \sum_{i=1}^n \phi(e_i)^2$, and $\phi(x)^2 = |\phi(x)|^2 \leqslant \|x\|^2$, therefore $$ \left(\sum_{i=1}^n \phi(e_i)^2\right)^2 \leqslant \sum_{i=1}^n \phi(e_i)^2, $$ therefore $\sum_{i=1}^n \phi(e_i)^2 \leqslant 1$? I'll get me coat.

Answer 2

We may write $\phi = \sum_{i=1}^n \phi_i e_i^*$, where $e_i^*$ is the dual basis defined by $e_i^*(a_1 e_1 + \cdots + a_n e_n) = a_i$. Observe that $$\phi_i = \phi(e_i).$$ (This can be seen by applying $\phi$ to $e_j$.)

Suppose that $| \phi(x) | \leq \| x\|$ for all $x \in V$. If you're farmiliar with functional analysis, you'll notice that another way to say this is that $\|\phi\| \leq 1$, where $\|\phi\| = \sup_{\|x\|=1} |\phi(x)|$ is the operator norm on $V^*$. Using our expansion $\phi = \sum_{i=1}^n \phi_i e_i^*$, we have $$ \|\phi\| = \sup_{\|x\|=1} \left| \sum_{i=1}^n \phi_i e_i^*(x)\right| = \sup_{\|\vec x\|=1} \left| \vec{\phi} \cdot \vec x \right| = \left| \vec \phi \right|, $$ where $\vec \phi = (\phi_1, \dots, \phi_n)$ and $\vec x = (x_1, \dots, x_n)$ are vectors in $\mathbb R^n$ formed by the coefficients of the vectors $\phi$ and $x$ in $V$ with respect to the basis $e_1, \dots, e_n$. The last step follows from taking $\vec x = \frac{\vec \phi}{\left| \vec \phi \right|}$ according to the case of equality in the Cauchy-Schwarz inequality. Since $\|\phi\| \leq 1$ and the above shows $\|\phi\| = \left|\vec \phi\right|$, we must have $\left| \vec \phi \right| \leq 1$. Expanding this yields $$ \left| \vec \phi \right| = \sqrt{\sum_{i=1}^n \phi_i^2} = \sqrt{\sum_{i=1}^n \phi(e_i)^2} \leq 1. $$ Squaring gives the desired result.