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Here is Theorem 4 and its proof (from Royden 4th edition "Real Analysis"):

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But I do not understand why in the second part in finding $c_2,$ in the second paragraph from below: $\{e_{1}, ... , e_{n}\}$ being linearly independent leads to that $f$ takes positive values on the boundary of the unit ball, could anyone explain this for me ,please?

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    $\begingroup$ (Note that $f \ge 0$.) Because if $f(x) = 0$ then $\sum_k x_k e_k = 0$ and so, by linear independence, $x_k = 0$ for all $k$ contradicting $\sum_k x_k^2 = 1$. $\endgroup$
    – copper.hat
    Mar 24, 2020 at 3:38

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Basically linear independence here guarantees that $\sum_{i=1}^nx_ie_i$ is not $0$.

By definition of norm, $||\sum_{i=1}^n x_ie_i||=0$ iff $\sum_{i=1}^n x_ie_i=0$, and by linear independence, that happens only if $x_i=0$ for all $i$. Since $x$ is on the unit ball, this cannot happen, so $\sum_{i=1}^n x_ie_i\neq 0$, and $||\sum_{i=1}^n x_ie_i||>0$.

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