My question:
Why can we conclude in the solution below that $(M + \varepsilon)||\sum \beta_i f_i|| \leq \sum \beta_i \alpha_i$?
The following is Exercise 1.11 in Brezis Functional Analysis, Sobolev Spaces and Partial Differential Equations:
Let $E$ be a normed vector space and let $M > 0$. Fix $n$ elements $(f_i)_{1 \leq i \leq n}$ in $E^*$ and $n$ real numbers $(\alpha_i)_{1 \leq i \leq n}$. Prove that the following properties are equivalent>
$$
(A) \quad \forall \varepsilon > 0 \ \exists x_\varepsilon \in E \text{ such that } ||x_\varepsilon|| \leq M + \varepsilon \text{ and } \langle f_i, x_\varepsilon \rangle = \alpha_i \ \forall i.
$$
$$
(B)\left|\sum_1^n \beta_i \alpha_i \right| \leq M ||\sum_1^n \beta_i f_i|| \ \forall \beta_1, \ldots, \beta_n \in \Bbb{R}.
$$
The book provides the following solution:
We know that
$$
\left\langle\sum \beta_i f_i, x \right\rangle \leq ||x|| \ ||\sum \beta_i f_i|| \leq (M+\varepsilon) ||\sum \beta_i f_i|| \quad \forall x \in C,
$$
but why does it hold that $(M + \varepsilon)||\sum \beta_i f_i|| \leq \sum \beta_i \alpha_i$
Thanks in advance and kind regards.
Best Answer
It follows from the fact (prove it using linearity of the map, homogeneity of the norm and the definition of $C$)) that $$ || g || = \sup_{x\in C} \frac{g}{||x||} = \sup_{x\in C} \frac{g}{M+\epsilon} $$ for any linear functional $g$ and so in particular for $\sum \beta_i f_i$.