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:

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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.


1 Answer 1


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 $\sum \beta_i f_i$.


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