# Brezis' Exercise 1.11

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.

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 $$\sum \beta_i f_i$$.