Lemma 3.9 Rudin's functional analysis

Question

I need help in understanding the proof of the following

Suppose $\Lambda_1,\ldots,\Lambda_n$ and $\Lambda$ are linear functionals on a vector space $X$. Let $$ N = \left\{x : \Lambda_1x = \ldots = \Lambda_n = 0 \right\} $$ The following three are equivalent

a) There are scalars $\alpha_1,\ldots,\alpha_n$ such that $$ \Lambda = \alpha_1 \Lambda_1 + \ldots + \alpha_n \Lambda_n $$

b) There's $\gamma < \infty$ such that $$ |\Lambda x | \leq \gamma > \max_{1\leq i \leq n} |\Lambda_i x| \;\;,(x\in X) $$

c) $\Lambda x = 0$ for every $x \in N$.

Proof: It is clear that (a) implies (b) and that (b) implies (c)...etc.

That's the bit I don't get. How does (a) implies (b)? I believe I get why (b) implies (c). Since if $x \in N$ then for each $i$ we have $\Lambda_i x = 0$, but then by (b) we have $|\Lambda x = 0|$ and this implies $\Lambda x = 0$. (correct?)

So again

Why (a) implies (b)? Is my proof of (b) implies (c) correct?

Answer 1

Your proof of (b) implies (c) is correct.

As for (a) implies (b):

We have $$ | \Lambda x | = | \alpha_1\Lambda_1 x +\dots + \alpha_n\Lambda_n x | \\ \leq | \alpha_1|\cdot |\Lambda_1 x | +\dots + |\alpha_n|\cdot |\Lambda_n x| \\ \leq (|\alpha_1|+\dots + |\alpha_n|) \max_{1\leq i \leq n} |\Lambda_i x|. $$ This means we can choose $\gamma=|\alpha_1|+\dots + |\alpha_n|$.

Answer 2

a) implies b): $|\Lambda x| \leq \sum |\alpha_i| | \Lambda_i x|\leq \max\{|\Lambda_i(x)|\} \sum |\alpha_i|$ so you can take $\gamma =\sum |\alpha_i|$ Your argument for b) implies c) is correct.