# Mathematics Prelims

## July 16, 2008

### Lemma Concerning “Minimum Norms”

Filed under: Analysis,Functional Analysis — cjohnson @ 2:24 pm

Suppose $X$ is an n-dimensional normed space over $\mathbb{R}$ (or $\mathbb{C}$) with basis $\{ x_1, x_2, ..., x_n \}$.  There exists a $c > 0$ such that

$\displaystyle \| \alpha_1 x_1 + ... + \alpha_n x_n \| \geq c (|\alpha_1| + ... + |\alpha_n|)$

for any selection of $\alpha_1, ..., \alpha_n$ in the field.

Proof: Let $s = |\alpha_1| + ... + |\alpha_n|$ and let $\beta_i = \frac{\alpha_i}{s}$.  Now our original inequality is equivalent to

$\displaystyle \| \beta_1 x_1 + ... + \beta_n x_n \| \geq c$

for all $\beta_1, ..., \beta_n$ that satisfy $\sum_{i=1}^n |\beta_i| = 1$.  Suppose no such $c$ exists.  Then we can construct a sequence $(y_m)$ where $y_m = \sum_{i=1}^n \beta_i^{(m)} x_i$, with $\sum_{i=1}^n |\beta_i^{(m)}| = 1$ for each $m$, and $\| y_m \| \to 0$.  Now note that since $\sum_{i=1}^n |\beta_i^{(m)}| = 1$ implies if we look at the sequence $(\beta_i^{(m)})_{m \in \mathbb{N}}$ (where $i$ is fixed), that we have a bounded sequence, and so we must have a convergent subsequence.  Apply this for $i = 1$, let the limit of that subsequence be $\beta_1$, and let $y_{1,m}$ be the associated subsequence of the original $(y_m)$ sequence.  On that subsequence, apply again for $i = 2$, then on that subsequence again on $i = 3$ and so on until we have

$\displaystyle y_{n,m} = \sum_{i=1}^n \beta_i^{(m)} x_i$

Now note that each $\beta_i^{(m)} \to \beta_i$ (where $\beta_i$ is the limit of the subsequence with the associated $i$ from earlier).  This implies $y_{n,m} \to \sum_{i=1}^n \beta_i x_i = y$.  Since we required $\sum_{i=1}^n |\beta_i^{(m)}| = 1$ for each $m$ earlier, we now have that $\sum_{i=1}^n |\beta_i| = 1$.  This means that $y \neq 0$, so the subsequence converged to a non-zero element, which means the original sequence can not converge to the zero element.  We could only do this because we assumed no such $c > 0$ existed, so a $c > 0$ with the desired property must exist.

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