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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3 Comments »

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