I am working some problems that I am stuck on, i'm finding it hard to verify that my solution is correct or make any progress, since I can't find similar problems in books. I would appreciate some help:
Space = real Hilbert space $l_2(\mathbb{R})$ elements in the space: real sequences $x = \{x_{i}\}$
find the norm of : $<f,x>$ where $$f(x) = \sum_{i=1}^{n}a_{i}x_{i}$$
My work: 1. I think the norm here is the Euclidian norm, if I'm correct, so I have norm of f = $\sqrt{\sum_{i=1}^{n}(a_{i}x_{i})^2}$
but I'm not sure where to go from here?
It's actually the operator norm, which can be expressed as: $$\|f\| = \sup_{\|x\| \le 1} |f(x)|.$$ Note that, if $\|x\| \le 1$, i.e. $\sum_{i=1}^\infty |x_i|^2 \le 1$, then by Cauchy-Schwarz on $\Bbb{R}^n$, $$|f(x)|^2 = \left|\sum_{i=1}^n a_i x_i\right|^2 \le \left(\sum_{i=1}^n |a_i|^2\right) \left(\sum_{i=1}^n |x_i|^2\right) \le \sum_{i=1}^n a_i^2.$$ Thus, we have an upper bound; we have shown $$\|f\| \le \sqrt{\sum_{i=1}^n a_i^2}.$$ On the other hand, we can show that this is the supremum by showing it is a maximum. Consider the sequence $$y = (a_1, a_2, \ldots, a_n, 0, 0, \ldots),$$ and $$x = \frac{y}{\|y\|} = \frac{y}{\sqrt{\sum_{i=1}^n a_i^2}}.$$ Then $\|x\| \le 1$ and, $$|f(x)| = \frac{|f(y)|}{\sqrt{\sum_{i=1}^n a_i^2}} = \frac{\sum_{i=1}^n a_i^2}{\sqrt{\sum_{i=1}^n a_i^2}} = \sqrt{\sum_{i=1}^n a_i^2},$$ achieving the upper bound. Thus the upper bound is a maximum and hence a supremum, i.e. $$\|f\| = \sqrt{\sum_{i=1}^n a_i^2}.$$