This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: k xy k H . 01? (d) Show that k x k k x k H n k x k . Show that both inequalities are taken on for some (dierent) choice of x . Hint: for the rst inequality, use x m = 1 2 Z 2 n X k =1 x k ejk ! e jm d . 7. Show that the norms k u k 1 = Z 1  u ( t )  dt k u k = sup { u ( t )  : 0 t 1 } are not equivalent on the vector space of continuous functions from [0 , 1] into IR . 8. Let V be the following vector space of sequences: V = { x [ k ] : x [ k ] IR, k ,  x [ k ]  < } . Dene k u k = sup k  u [ k ]  and k u k w = sup k w [ k ]  u [ k ]  for some given weighting sequence w [ k ]. (a) What conditions on w make k k w a norm on V ? (b) What conditions on w make k k w a norm on V equivalent to k k ?...
View
Full
Document
This note was uploaded on 01/24/2012 for the course EECS 250 taught by Professor Swindlehurst during the Fall '09 term at UC Irvine.
 Fall '09
 Swindlehurst
 Digital Signal Processing, Signal Processing

Click to edit the document details