hw4_sol

# hw4_sol - p 6 = 2 1 5 We have max ≤ i ≤ n | x i | p ≤...

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1. If every subsequence of { p n } converges to p then { p n } must converge to p , since it is a subsequence of itself. Conversely, if { p n } converges to p , and we let { n k } be an increasing sequence of natural numbers, then ² > 0 , N > 0 s.t. n N ⇒ | p n - p | < ² , and since n k k , k N n k N ⇒ | p n k - p | < ² , hence our subsequence converges to p . 2. DeFne a n = s 2 n = 1 4 + 1 2 a n - 1 and b n = s 2 n +1 = 1 2 + 1 2 b n - 1 . It is easy to show (by demonstrating monotonicity and boundedness) that { a n } and { b n } converge, and hence must converge to 1 / 2 and 1 respectively. Thus, the inFmum and supremum of the subsequential limits are 1 / 2 and 1 respectively as well. 3. Note that k x + y k 2 = k x k 2 + 2 h x , y i + k y k 2 and k x - y k 2 = k x k 2 - 2 h x , y i + k y k 2 , hence 1 4 ( k x + y k 2 - k x - y k 2 ) = h x , y i . 4. A norm comes from an inner product if and only if the parallelogram identity holds. That is, if x , y R n we have 2 k x k 2 + 2 k y k 2 = k x + y k 2 + k x - y k 2 . Take x = (1 , 0 , 0 , 0 , . . . ), y = (0 , 1 , 0 , 0 , . . . ), then k x + y k 2 p + k x + y k 2 p = 2 · 2 2 /p , and 2 k x k 2 p + 2 k y k 2 p = 4 , hence, if the p -norm comes from an inner product, we must have that 2 2 /p +1 = 4 2 /p + 1 = 2 p = 2 , hence the p -norm is not an inner product if

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Unformatted text preview: p 6 = 2. 1 5. We have max ≤ i ≤ n | x i | p ≤ n X | x i | p , hence k x k ∞ ≤ k x k p . Also, n X | x i | p ≤ n max ≤ i ≤ n | x i | p , hence k x k ∞ ≥ n-1 p k x k p . 6. Using the naming convention for constants in the previous problem, k x k q ≤ 1 c q k x k ∞ ≤ d p c q k x k p . Similarly, k x k q ≥ 1 d q k x k ∞ ≥ c p d q k x k p . 7. For z ∈ A d ( x , z ) ≤ d ( x , y ) + d ( y , z ) , hence inf z ∈ A d ( x , z ) ≤ d ( x , y ) + inf z ∈ A d ( y , z ) , which gives, by de±nition, d ( x , A ) ≤ d ( x , y ) + d ( y , A ) . Hence, d ( x , A )-d ( y , A ) ≤ d ( x , y ) . Similarly d ( y , A )-d ( x , A ) ≤ d ( x , y ) , hence | d ( x , A )-d ( y , A ) | ≤ d ( x , y ) . 2...
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hw4_sol - p 6 = 2 1 5 We have max ≤ i ≤ n | x i | p ≤...

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