Question

6. (Bonus question: extra marks) Here we will show that ||p difficult property is the triangle inequality: ||la +b||p < || ||, + ||b||p . Here are the main steps: ( )P is a norm over J R" for p > 1.The only k-1 1 + q for a, b0 and 1 1. Prove Young's inequality: ab < b 19)/for any ak, bk E R and 2. Prove Holder's inequality: 1 lab (=1 |akP)"/? TI 1. You + k=1 can assume that the p-norms are non-zero (otherwise the result is immediate) br( b1/a and apply Young's inequality to each term in 1. Let A la )-1/? and B = a( k=1 k=1 A4B}|. 1 = 1 to obtain , |A B}| <1. Rearrange this inequality to finally get Holder's inequality. 3. Use Holder to get the bound |la b| = -1 |af + bk \° < ( lak + b»[P_14)1/4(||a||p + ||b||p) 2. Use that k=1 4. Use that (p-1)q=p and rearrange to get the desired triangle inequality.

Answer 1

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