Generalizing the Cauchy Schwartz inequality We saw in lecture the Cauchy Schwartz inequality, which says that for any two vectors 3 ( R", y( R" for some positive scalar c. 1. Holder's inequality generalizes this to dual norms. That is, for any p and q where # + , = 1, we have where equality is reachable for a specific choice of a and y. (a) When p = 1, the corresponding q is q = too, and | xso is the max-norm, e.g. (x = max; )x;|. Prove Holder's inequality for this choice of p and q. That is, prove that for any z and any y, (b) For p = 1, q = too, given a, list the entire set of possible y where r'y = (xlilly| Hint: The rule for x = [0,0, ..., 0" is different than for a if all the values are nonzero. 2. We can further generalize Holder's inequality to include conic constraints For example, suppose a is restricted to the nonnegative orthant, e.g. > > 0 for all i. Then ry In maxy. (:) (a) Prove (*). (b) List the set of y such that, given a, (+) is true with equality. 3. The singular value decomposition of a matrix X e R" "" decomposes X to its singular values and vectors. It is usually written as X = \\ where ui ( R" are the left singular values, we ( R" are the right singular values, and a are positive scalars. Here, r is the rank of X. Each singular vector are orthonormal, e.g. 1 ifi=j ifi= j else, else. The trace norm of X is the sum of its singular values (denoted (JX| |.). The spectral norm of X is its largest singular value (denoted |X |2). (a) Prove the following generalization of the Cauchy Schwartz inequality for matrices X 6 RX" and Y( RX tr (X Y)