Problem 9.3 (3 points). (a) Show that the function f(x) =...

Problem 9.3 (3 points). (a) Show that the function $f(x) = \log(x) : \mathbb{R}^+ \to \mathbb{R}$ is concave, that is, $-f(x)$ is convex. (b) Use the concavity of $\log$, prove that for $p, q > 1$, $\frac{1}{p} + \frac{1}{q} = 1$, and any non-negative reals $x, y \in \mathbb{R}^+$, we have the inequality $xy \le \frac{x^p}{p} + \frac{y^q}{q}$. Hint: evaluate the logarithm of the above expression. (c) With the above result, show that for $p, q > 1$, $\frac{1}{p} + \frac{1}{q} = 1$, and any $x, y \in \mathbb{R}^n$, we have $\sum_{i=1}^n |x_i y_i| \le ||x||_p ||y||_q$, where $||x||_p = (\sum_{i=1}^n |x_i|^p)^{1/p}$ is the $l_p$ norm. Hint: apply the inequality in part (b) to the product $\frac{|x_i|}{||x||_p} \frac{|y_i|}{||y||_q}$.

Transcript

00:01 In this question we need to adapt the proof of concavity of the log determinant function and need to show that so the first part in this first part we need to show that f of x is equal to face of x inverse is a convex function so for this we have f of x is equal to we can take let's take another function g of t is equal to f of z plus tv where this v is a positive definite matrix so we will get using this transformation you will get trace of that plus tv and now if we can show that this trace of z plus tv it is x inverse if we can show that this trace of z plus tv whole inverse is a convex function so this will show that this trace of x inverse is also a convex function so now gt is equal to this so we can write it as trace of z plus t v we can multiply and by both inverse and the value of z inverse so this will be equal to trace of we can take it common that is identity matrix plus t v z inverse whole inverse and now using the property this will come this side see and now using the property that a b whole inverse is equal to b inverse a inverse we will get it equal to trace of z inverse identity matrix plus t v z inverse whole inverse now as z is a positive definite matrix so z inverse is all also positive definite and we can write z inverse as z minus 1 by 2 and z minus 1 by 2 now we will have this whole equation that is g of t will be equal to trace of z inverse i plus t of z in minus 1 by 2 v z minus 1 by 2 whole inverse now let us take z inverse by z in 1 minus 1 by 2 is equal to q and this as q inverse so what we will get trace of z inverse i plus t q q q inverse whole inverse now this will be equal to trace of z inverse q q inverse plus t q q inverse whole inverse and this will be then equal to trace of z inverse q times q plus t taking you common we will get q inverse that is t q and this inverse and this will be equal to trace of z inverse q q plus t q this as if q q transpose is equal to is equal to i then q transpose whole inverse will be equal to q only.

04:39 So we will get this term.

04:41 And now this will be equal to trace of qz inverse q identity matrix plus t whole inverse...

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