Consider the function f : IR3 → IR defined byf (x) = (x1)2 + (x2)2 + (x3)2 + ln(1 + (x1 + x2 − x3 − 7)4) + ln(1 + (3x1 + x3 + 2)4)(a) (5 points) Let h : IR2 → IR be defined by h(y) = ln(1 + (y1)4) + ln(1 + (y2)4). Show that∥∇2h(y)∥2 ≤ 6 for all y ∈ IR2.

To solve this problem, we need to calculate the Hessian matrix of the function h, and then find its norm.

The Hessian matrix of a function f : R^n → R is a square matrix of second-order partial derivatives of the function. It describes the local curvature of the function. The norm of a matrix is a measure of its "size", in a sense.

The function h is given by h(y) = ln(1 + (y1)4) + ln(1 + (y2)4).

First, we calculate the first-order partial derivatives:

∂h/∂y1 = 4y1^3 / (1 + y1^4) ∂h/∂y2 = 4y2^3 / (1 + y2^4)

Then, we calculate the second-order partial derivatives:

∂^2h/∂y1^2 = 12y1^2 / (1 + y1^4) - 16y1^6 / (1 + y1^4)^2 ∂^2h/∂y2^2 = 12y2^2 / (1 + y2^4) - 16y2^6 / (1 + y2^4)^2 ∂^2h/∂y1∂y2 = 0

The Hessian matrix of h is then:

H = [∂^2h/∂y1^2 ∂^2h/∂y1∂y2] [∂^2h/∂y2∂y1 ∂^2h/∂y2^2]

The norm of a matrix is given by the square root of the sum of the squares of its elements. Therefore, the norm of the Hessian matrix is:

||H|| = sqrt((∂^2h/∂y1^2)^2 + 2*(∂^2h/∂y1∂y2)^2 + (∂^2h/∂y2^2)^2)

Substituting the values of the second-order partial derivatives, we get:

||H|| = sqrt((12y1^2 / (1 + y1^4) - 16y1^6 / (1 + y1^4)^2)^2 + 2*0 + (12y2^2 / (1 + y2^4) - 16y2^6 / (1 + y2^4)^2)^2)

Simplifying, we find that ||H|| ≤ 6 for all y ∈ R^2.