5 points) Let h : IR2 → IR be defined by h(y) = ln(1 + (y1)2) + 1.1(y2)2. Show that∥∇2h(y)∥2 ≤ 2.2 for all y ∈ IR2
Question
5 points) Let h : IR2 → IR be defined by h(y) = ln(1 + (y1)2) + 1.1(y2)2. Show that∥∇2h(y)∥2 ≤ 2.2 for all y ∈ IR2
Solution
The question is asking to prove that the norm of the Hessian matrix of the function h, denoted as ∥∇²h(y)∥², is less than or equal to 2.2 for all y in the set of real numbers IR².
The function h is given as h(y) = ln(1 + (y₁)²) + 1.1(y₂)².
First, we need to compute the Hessian matrix of h, which is the matrix of second derivatives.
The partial derivatives of h are:
∂h/∂y₁ = 2y₁ / (1 + (y₁)²) ∂h/∂y₂ = 2.2y₂
The second partial derivatives are:
∂²h/∂y₁² = 2 / (1 + (y₁)²)² ∂²h/∂y₁y₂ = 0 (since h does not contain a term with both y₁ and y₂) ∂²h/∂y₂y₁ = 0 (same reason as above) ∂²h/∂y₂² = 2.2
So, the Hessian matrix ∇²h is:
∇²h = [2 / (1 + (y₁)²)², 0] [0, 2.2]
The norm of a matrix is the square root of the sum of the squares of its elements. So, the square of the norm of the Hessian matrix is:
∥∇²h(y)∥² = (2 / (1 + (y₁)²)²)² + (0)² + (0)² + (2.2)² = 4 / (1 + (y₁)²)⁴ + 4.84 ≤ 4 + 4.84 = 8.84
So, we have ∥∇²h(y)∥² ≤ 8.84 for all y in IR², which is not less than or equal to 2.2.
Therefore, the statement in the question is not correct.
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