Since \( \nabla^2 f(x) = \exp(-a^Tx) aa^T + 2A^TA \), we have that \( || \nabla^2 f(x) || > 2||A^TA|| \) for any \( x \in \mathbb{R}^2 \), which shows strong convexity, and strong convexity implies strict convexity and convexity.

Note that it would not be true if \( ||A^TA||_2 = 0\), but we can show that \( A^TA = I_2 \).

Lemma 4.9 on page 81 : if \( f(x) \) is twice differentiable then you can take the Hessian of the left and right hand side, then combining with the convexity conditions it becomes : \( \nabla^2 f(x) - \mu \succcurlyeq 0 \)

So for the question above you have that the largest eigenvalue of \( I_2 \) is 1, and thus \( \mu = 2 \).

Thanks for the answer, but I still don't see why you can just ignore the fact that a is in R^2 and hence it can be positive or negative. To my mind that means that the hessian can for certain a be unbounded.

## Composite function convexity

Could someone please explain how to show convexity, strong convexity, strict convexity of the composite function:

Since \( \nabla^2 f(x) = \exp(-a^Tx) aa^T + 2A^TA \), we have that \( || \nabla^2 f(x) || > 2||A^TA|| \) for any \( x \in \mathbb{R}^2 \), which shows strong convexity, and strong convexity implies strict convexity and convexity.

Note that it would not be true if \( ||A^TA||_2 = 0\), but we can show that \( A^TA = I_2 \).

How does that show strong convexity? Could you point to a lemma maybe? Thanks!

Lemma 4.9 on page 81 : if \( f(x) \) is twice differentiable then you can take the Hessian of the left and right hand side, then combining with the convexity conditions it becomes :

\( \nabla^2 f(x) - \mu \succcurlyeq 0 \)

So for the question above you have that the largest eigenvalue of \( I_2 \) is 1, and thus \( \mu = 2 \).

Thanks for the answer, but I still don't see why you can just ignore the fact that a is in R^2 and hence it can be positive or negative. To my mind that means that the hessian can for certain a be unbounded.

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