Let $f(z) = \operatorname{log} \sum_{i=1}^n z_i = \operatorname{log} 1^Tz$.

This problem comes from the following famous theorem:

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My work:
The following step is just consider one entry of the Hessian matrix:

$$\nabla^2 f(z)_{ij} = \frac{\partial f(z)}{\partial z_i\partial z_j} = \frac{\partial}{\partial z_i}(\frac{\partial f(z)}{\partial z_j}) = \frac{\partial}{\partial z_i}(\frac{1}{1^Tz})=-\frac{1}{(1^Tz)^2}$$
I have no idea how to obtain the desired form. How to obtain the other term such as $\operatorname{diag(z)}$. Moreover, the numerator here is $1$.


1 Answer 1


Note that you have to take the derivative with respect to $x$ if you want to check convexity in $x$. This gives some more terms form the chain rule: We have $$ \frac{\partial}{\partial x_j} f(x) = \frac{\partial}{\partial z_j} \log 1^tz \cdot \frac{\partial z_j}{\partial x_j} = \frac 1{1^tz} z_j $$ Hence, \begin{align*} \frac{\partial^2 f}{\partial x_i \partial x_j} &= \frac{\partial}{\partial z_i} \left(\frac{z_j}{1^tz}\right) \cdot \frac{\partial z_i}{\partial x_i}\\ &=\frac{\delta_{ij}1^tz - z_j}{(1^t z)^2} \cdot \exp x_i\\ &= \frac{\delta_{ij}z_i\cdot 1^tz - z_iz_j}{(1^t z)^2}\\ &= \frac{\delta_{ij}z_i}{1^tz} - \frac{z_iz_j}{(1^tz)^2}\\ &= \left(\frac 1{1^tz} \mathop{\rm diag}(z) - \frac{1}{(1^tz)^2} zz^t\right)_{i,j} \end{align*}

  • 1
    $\begingroup$ Can you explain a bit about $\delta_{ij}$? $\endgroup$
    – user436661
    Feb 6, 2020 at 8:38
  • $\begingroup$ The so called Kronecker symbol, a shorthand for $\delta_{ij} = 1$ iff $i=j$, otherwise $\delta_{ij} = 0$. $\endgroup$
    – martini
    Jun 12, 2021 at 9:05

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