KKT for second order approximation of $f(x)$

Question

Let $f: \mathbb{R}^n \rightarrow \mathbb{R}.$ Consider second order approximation $f(x) \approx f_0(x)$ where $$f_0(x) = f(x_0) + \nabla f(x_0)^T (x-x_0) + (\mathrm{H}f(x_0)(x - x_0))^T(x - x_0)$$ where $\mathrm{H}$ is a hessian matrix. Write KKT conditions for problem of minimizing $f_0$ with constraint $Ax + b$ = 0.

It seems like there are only equality constraints, I have to create a Lagrangian and for that I need a gradient of $f_0$.

I treat $x$ as a column vector.

My attempt: Let $f(x_0) = c_0, \nabla f(x_0) = c, \mathrm{H}f(x_0) = D$

then $\nabla f_0(x) = c + (D + D^T)x$

Basically I treat value, gradient and hessian at $x_0$ as constants.

Is this correct way of calculating the gradient of $f_0(x)$?

That means the lagrangian is $$c + (D + D^T)x + \lambda^T(Ax - b)$$ with constraint that $Ax - b = 0$.

So the KKT conditions are in this case $$c + (D + D^T)x + \lambda^T(Ax - b) = 0 \\ Ax-b = 0$$

Can someone agree or say what is wrong?

Answer 1

The gradient and Hessian of $f$ at $x=x_0$ are constants (vector and matrix respectively), so $f_0(x)$ is a quadratic function of $x$. If it helps, write it as $f_0(x) = c_0 + c^Tx + x^T D x$. Hopefully you know the gradient of that. Then figure out what $c_0$, $c$ and $D$ are in terms of $x_0$, $f(x_0)$, $\nabla f(x_0)$ and $Hf(x_0)$, substitute and maybe simplify a little.