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i'm reading a book of "Ulbrich" about Nonlinear Optimization. In the chapter about convex optimization he says, given a NLP $$\min f(x), s.t. g(x) \leq 0, h(x) = 0,$$ the NLP is convex if $f,g_i$ are convex functions and $h$ is affine ly linear.

After that he shows that the feasible set $X := \{x \in \mathbb{R}^n \mid g(x) \leq 0, h(x) = 0\}$ is then convex via: $\forall x,y \in X, \lambda \in [0,1]$ it holds $$g(x + \lambda (y-x)) \leq g(x) + \lambda (g(y) - g(x)) \leq 0 \text{ clear!}$$

and

$$h(x + \lambda (y-x)) \overset{?}{=} h(x) + \lambda (h(y) - h(x)) = 0.$$

The equality is wrong, right? I mean affinely linear does not mean that we behave linearly.

But we can say that the first derivative does so, and additionally is constant, right?

Greetings,

Dom

Dom
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  • The equation marked with ? does not imply linearity. It is a characterization of an affine function (for all relevant $x,y,\lambda$) – user251257 Dec 04 '19 at 19:13
  • $h$ is afine iff for all $t$ we have $h(tx+(1-t)y) = th(x)+(1-t)h(y)$. – copper.hat Dec 04 '19 at 19:32
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    You are right, i was dumb. This special equality does indeed hold, since by $h(x) = Ax + b$, you simply add $-\lambda b + \lambda b$ after resolving and then you get the formula. It seemed like he used the typical linearity properties $h(ax + by) = a h(x) + b h(y).$ – Dom Dec 04 '19 at 19:43

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