<p>Boyd and Vandenberghe say that a convex optimization problem is one of the form:</p>

<blockquote>
  <p>minimize $f_0(x)$ subject to 
  $$f_i(x)\le 0,  i=1,\ldots m$$
                $$a_i^\top x=b_i, i=1,\ldots p$$</p>
</blockquote>

<p>where $f_0, \ldots,f_m$ are convex functions and the equality constraint functions must be affine.</p>

<p>My question is, what stops me from converting the equality constraint functions into inequalities so that I am no longer restricted to affinity with those functions? (any equality $a=b$ can be expressed as $a\le b$ and $a \ge b$).</p>

CC BY-SA 3.0
2014-04-27T20:20:14.780Z
An equality constraint of the form $f(x) = 0$ where $f$ is nonlinear is not convex. Imagine if the constraint was that the points should lie on the surface of a sphere. Is this what you're getting at ?
Suresh Venkat
CC BY-SA 3.0
2014-04-27T20:54:49.367Z
What Suresh said. The problem is that for a convex $f$, to express $f(x) = 0$ as a pair of inequalities, you need to write $f(x) \leq 0$ (which is allowed in a convex optimization problem), as well as $f(x) \geq 0$ (which is not allowed)
Aaron Roth
CC BY-SA 3.0
2014-04-27T21:18:25.020Z
Thanks for the comments, this was the kind of comments that I was looking for.
ASDF
CC BY-SA 3.0
2014-04-28T01:30:16.013Z
I don't think anything technically stops you from having all constraints be of the form $f_i(x) \leq 0$ for convex $f_i$: $a_i^\top x - b_i$ is both convex and concave so it is fine to write two constraints in the way you described.
Sasho Nikolov
CC BY-SA 3.0
2014-04-28T04:49:31.687Z
However, it is often convenient to treat the affine constraints separately. For example, by Slater's condition strong duality holds if there exists a primal feasible solution which is strictly feasible for all non-affine constraints.
Sasho Nikolov
<cc.complexity-theory><linear-programming><convex-optimization>
Definition of convex optimization problem by Stephen Boyd and Lieven Vandenberghe
ASDF
# Definition of convex optimization problem by Stephen Boyd and Lieven Vandenberghe

(Asked by: username_0 on 2014-04-27)

Boyd and Vandenberghe say that a convex optimization problem is one of the form:

> minimize $f\_0(x)$ subject to
> $$f\_i(x)\le 0, i=1,\ldots m$$
> $$a\_i^\top x=b\_i, i=1,\ldots p$$

where $f\_0, \ldots,f\_m$ are convex functions and the equality constraint functions must be affine.

My question is, what stops me from converting the equality constraint functions into inequalities so that I am no longer restricted to affinity with those functions? (any equality $a=b$ can be expressed as $a\le b$ and $a \ge b$).

**Tags:** cc.complexity-theory, linear-programming, convex-optimization

> **Comments:**
> **username_1**: An equality constraint of the form $f(x) = 0$ where $f$ is nonlinear is not convex. Imagine if the constraint was that the points should lie on the surface of a sphere. Is this what you're getting at ?  
> **username_2**: What Suresh said. The problem is that for a convex $f$, to express $f(x) = 0$ as a pair of inequalities, you need to write $f(x) \leq 0$ (which is allowed in a convex optimization problem), as well as $f(x) \geq 0$ (which is not allowed)  
> **username_0**: Thanks for the comments, this was the kind of comments that I was looking for.  
> **username_3**: I don't think anything technically stops you from having all constraints be of the form $f_i(x) \leq 0$ for convex $f_i$: $a_i^\top x - b_i$ is both convex and concave so it is fine to write two constraints in the way you described.  
> **username_3**: However, it is often convenient to treat the affine constraints separately. For example, by Slater's condition strong duality holds if there exists a primal feasible solution which is strictly feasible for all non-affine constraints.  

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## 0 answers

