8
$\begingroup$

In lecture 3 of the course Convex Optimization conducted by Stephen Boyd at 21 minutes mark he says that a function is convex if its convex when we restrict it to a line. What does he mean by restricting a function to a line?

Secondly his exact statement is that a function $f:R^n\to R$ is convex if $g:R\to R$ is convex where $$ g(t) = f(x+tv), \{x \in dom(f), t \in dom(g),v \in R^n\} $$

My doubt regarding this is that won't any function on real space be convex since it would just be a line?

Please correct me if I am wrong.

$\endgroup$
2
  • 1
    $\begingroup$ Try an example: Let $f(x)=x_1^2+x_2^2$. Start your line from $x=(1,0)$ in the direction $v=(0,1)$. What is $g(t) = f(x+tv)$? Do you think it is a line? $\endgroup$
    – user856
    Feb 2, 2013 at 9:17
  • $\begingroup$ @R^n: I get your point now. Thank you.... $\endgroup$
    – shaunshd
    Feb 5, 2013 at 4:33

3 Answers 3

8
$\begingroup$

1] By restricting it to a line means, basically, you draw line in the domain of the function; then you evaluate your function only along that line.

2] Imagine a paraboloid $f:\mathbb{R}\times\mathbb{R} \mapsto \mathbb{R}$ defined by $f(x,y) = \frac{x^2}{a^2} + \frac{y^2}{b^2}$.

$\hskip2in$ Paraboloid

Now, if you draw a line in the domain and evaluate this paraboloid only along that line, it would look like a parabola. Analytically, if you want to check how the function would be along the x-axis, then substitute y = 0 in the equation above and you get $f(x,y) = \frac{x^2}{a^2}$ which you might know is the equation for the parabola. Now, a parabola is convex and since every line in the domain here would give you a parabola, a paraboloid is convex. On the other hand, if you take a hyperbolic paraboloid:

$\hskip2in$ hyperbolicparaboloid

You draw a line in the domain in one direction, it would look like a parabola and you draw a line in the domain in another direction, it would look like an inverted parabola. Now, inverted parabolas are concave and not convex. Therefore, hyperbolic paraboloids are not convex.

  • Images have been borrowed from the internet.
$\endgroup$
2
$\begingroup$

The function $g(t)$ represent a "section" of the function $f$ along a line. What Boyd is saying is that a function $f$ (a function of several variables) is convex if every restriction on a line is a convex function of one variable.

$\endgroup$
0
$\begingroup$

I would like to add another point which tells us why it is helpful to restrict a function to a line!

Doing so will tell us if a function is convex or concave without evaluating the Hessian (page 73, in Boyd's book). However, as explained in @TenaliRaman answer, two perpendicular lines, that cross at the saddle point, will tell us two different stories about the function, since each line carries a different "profile" of the function, one convex and the other concave, this is where the Hessian is indefinite.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .