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In "Ming-Jun Lai, Larry L. Schumaker. Spline Functions on Triangulations. Cambridge University Press, 2007, p.72." we have:

A function $f$ defined on a triangle $T$ is said to be convex in the direction $u$ provided
$$\frac{f(w_3)-f(w_2)}{|w_3-w_2|} \ge \frac{f(w_2)-f(w_1)}{|w_2-w_1|}$$
for all ordered sets of points $w_1, w_2, w_3$ in $T$ lying on a line pointing in the direction of $u$. We say that $f$ is convex on $T$ provided it is convex in all directions. As is well known from calculus, if $f$ has two derivatives in the direction $u$, then this definition of convexity in the direction $u$ is equivalent to $D_u^2f(v)\ge 0$, all $v\in T$.

But All thing that we have about function convexity briefly is here: http://en.wikipedia.org/wiki/Convex_function.

I can't make a connection between definition of convex function and convexity of a function on a certain direction, also both of them with second directional derivative of function on a direction. How we can proof that usual definition of convex function and "We say that $f$ is convex on $T$ provided it is convex in all directions." are equivalence? What's proof of "If $f$ has two derivatives in the direction $u$, then this definition of convexity in the direction $u$ is equivalent to $D^2_uf(v)\ge 0$, all $v\in T$."?

A link to a comprehensive source is pleasured.

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You have two different questions.

Question 1 :

($f$ is convex) $\Leftrightarrow$ ($f$ is convex all the directions)

Question 2 :

If $f$ has two derivatives, then ($f$ convex in the direction $u$) $\Leftrightarrow$ ($D^{2}_{u} (f) (v) \geq 0$, $\forall v \in T$)

For the first question.

One only needs to recall the definition of convexity.

$f$ is called convex if $\forall x_{1} ,x_{2} \, \forall t \in [0,1] : f (t x_{1} + (1-t)x_{2}) \leq t \, f (x)_{1} + (1-t) \, f (x_{2})$

$(\Rightarrow)$ : We choose an arbitrary direction $u$ and three points $w_{1}$, $w_{2}$ and $w_{3}$ aligned in this direction and in this order. We define the number $t$ as

$$ t = \frac{|w_{3} - w_{2}|}{|w_{1} - w_{3}|} $$ Since the points are aligned in the order $w_{1}$, $w_{2}$ and $w_{3}$, we have $t \in [0,1]$. Moreover, we note that we have $w_{2} = t \, w_{1} + (1-t) \, w_{3}$. Applying the usual definition of convexity, we obtain:

$$ f (w_{2}) \leq \frac{|w_{3} - w_{2}|}{|w_{1} - w_{3}|} f (w_{1}) + \frac{|w_{1} - w_{2}|}{|w_{1} - w_{3}|} f (w_{3}) $$

so that we have

$$ \frac{|w_{1} - w_{3}|}{|w_{1} - w_{2}| |w_{3} - w_{2}|} \, f (w_{2}) \leq \frac{f(w_{1})}{|w_{1} - w_{2}|} + \frac{f (w_{3})}{|w_{3} - w_{2}|} $$

The final step of the calculation is to note that $|w_{1} - w_{3}| = |w_{1} - w_{2}| + |w_{2} - w_{3}|$ since the points are aligned in the good order, and after rearrangement, we find back the expression for the definition of the convexity in the direction $u$.

($\Leftarrow$) : Completely similar. One just needs to redo this reasoning in the other way. (Is it clear or do you need more details ?)

For the second question.

The idea is that as soon as you have given yourself a direction $u$, you are back to the case of a $1D$ function. Indeed, for a given direction $u$ and a given starting point $v_{0}$, you need to study the convexity of the function

$$\begin{aligned} f_{u , v_{0}} : &\mathbb{R} \to X \\ & t \mapsto f (v_{0} + t \, u) \end{aligned} $$ The usual criteria of convexity for twice differentiable function imposes you $f_{u , v_{0}} '' \geq 0$. The final remark is to note that $f_{u , v_{0}} '' (t) = D^{2}_{u} f \, (v_{0} + t u) $, so that the equivalence of the two properties is immediate.

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  • $\begingroup$ OK, I got it, thanks @jibe. $\endgroup$ – meysam Aug 5 '14 at 18:05

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