0
$\begingroup$

I know the definition of the convex function $$f:X \rightarrow \Bbb R $$ $$\forall x_1, x_2 \in X, \forall t \in [0, 1]: \qquad f(tx_1+(1-t)x_2)\leq t f(x_1)+(1-t)f(x_2)$$

Now , I want to know how we can get it using equations of $f(x)$ and the line that connects two points of $f(x)$ (consider $g(x)$) and then solve $g(x) \ge f(x)$

  • 0
    Your question isn't clear. If you are asking how to prove a particular function $f$ is convex you have to tell us what function you are interested in. – 2017-02-10
  • 0
    @EthanBolker No , In general I want to know how we did get this definition. In other words , You assume convex function as a function when we connect two points of it then the line is above the function and then conclude this definition. – 2017-02-10
  • 0
    The one you wrote IS the definition. The 'definition' that says that the line connecting two points of the graph lies above the graph is useful in 1, maybe 2 dimensions, but it's definitely not easily portable in a general framework. This definition only requires that the codomain of $f$ is ordered, so that you can define the $\leq$ operator. In your case it's $\mathbb{R}$, and we can also picture the graph of the function. But in a general framework, it may be difficult to picture the line connecting two points. That's why this definition is much more easily portable. – 2017-02-10
  • 0
    @bartgol I only need to two dimension proof – 2017-02-10

1 Answers 1

1

Let us choose two points $x_1, x_2\in X$. Every point of $[x_1, x_2]$ can be written as $tx_1+(1-t)x_2$ with $t\in[0,1]$. Now, the line that goes through the points $(x_1, f(x_1))$ and $(x_2, f(x_2))$ has the equation

$$(x,y)=(x_1, f(x_1)) + t[(x_2, f(x_2))-(x_1, f(x_1))]$$

or, in other words

$$y= \frac{f(x_2)-f(x_1)}{x_2-x_1}x + \frac{x_2f(x_1)-x_1f(x_2)}{x_2-x_1}$$

and if you pick $t\in[0,1]$, you obtain the points of the segment between $(x_1, f(x_1))$ and $(x_2, f(x_2))$.

As you know that the line is above the graph of $f$, then the $y$ coordinate of the point $(x,y)=t(x_1, f(x_1)) + (1-t)(x_2, f(x_2))$ has to be greater than $f(tx_1+(1-t)x_2$. This way, you obtain the inequality of the definition:

$$f(tx_1+(1-t)x_2)\leq tf(x_1)+(1-t)f(x_2)$$

Note that, when you fix two points $x_1$ and $x_2$, the inequality is only valid for the points between $x_1$ and $x_2$. For example, let us call $f(x)=x^2$, which is convex. If you fix $-1$ and $1$, for the points outside $[-1,1]$ the inequality does not hold. enter image description here

Another example with the points $-2$ and $1$. Now the line is obviously different, but the inequality is again true only for the points in $[-2,1]$.enter image description here

The thing is that you can choose $x_1$ and $x_2$ in all $\mathbb R$, you have no restriction in that. But, once chosen, the inequality would be only valid for the points between.

  • 0
    We can choose any $x$ in domain but you restricted the domain. – 2017-02-10
  • 0
    The definition of a convex function is that, given $x_1, x_2\in X$, the image of any point $x\in[x_1, x_2]$ is below the line that goes through $(x_1, f(x_1))$ and $(x_2, f(x_2))$. It is not valid for the points outside $[x_1, x_2]$. Each time you fix two points $x_1$ and $x_2$, you are fixing the line that goes through $(x_1, f(x_1))$ and $(x_2, f(x_2))$. – 2017-02-10
  • 0
    Can you draw a picture for better understanding ? – 2017-02-10
  • 0
    Okay , Thank you a lot – 2017-02-10
  • 0
    Thank you but I have a problem still . In your example you choose $-2$ , $1$ which they are not in $[0 , 1]$ interval – 2017-02-10
  • 0
    Note that $t$ is what you choose in $[0,1]$, not $x_1$ and $x_2$. Those points can be chosen in $X$ (in the example, $X=\mathbb R)$. Choosing $t$ is the same as choosing a point between $x_1$ and $x_2$. For example, with $t=\frac12$ you get the middle point, $\frac12 x_1 + \frac12 x_2$. – 2017-02-10
  • 0
    Okay but in your answer in the first line , you said "Let us choose two points $x_1, x_2\in [0,1]$" – 2017-02-10
  • 0
    Can you explain more about the equation that you wrote ? What is the $s$ and $t$ ? – 2017-02-10
  • 0
    I have rewrite it in another way. I hope it is clear now. – 2017-02-10
  • 0
    Can you rewrite in $y=mx+h$ form ? I'm very sorry for many questions that I've asked. – 2017-02-10
  • 0
    I will rewrite in a moment. It's OK, no problem – 2017-02-10