Convexity of a function

The sign of the second derivative indicates convexity

Let \(f\) be a function of class \( \mathbb{C}^{2}\) on its definition set and its tangent at a fixed point \(a\).

Let us seek to determine hte position of its tangent according to the nature of \(f\).

Let be two mobile points \(M(x;\ f(x))\) and \(M_T(x; \ T_a(x))\) depending on \(x\), respective images of \(f\) and the one of \(T_a\) at point \(x\).

The tangent equation \(T_a\) is worth:

So, point \(M_T\) will have the coordinates:

Let us now consider a new function \(g\), being worth the difference between \(f\) and \(T_a\):

To determine which of the two is above the other one, we must study the sign of \(g\) at the neighborhood of \((x=a)\), value for which the function g vanishes.

Function \(f\) being derivable twice by hypothesis and \(T_a\) being of class \( \mathbb{C}^{\infty}\), function \(g\) is also derivable twice, and:

With \((g)\), \((g')\) and \((g'')\), we then have:

1. Assumption of \( f''(x) > 0\) to the left and right of \(a\)

   Let us assume that it is possible to find a positive number \(\eta\) such as:

   $$ 0 < |x-a| < \eta \Longrightarrow f''(x) > 0 \qquad(S_+) $$

   Which means that \(f''\) is supposed to be positive for the values to the left and right of \(a\). Except for \(x=a\), as clearly indicated \((S_+)\) because there may be cases where \(f''(a)\) will be infinite or even not defined at all.

   This assumption made, we see with \((g'')\) that, this also entails \(g'' > 0\) to the left and right of \(a\) with \(g''(a) = 0\).

   Then the derivative \(g'\) will be increasing to the left and right of \(a\), with \(g'(a) = 0\).

   We will then have to the left and right of \(a\):

   $$ \left \{ \begin{gather*} \Bigl[ \ x < a \ \Longleftrightarrow \ g'(x) < 0 \ \Bigr] \ \Longleftrightarrow \ g \text{ is decreasing for } (x < a) \\ \Bigl[ \ x > a \ \Longleftrightarrow \ g'(x) > 0 \ \Bigr] \ \Longleftrightarrow \ g \text{ is increasing for } (x > a) \end{gather*} \right \} $$
   

   Which leads to \(g\) is decreasing on the left and increasing on the right of \(a\).

   But \( g(a) = 0\), so \(g > 0\) left and right in the vicinity of \(a\).

   We can conclude that for any interval \(I \):

   $$ \forall x \in I, \ f''(x) \geqslant 0 \ \Longleftrightarrow f \text{ is convex on } I $$

2. Assumption of \( f''(x) < 0\) to the left and right of \(a\)

   It is possible to make the same demonstration by making the opposite supposition of \((S_+)\), namely:

   $$ 0 < |x-a| < \eta \Longrightarrow f''(x) < 0 \qquad (S_-) $$

   So in this case \(g\) will be increasing to the left and decreasing to the right of \(a\), and we will have in the same way \(g < 0\) left and right in the vicinity of \(a\).

   $$ \forall x \in I, \ f''(x) \leqslant 0 \ \Longleftrightarrow f \text{ is concave on } I $$

3. Assumption of a different sign for \( f''(x)\) to the left and right of \(a\)

   In the case where \( f''\) would have a different sign to the right and left of \(a\), there are then two cases.

   Let us assume a first case, namely that \( f'' < 0\) to the left and \( f'' > 0\) on the right of \(a\). Then,

   $$g''(a) = g'(a) = g(a) = 0 $$
   

   \(g'\) will be decreasing on the left then increasing on the right, which results in \(g'(x) > 0\).

   So \(g\) will always be increasing and as \(g(a) = 0\), we will have \(g < 0\) to the left and \(g > 0\) on the right of \(a\).

   We will then have an inflection point at point \(a\); which means \(f\) will change convexity.

   
   $$f''(x) \text{ changes sign before and after } (x=a) \Longleftrightarrow f \text{ admits an inflection point at } (x=a)$$

The inequality of tangents

As we have just shown previously that if the function \(f\) is convex, then the curve of the function is always above its tangents, i.e.:

Then,

The inequality of convexity

Let \( f \) be a continuous and convex function on an interval \( I = \bigl[a,b \bigr] \), and a point \( c \in \bigl[a,b \bigr] \) of this interval, such as the following figure:

We introduce a real number \(\lambda\), proportion between points \( c \) and \( b \) according to the interval \( \bigl[a,b \bigr] \):

Consequently, we will have the other part which will be equal to:

In the following figure, we introduce a rope going from \( a \) to \( b\):

We will be able to locate the position of the rope in relation to the curve of \( f \) at point \( c \).

We then have:

1. For the curve of \( f \)

   With the equation \( (\lambda)\), we do have:

   $$ \lambda = \frac{b-c}{b-a} \qquad (\lambda) $$

   By performing a cross product :

   $$ \lambda (b-a) = b-c $$
   $$ c = b - \lambda (b-a) $$
   $$ c = b - \lambda b + \lambda a $$
   $$ c = \lambda a + (1 - \lambda) b $$

   Therefore, \( f(c) \) is worth:

   $$ f(c) = f\bigl (\lambda a + (1 - \lambda) b \bigr )$$

2. For the rope going from \( a \) to \( b\)

   By calculating the slope of the rope:

   $$ \frac{f(b) - f(a)}{b-a} = \frac{y_c - f(a)}{c-a} $$
   $$ (c-a) \frac{f(b) - f(a)}{b-a} = y_c - f(a) $$

   But:

   $$ 1 - \lambda = \frac{c-a}{b-a} \qquad (1 - \lambda)$$

   So, replacing it by its value:

   $$ (1 - \lambda) (f(b) - f(a)) = y_c - f(a) $$
   $$ f(b) -f(a) - \lambda f(b) + \lambda f(a) = y_c - f(a) $$
   $$ y_c = f(b) - \lambda f(b) + \lambda f(a) $$
   $$ y_c = \lambda f(a) + (1 - \lambda) f(b) $$

If the function \( f \) is convex, this means that for any point \( c \in \bigl[a, b\bigr] \), its image will always be below the point of the rope \(y_c\).

And finally,

We will say that \( f \) is concave if \( (-f) \) is convexe. Thus:

By multiplying both parts by \( -1 \):