Convexity of a function

We call the convexity of a function $f$ its general shape, rather bowl-shaped (convex) or rather cellar-shaped (concave).

We define it more precisely according to the position of its tangents and the chords of the function.

Let $f$ be a function of class $ \mathbb{C}^{2}$ on its definition set $ D_f $ and any interval $I$.

The sign of the second derivative indicates convexity

Just as the derivative indicates the direction of variation of a function, the sign of the second derivative indicates convexity.

$$ \forall x \in I, $$

$$ f''(x) \geqslant 0 \ \Longleftrightarrow f \ convex \ on \ I $$

$$ f''(x) \leqslant 0 \ \Longleftrightarrow f \ concave \ on \ I $$

And,

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

The inequality of tangents

A function $ f $ is said to be convex on an interval $ I $, if any tangent at a point lies below the curve. A contrario, it is concave if any tangent is located above the curve.

Illustration of convexity by any tangent's position

$$\forall (a, x) \in I^2, $$

$$ f \enspace convex \enspace on \enspace I \Longleftrightarrow f(x) \geqslant f'(a)(x - a) + f(a) $$

$$f \enspace concave \enspace on \enspace I \Longleftrightarrow f(x) \leqslant f'(a)(x - a) + f(a) $$

The inequality of convexity

A function $ f $ is said to be convex on an interval $ I $, if any rope which connects two points of this interval lies above the curve. A contrario, it is concave if any rope is located below the curve.

Illustration of convexity by any rope's position

$$ f \enspace convex \enspace on \enspace I \Longleftrightarrow \forall (a,b) \in I^2, \enspace \forall \lambda \in [0,1], \enspace f\bigl(\lambda a + (1- \lambda)b \bigr) \hspace{0.2em} \leqslant \hspace{0.2em} \lambda f(a) + (1 - \lambda)f(b) $$

$$f \enspace concave \enspace on \enspace I \Longleftrightarrow \forall (a,b) \in I^2, \enspace \forall \lambda \in [0,1], \enspace f\bigl(\lambda a + (1- \lambda)b\bigr) \hspace{0.2em} \geqslant \hspace{0.2em} \lambda f(a) + (1 - \lambda)f(b) $$

Demonstrations

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 $a$.

A function with 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:

$$ T_{a}(x) = f'(a)(x - a) + f(a) $$

So, point $M_T$ will have the coordinates:

$$ M_T(m ; \ f'(a)(m - a) + f(a)) $$

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

$$ g(x) = f(x) - T_{a}(x) $$

$$ g(x) = f(x)- f(a) - f'(a)(x - a) \qquad(g) $$

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:

$$ \Biggl \{ \begin{align*} g'(x) = f'(x) - f'(a) \qquad(g') \\ g''(x) = f''(x) \hspace{5.2em}(g'') \end{align*} $$

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

$$ (1) \ \Biggl \{ \begin{align*} g(a) = 0 \\ g'(a) = 0 \\ g''(a) = 0 \ \Longleftrightarrow \ f''(a) = 0 \end{align*} $$

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 $a$:

$$ \Biggl \{ \begin{align*} \Bigl[ \ x < a \ \Longleftrightarrow \ g'(x) < 0 \ \Bigr] \ \Longleftrightarrow \ g \ decreasing \ for \ (x < a) \\ \Bigl[ \ x > a \ \Longleftrightarrow \ g'(x) > 0 \ \Bigr] \ \Longleftrightarrow \ g \ increasing \ for \ (x > a) \end{align*} $$

Function g with its both derivative and second derivative

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 $:

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

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 $a$.

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

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,

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

Function g with its both derivative and second derivative - case with different signs

$g'$ 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.

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

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.:

$$ g(x) = f(x)- f(a) - f'(a)(x - a) > 0$$

Then,

$$\forall (a, x) \in I^2, $$

$$ f \enspace convex \enspace on \enspace I \Longleftrightarrow f(x) \geqslant f'(a)(x - a) + f(a) $$

$$f \enspace concave \enspace on \enspace I \Longleftrightarrow f(x) \leqslant f'(a)(x - a) + f(a) $$

The inequality of convexity

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

Inequality of convexity - introduction of lambda

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

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

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

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

In the following figure, we introduce a rope going from $ a $ to $ b$:

Inequality of convexity - rope 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:

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:

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

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 [a, b] $, its image will always be below the point of the rope $y_c$.

$$ f(c) \leqslant y_c $$

And finally,

$$ f \enspace convex \enspace on \enspace [a,b] \Longleftrightarrow \forall (a,b), \enspace \forall \lambda \in [0,1], \enspace f\bigl(\lambda a + (1- \lambda)b \bigr) \hspace{0.2em} \leqslant \hspace{0.2em} \lambda f(a) + (1 - \lambda)f(b) $$

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

$$\enspace \exists \lambda \in [0,1], \enspace -f\bigl(\lambda a + (1- \lambda)b \bigr) \hspace{0.2em} \leqslant \hspace{0.2em} -\lambda f(a) + ( \lambda - 1)f(b) $$

By multiplying both parts by $ -1 $:

$$f\bigl(\lambda a + (1- \lambda)b \bigr) \hspace{0.2em} \geqslant \hspace{0.2em} \lambda f(a) + ( 1 -\lambda)f(b) $$

$$ f \enspace concave \enspace on \enspace [a,b] \Longleftrightarrow \forall (a,b), \enspace \forall \lambda \in [0,1], \enspace f\bigl(\lambda a + (1- \lambda)b\bigr) \hspace{0.2em} \geqslant \hspace{0.2em} \lambda f(a) + (1 - \lambda)f(b) $$

Examples

The inequality of convexity

Let's determine the convexity of the function $ f : x \longmapsto x^2 $ inside the interval $ [-1,1] $.

We will determine if there is any interval $ \lambda \in [0,1] $ which verifies one or the other convexity inequality.

Side of the curve of $ f $

We calculate in the interval $ [a, b] = [-1,1] $:

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

So,

$$ g(\lambda) = (-\lambda + (1- \lambda))^2 = \lambda -2\lambda(1- \lambda) + (1- \lambda)^2 $$

$$g(\lambda) = \lambda -2\lambda + 2\lambda ^2 + 1 -2\lambda + \lambda ^2$$

$$ g(\lambda) =3\lambda^2 - 3\lambda+ 1 $$

Side of the rope between $ -1 $ and $ 1 $

We calculate in the interval $ [a, b] = [-1,1] $:

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

So,

$$ h(\lambda) = \lambda (-1)^2 + (1- \lambda) 1^2 = \lambda + 1 -\lambda $$

$$ h(\lambda) = 1 $$

Comparison of $ g(\lambda) $ with $ h(\lambda)$

To determine the relative position of these two functions, let's take their difference.

$$ (g-h)(\lambda) = 3\lambda^2 - 3\lambda+ 1 - 1 $$

$$ (g-h)(\lambda) = 3\lambda^2 - 3\lambda $$

$$ (g-h)(\lambda) = 3\lambda (\lambda -1) $$

We can directly make a table of signs.

$$ \lambda $$	$$ -\infty $$	$$ \hspace{3em}$$	$$ 0 $$	$$ \hspace{3em}$$	$$1$$	$$ \hspace{3em}$$	$$ +\infty $$
$$ 3 \lambda $$	$$ - $$	$$ - $$	$$ 0 $$	$$ + $$	$$ + $$	$$ + $$	$$ + $$
$$ \lambda - 1$$	$$ - $$	$$ - $$	$$ - $$	$$ - $$	$$ 0 $$	$$ + $$	$$ - $$
$$ (g-h)(\lambda) $$	$$ + $$	$$ + $$	$$0 $$	$$ - $$	$$ 0 $$	$$ + $$	$$ -$$

We definitely have for all $ \lambda \in [0, 1], \enspace (g-h)(\lambda) \leqslant 0$.

It means that:

$$ g(\lambda) \leqslant f(\lambda) \Longleftrightarrow \bigl[point \enspace de \enspace la \enspace courbe\bigr] \leqslant \bigl[point \enspace de \enspace la \enspace corde \bigr] $$

The ropes are above the curve, the function is convex.