We observed the following in communication with M. Albijanic. Suppose that f is continuous function on closed interval [a, b] and differentiable on open interval (a, b). Then f is strictly convex iff for every points y, z in [a, b] there is an unique c such that f (z) -f (y) = f' (c) (z-y). This statement has visual geometric interpretation. Perhaps, anyone knows the right reference for this statement. For elementary properties of convex functions see for example: Miodrag Mateljevic, Marek Svetlik, Miloljub Albijanic, Nebojv sa Savic, generalizations of the Lagrange mean value theorem and applications, Filomat 27: 2 (2013).
I only outline the problem. Peter T Breuer is right. The force of argument is the additional "unique c" requirement?
It forces convexity or concavity.
It seemss that the following statement is appropriate. Suppose that f is continuous function on closed interval [a, b] and differentiable on open interval (a, b). Then f is strictly convex or f is strictly concave iff (B): for every points y, z in [a, b] there is an unique c such that f (z) -f (y) = f' (c) (z-y).
For $x< y$ define k_f(x,y)=\frac{f(y)-f(x)}{y-x}. Suppose that (i) f is continuous convex function on closed interval [a, b], then
(ii) k_f(x,y)\leq k_f(x,y) if x
Peter T Breuer gave interesting comments.
Really, a twice differentiable function of one variable is convex on an interval if and only if its second derivative is non-negative there; this gives a practical test for convexity. Visually, a twice differentiable convex function "curves up", without any bends the other way (inflection points). If its second derivative is positive at all points then the function is strictly convex, but the converse does not hold.
Suppose that f is continuous function on closed interval $[a, b]$ and differentiable on open interval $(a, b)$. Then (A): f is strictly convex or f is strictly concave iff (B): for every points $y, z$ in $[a, b]$ there is an unique c such that $f (z) -f (y) = f' (c) (z-y)$.
(C) every line intersect the graph at most two points.
(D) The line segment between any two points on the graph lies above (or below) the graph.
(B) implies (C)
On the contrary suppose that (C) does not hold.
Hence there a line (with the slope k) which intersects the graph G_f of f at least three points M_i=(x_i,y_i), 1=1,2,3, x_1
Suppose that f is continuous function on closed interval [a, b] and differentiable on open interval (a, b). Then f is strictly convex iff f' is increasing.
(B): for every points $y, z$ in $[a, b]$ there is an unique c such that $f (z) -f (y) = f' (c) (z-y)$.
One can prove.
(B) implies (B'): f' is increasing or decreasing.
Is there connection with Darboux's theorem?
Can we use Darboux's theorem to conclude (B) implies (B'). .
Darboux's theorem is a theorem in real analysis, named after Jean Gaston Darboux. It states that all functions that result from the differentiation of other functions have the intermediate value property: the image of an interval is also an interval.
When f is continuously differentiable (f in C1([a,b])), this is a consequence of the intermediate value theorem. But even when f′ is not continuous, Darboux's theorem places a severe restriction on what it can be.
Dear followers,
We give more details related to the posed question.
The main contribution is outline of proof of the following :
Suppose that f is continuous function on closed interval $[a, b]$ and differentiable on open interval $(a, b)$. If in addition f has Lagrange unique property. Then f’ is strictly decreasing or increasing.
We expect that this is known, but we could not find an exact reference for this statement.
Suppose that f is continuous function on closed interval $[a, b]$ and differentiable on open interval $(a, b)$. Then
(A): f is strictly convex or f is strictly concave iff
(B): for every points $y, z$ in $[a, b]$ there is an unique c such that $f (z) -f (y) = f' (c) (z-y)$ (Lagrange unique property).
(C) every line intersect the graph at most two points (two points property).
(D) The line segment between any two points on the graph lies above (or below) the graph.
(B) implies (C)
On the contrary suppose that (C) does not hold.
Hence there a line (with the slope k) which intersects the graph G_f of f at least three points M_i=(x_i,y_i), 1=1,2,3, x_1
Dear followers,
We give more details related to the posed question.
We give a short proof of the following.
If f has Lagrange unique property then f is strictly convex or concave.
Suppose that f is continuous function on closed interval $[a, b]$ and differentiable on open interval $(a, b)$.
Miloljub Albijanic suggested a different proof.
We expect that this is known, but we could not find an exact reference for this statement.
The proof is based on
Lemma 1. Let x_1
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