To be concave on $[0,1]$, $f(t)\leq f(0)+tf'(0)$ is enough?

Question

Suppose $f(t)>0$.

$f(t)\leq f(0)+tf'(0)$ if and only if $f$ is concave over $[0,1]$.

Is the above stetement true? There must be a counterexample in my opinion.

Answer 1

An obvious counterexample would be $f(t)=\sin(2\pi t)+1$, which is neither convex nor concave but fulfills your inequality.

Answer 2

$f(x)=\text{arctan}(2x-1) + \pi$ is above its tangent line in $0$ over all $[0, 1]$, but it is concave in $[0, 1/2)$ and convex in $(1/2, 1]$.

Answer 3

If you think about it geometrically, it is clear that the statement is false. In fact, in geometrical terms it says that the function $f(t)$ in $[0,1]$ must lie below its tangent at origin. To realize this, it is useful to change the inequality (for $t \in (0, 1]$) to the form:

$$\frac{f(t) - f(0)}{t} \leq f'(t)$$

and recall that the expression on the left is the slope of the line between $(0, f(0))$ and $(t, f(t))$.

As a intituitive counterexample, you can draw something concave at first, with a an inflection point that makes it convex before $t=1$. The hypothesis $f(t) > 0$ is not a big constraint (geometrically it tells you that you must draw in the trapezoid given by the x and y axis, the tangent in the origin and the line $x=1$).

If you want a mathematical expression, you can use the fine example of Toffomat: $$\sin(2\pi t) + \alpha$$

valid for any $\alpha > 1$.

If you prefer to go polynomial, consider the following translation of the standard cubic with 3 zeros:

$$(t-\alpha + 1)(t-\alpha)(t-\alpha -1)$$

This is a counterexample for all $\alpha \in (-1, \beta)$, for a $\beta$ that can be prove to exist in $(-1/2, 0)$.