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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.

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

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  • $\begingroup$ The problem mentions that $f(t)>0$ $\endgroup$ – GoodDeeds Sep 6 '18 at 7:57
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    $\begingroup$ @GoodDeeds: Oh, missed that. Doesn't change much though... $\endgroup$ – Toffomat Sep 6 '18 at 8:05
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$f(x)=\text{arctan}(2x-1) + \pi$ is above its tangent line in $0$ over all $[0, 1]$, but it is concave in $[0, 1/2)$ and convex in $(1/2, 1]$.

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    $\begingroup$ Why is a straight line not concave? $\endgroup$ – Calvin Khor Sep 6 '18 at 7:47
  • $\begingroup$ Because a definition of concavity is the graph of the function being completely below the line connecting two points of the graph, of course except the two points themselves. Also, another definition is whenever the second derivative is positive and not zero. At least, this is what I recall; is a straight line also convex? $\endgroup$ – Niki Di Giano Sep 6 '18 at 7:51
  • $\begingroup$ The problem mentions that $f(t)>0$ $\endgroup$ – GoodDeeds Sep 6 '18 at 7:57
  • $\begingroup$ math.stackexchange.com/questions/611633/… I would say it is convex(and concave), but not strictly convex. $\endgroup$ – Calvin Khor Sep 6 '18 at 7:59
  • $\begingroup$ Well, it's just semantics then. I have addressed both commenters' observations in my last edit. $\endgroup$ – Niki Di Giano Sep 6 '18 at 8:03
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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 $(-1/2, 0)$.

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