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If I have the inequality $f \leq g$, does this imply $f'\leq g'$?

The reason I am asking is because I am using this fact to prove a question with Taylor theorem.

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    $\begingroup$ On a somewhat related note, does $a \le b, \,c \le d$ imply $a-c \le b-d\,$? $\endgroup$ – dxiv May 24 '17 at 3:49
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    $\begingroup$ Can you think of a increasing function whose graph is under the x-axis? Yes, you can (just take any bounded increasing function and apply a translation). So, the answer is no because your questions is equivalent to "is a function which attains only non-positive values necessarily non-increasing?" (see my post for details). $\endgroup$ – Pedro May 24 '17 at 4:46
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    $\begingroup$ @AspiringMat: Oh... please prioritize sleep higher in your life :( it does wonders... $\endgroup$ – user541686 May 24 '17 at 8:53
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    $\begingroup$ No, but inequalities are preserved under integration over a positive directed interval, for example $x \leq e^x \implies \int_0^x x'{\rm d}x' \leq \int_0^x e^{x'}{\rm d}x'$ or $\frac{x^2}{2} \leq e^x-1$ for $x>0$. $\endgroup$ – Winther May 24 '17 at 12:17
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    $\begingroup$ @Mehrdad Strange things frighten you. Let's avoid being condescending with people trying to learn. $\endgroup$ – thumbtackthief May 24 '17 at 18:25
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Consider $x^2\leq x $ in $[0,1]$. Does that imply that $2x\leq 1$ for all x in $[0,1] $? ( A counterexample ).

Note you can never differentiate with an inequality. Instead, the general idea for checking inequalities with differentiation is that we take $h (x)=f (x)-g (x) $ and then try the derivative test to see whether function is increasing or decreasing. That way, if the inequality $h(a) \geq 0$ holds at a particular point a, we can prove it holds for all $x \geq a$ if $h$ is increasing and for all $x \leq a$ if $h$ is decreasing.

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    $\begingroup$ does not imply? $\endgroup$ – mathematician May 24 '17 at 5:14
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    $\begingroup$ No I wanted to make it rhetorical. $\endgroup$ – Archis Welankar May 24 '17 at 6:04
  • $\begingroup$ Thank you. It's 5 am here and I was really confused. Don't even know why i asked this question in the first place $\endgroup$ – AspiringMat May 24 '17 at 8:51
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    $\begingroup$ No problem we are humans! $\endgroup$ – Archis Welankar May 24 '17 at 9:16
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Your question is equivalent to is a function which attains only non-positive values necessarily non-increasing? So, the answer is no because there exist increasing functions whose graphs are under the $x$-axis:

enter image description here

(As a general remark, the rates of change of a function has "nothing" to do with the image of the function in the sense that: (i) given a set $X$, we can find functions having $X$ as image and different rates of change; (ii) given a function, there are other functions having the same rates of change and different images.)

Details: As differentiation is linear, the question

Does $f\leq g$ imply $f'\leq g'$?

is the same as

Does $h\leq 0$ imply $h'\leq 0$?

which can be rewritten as

Is a function which attains only non-positive values necessarily non-increasing?

From this point of view, the answer becomes clear: No because there are (a lot of) increasing functions whose graphs are under the $x$-axis (take any bounded increasing function and apply a translation). Probably, the simplest example is a line segment (see picture above).

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    $\begingroup$ +1 But I wouldn't say that it has nothing to do with the image of the function though $\endgroup$ – Ant May 24 '17 at 7:56
  • $\begingroup$ I would go with $0 \leq h$ and $0 \leq h'$, and a decreasing positive function. The first quadrant is the most usual... :-) $\endgroup$ – Pablo H May 24 '17 at 13:23
  • $\begingroup$ @Ant I've edited that sentence to make clear my point. $\endgroup$ – Pedro May 24 '17 at 13:42
  • $\begingroup$ @PabloH The setting you suggest is indeed good. It is simpler in the sense that we do not even need to think in translations. My setting is a consequence of the habit of isolating things at left when dealing with algebraic expressions. $\endgroup$ – Pedro May 24 '17 at 13:58
  • $\begingroup$ It took me a second, but I thought of a function that converges at a limit as an example of a n always-increasing function that never achieves a positive value. $\endgroup$ – DeadMG May 24 '17 at 22:34
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Take two functions whose derivative have whatever inequality you wish (ie., $f^\prime > g^\prime$ or vice versa). Then just add or subtract a constant to one of the functions to make it larger or smaller than the other.

In short, you can very easily create pairs of functions whose values obey one inequality relation and their derivatives hold another. Or the same.

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