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I want to know if any real function can be expressed as: $f(x)=g(x)+h(x)$ such as $g(x)$ is an increasing function and $h(x)$ is a decreasing function? thanks

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    Could you share your thoughts? For example, How would you represent $f(x)=x^2$ like this? – S.C.B. Mar 10 '17 at 09:38
  • I don't understand could you please explain !? –  Mar 10 '17 at 09:50
  • What did you attempt to do to solve this problem? – S.C.B. Mar 10 '17 at 09:51
  • to be honest, i think we can't express a function like so, so i went to "argument to absurdity" i said let's supose f(x)=g(x)+h(x)--- g is increasing so if x1<x2 then g(x1)<g(x2) so g(x1)-g(x2)<0--- h is decreasing so if x1<x2 then h(x1)>h(x2) so h(x1)-h(x2)>0--- then i suposed f increasing, in order to end up with something wrong so if x1<x2 then f(x1)<(x2) --- g(x1)+h(x1)<g(x2)+h(x2)--- g(x1)-g(x2)<h(x2)-h(x1)--- ... that's what i tried to do :/ –  Mar 10 '17 at 10:08
  • Related: http://math.stackexchange.com/questions/141338/bounded-variation-difference-of-two-increasing-functions – Hans Lundmark Mar 10 '17 at 10:15
  • To start with, in order to have a precise question, you should define what you mean by "increasing" and "decreasing". Do you mean that $x\lt y\implies g(x)\lt g(y),$ or do you mean $x\lt y\implies g(x)\le g(y)$? Anyway the answer is no. For one thing, $g(x)+h(x)$ can have at most countably many discontinuities. – bof Mar 10 '17 at 11:30
  • On the other hand, it's true for functions of bounded variation. – bof Mar 10 '17 at 11:33
  • @S.C.B. If $f(x)=x^2$ you can take, for example, $g(x)=x+x^2H(x)$ (where $H(x)$ is the Heaviside unit step function; then $g(x)$ is strictly increasing and $h(x)=f(x)-g(x)$ is strictly decreasing. – bof Mar 10 '17 at 13:11
  • Nice comments though,any example of a function which cannot be written as the sum of an increasing function and an decreasing function ?especially by @bof relating to countable discontinuity ? – BAYMAX Mar 11 '17 at 01:26
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    @BAYMAX The Dirichlet function, being discontinuous everywhere, cannot be written as the sum of two monotone functions. – bof Mar 11 '17 at 01:29
  • @bof ok as in the dirichlet function there are uncountable no. of discontinuities but in the sum of continuous functions being continuous has only countable number of discontinuities and hence giving contradiction..ok can we give an example of continuous function ? – BAYMAX Mar 11 '17 at 01:32
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    $f(x)=x\sin\frac1x$ with $f(0)=0.$ – bof Mar 11 '17 at 01:49

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The difference of two increasing functions is a function of Bounded Variation. Note that this answers your question.

robjohn
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