If two functions are increasing is their product increasing?












2














The problem is:




If $f$ and $g$ are increasing, then is $f cdot g$ also increasing?




First, I started out by working some basic definitions and assumptions:




  • Assume $a < b$

  • Increasing means $f(a) < f(b)$


Then I tried using case analysis:



Case 1: Assume $f(a), f(b), g(a), g(b)$ are all positive. Then:



$$f(a)g(a) < f(b)g(b)$$



which means $f cdot g$ is increasing.



Case 2: Assume $f(a), f(b), g(a), g(b)$ are all negative. Then:



$$f(a)g(a) > f(b)g(b)$$



which means $f cdot g$ is not increasing.



Case 3: Assume $f(a), f(b)$ are positive and $g(a), g(b)$ are negative. Then:



$$f(a)g(a) > f(b)g(b)$$



Means $f cdot g$ is not increasing.




A few problems that I don't like with my approach:




  • The case analysis is quite cumbersome, and I still miss cases where the functions are equal to $0$.


  • There is no real proof here, I mostly tried with numerical examples and general logic. Is there an algebraic way to start with the inequality $f(a) < f(b)$ and then modify it so that we end up with $f(a)g(a) < f(b)g(b)$? I couldn't get anywhere with that approach though.







real-analysis calculus proof-writing monotone-functions







share|cite|improve this question





























    2














    The problem is:




    If $f$ and $g$ are increasing, then is $f cdot g$ also increasing?




    First, I started out by working some basic definitions and assumptions:




    • Assume $a < b$

    • Increasing means $f(a) < f(b)$


    Then I tried using case analysis:



    Case 1: Assume $f(a), f(b), g(a), g(b)$ are all positive. Then:



    $$f(a)g(a) < f(b)g(b)$$



    which means $f cdot g$ is increasing.



    Case 2: Assume $f(a), f(b), g(a), g(b)$ are all negative. Then:



    $$f(a)g(a) > f(b)g(b)$$



    which means $f cdot g$ is not increasing.



    Case 3: Assume $f(a), f(b)$ are positive and $g(a), g(b)$ are negative. Then:



    $$f(a)g(a) > f(b)g(b)$$



    Means $f cdot g$ is not increasing.




    A few problems that I don't like with my approach:




    • The case analysis is quite cumbersome, and I still miss cases where the functions are equal to $0$.


    • There is no real proof here, I mostly tried with numerical examples and general logic. Is there an algebraic way to start with the inequality $f(a) < f(b)$ and then modify it so that we end up with $f(a)g(a) < f(b)g(b)$? I couldn't get anywhere with that approach though.







    real-analysis calculus proof-writing monotone-functions







    share|cite|improve this question



























      2












      2








      2







      The problem is:




      If $f$ and $g$ are increasing, then is $f cdot g$ also increasing?




      First, I started out by working some basic definitions and assumptions:




      • Assume $a < b$

      • Increasing means $f(a) < f(b)$


      Then I tried using case analysis:



      Case 1: Assume $f(a), f(b), g(a), g(b)$ are all positive. Then:



      $$f(a)g(a) < f(b)g(b)$$



      which means $f cdot g$ is increasing.



      Case 2: Assume $f(a), f(b), g(a), g(b)$ are all negative. Then:



      $$f(a)g(a) > f(b)g(b)$$



      which means $f cdot g$ is not increasing.



      Case 3: Assume $f(a), f(b)$ are positive and $g(a), g(b)$ are negative. Then:



      $$f(a)g(a) > f(b)g(b)$$



      Means $f cdot g$ is not increasing.




      A few problems that I don't like with my approach:




      • The case analysis is quite cumbersome, and I still miss cases where the functions are equal to $0$.


      • There is no real proof here, I mostly tried with numerical examples and general logic. Is there an algebraic way to start with the inequality $f(a) < f(b)$ and then modify it so that we end up with $f(a)g(a) < f(b)g(b)$? I couldn't get anywhere with that approach though.







      real-analysis calculus proof-writing monotone-functions







      share|cite|improve this question















      The problem is:




      If $f$ and $g$ are increasing, then is $f cdot g$ also increasing?




      First, I started out by working some basic definitions and assumptions:




      • Assume $a < b$

      • Increasing means $f(a) < f(b)$


      Then I tried using case analysis:



      Case 1: Assume $f(a), f(b), g(a), g(b)$ are all positive. Then:



      $$f(a)g(a) < f(b)g(b)$$



      which means $f cdot g$ is increasing.



      Case 2: Assume $f(a), f(b), g(a), g(b)$ are all negative. Then:



      $$f(a)g(a) > f(b)g(b)$$



      which means $f cdot g$ is not increasing.



      Case 3: Assume $f(a), f(b)$ are positive and $g(a), g(b)$ are negative. Then:



      $$f(a)g(a) > f(b)g(b)$$



      Means $f cdot g$ is not increasing.




      A few problems that I don't like with my approach:




      • The case analysis is quite cumbersome, and I still miss cases where the functions are equal to $0$.


      • There is no real proof here, I mostly tried with numerical examples and general logic. Is there an algebraic way to start with the inequality $f(a) < f(b)$ and then modify it so that we end up with $f(a)g(a) < f(b)g(b)$? I couldn't get anywhere with that approach though.







      real-analysis calculus proof-writing monotone-functions




      real-analysis calculus proof-writing monotone-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited 2 days ago









      amWhy

      192k28224439




      192k28224439










      asked Jan 3 at 0:36









      Max

      636519




      636519






















          2 Answers
          2






          active

          oldest

          votes


















          5














          The statement is false and a single example will prove it. Take $fcolonmathbb{R}longrightarrowmathbb R$ defined by $f(x)=-e^{-x}$ and take $g=f$. Then $f$ and $g$ are increasing. But $(f.g)(x)=e^{-2x}$ and therefore $f.g$ is decreasing.


          Of course, this is an extreme example: both $f$ and $g$ are increasing, but $f.g$ is decreasing. If you consider the simpler example in which $f(x)=g(x)=x$, then $f.g$ is neither increasing nor decreasing. (Of course, as I did before, I am assuming that the domain of $f$ and of $g$ is $mathbb R$).




          share|cite|improve this answer























          • Ok fair enough, I should've considered just a single example to show the statement is false instead of jumping through the hoops of case analysis and trying to prove it's true.
            – Max
            Jan 3 at 0:51



















          2














          Consider $f(x)=x-1$ and $g(x)=x+1$. Both are increasing; on the other hand,
          $$
          f(x)g(x)=x^2-1
          $$
          is decreasing over $(-infty,0]$ and increasing over $[0,infty)$.



          What's the problem? In the interval $(-infty,-1)$, both functions are negative, so their product is decreasing: if $a<b<-1$, we have $f(a)<f(b)<0$ and $g(a)<g(b)<0$, so
          $$
          f(a)g(a)>f(b)g(b)
          $$
          Similarly, the product is certainly increasing where both functions are positive.



          In the part where one function is positive and one is negative, anything can happen. In this particular case, $fg$ is decreasing over $(-1,0)$ and increasing over $(0,1)$.



          You can try and look for examples where weird things happen. For instance, with $f(x)=x^3$ and $g(x)=x+1$, over the interval $(-1,0)$ the function has a minimum.






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5














            The statement is false and a single example will prove it. Take $fcolonmathbb{R}longrightarrowmathbb R$ defined by $f(x)=-e^{-x}$ and take $g=f$. Then $f$ and $g$ are increasing. But $(f.g)(x)=e^{-2x}$ and therefore $f.g$ is decreasing.


            Of course, this is an extreme example: both $f$ and $g$ are increasing, but $f.g$ is decreasing. If you consider the simpler example in which $f(x)=g(x)=x$, then $f.g$ is neither increasing nor decreasing. (Of course, as I did before, I am assuming that the domain of $f$ and of $g$ is $mathbb R$).




            share|cite|improve this answer























            • Ok fair enough, I should've considered just a single example to show the statement is false instead of jumping through the hoops of case analysis and trying to prove it's true.
              – Max
              Jan 3 at 0:51
















            5














            The statement is false and a single example will prove it. Take $fcolonmathbb{R}longrightarrowmathbb R$ defined by $f(x)=-e^{-x}$ and take $g=f$. Then $f$ and $g$ are increasing. But $(f.g)(x)=e^{-2x}$ and therefore $f.g$ is decreasing.


            Of course, this is an extreme example: both $f$ and $g$ are increasing, but $f.g$ is decreasing. If you consider the simpler example in which $f(x)=g(x)=x$, then $f.g$ is neither increasing nor decreasing. (Of course, as I did before, I am assuming that the domain of $f$ and of $g$ is $mathbb R$).




            share|cite|improve this answer























            • Ok fair enough, I should've considered just a single example to show the statement is false instead of jumping through the hoops of case analysis and trying to prove it's true.
              – Max
              Jan 3 at 0:51














            5












            5








            5






            The statement is false and a single example will prove it. Take $fcolonmathbb{R}longrightarrowmathbb R$ defined by $f(x)=-e^{-x}$ and take $g=f$. Then $f$ and $g$ are increasing. But $(f.g)(x)=e^{-2x}$ and therefore $f.g$ is decreasing.


            Of course, this is an extreme example: both $f$ and $g$ are increasing, but $f.g$ is decreasing. If you consider the simpler example in which $f(x)=g(x)=x$, then $f.g$ is neither increasing nor decreasing. (Of course, as I did before, I am assuming that the domain of $f$ and of $g$ is $mathbb R$).




            share|cite|improve this answer














            The statement is false and a single example will prove it. Take $fcolonmathbb{R}longrightarrowmathbb R$ defined by $f(x)=-e^{-x}$ and take $g=f$. Then $f$ and $g$ are increasing. But $(f.g)(x)=e^{-2x}$ and therefore $f.g$ is decreasing.


            Of course, this is an extreme example: both $f$ and $g$ are increasing, but $f.g$ is decreasing. If you consider the simpler example in which $f(x)=g(x)=x$, then $f.g$ is neither increasing nor decreasing. (Of course, as I did before, I am assuming that the domain of $f$ and of $g$ is $mathbb R$).





            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered Jan 3 at 0:41









            José Carlos Santos

            151k22123224




            151k22123224












            • Ok fair enough, I should've considered just a single example to show the statement is false instead of jumping through the hoops of case analysis and trying to prove it's true.
              – Max
              Jan 3 at 0:51


















            • Ok fair enough, I should've considered just a single example to show the statement is false instead of jumping through the hoops of case analysis and trying to prove it's true.
              – Max
              Jan 3 at 0:51
















            Ok fair enough, I should've considered just a single example to show the statement is false instead of jumping through the hoops of case analysis and trying to prove it's true.
            – Max
            Jan 3 at 0:51




            Ok fair enough, I should've considered just a single example to show the statement is false instead of jumping through the hoops of case analysis and trying to prove it's true.
            – Max
            Jan 3 at 0:51











            2














            Consider $f(x)=x-1$ and $g(x)=x+1$. Both are increasing; on the other hand,
            $$
            f(x)g(x)=x^2-1
            $$
            is decreasing over $(-infty,0]$ and increasing over $[0,infty)$.



            What's the problem? In the interval $(-infty,-1)$, both functions are negative, so their product is decreasing: if $a<b<-1$, we have $f(a)<f(b)<0$ and $g(a)<g(b)<0$, so
            $$
            f(a)g(a)>f(b)g(b)
            $$
            Similarly, the product is certainly increasing where both functions are positive.



            In the part where one function is positive and one is negative, anything can happen. In this particular case, $fg$ is decreasing over $(-1,0)$ and increasing over $(0,1)$.



            You can try and look for examples where weird things happen. For instance, with $f(x)=x^3$ and $g(x)=x+1$, over the interval $(-1,0)$ the function has a minimum.






            share|cite|improve this answer


























              2














              Consider $f(x)=x-1$ and $g(x)=x+1$. Both are increasing; on the other hand,
              $$
              f(x)g(x)=x^2-1
              $$
              is decreasing over $(-infty,0]$ and increasing over $[0,infty)$.



              What's the problem? In the interval $(-infty,-1)$, both functions are negative, so their product is decreasing: if $a<b<-1$, we have $f(a)<f(b)<0$ and $g(a)<g(b)<0$, so
              $$
              f(a)g(a)>f(b)g(b)
              $$
              Similarly, the product is certainly increasing where both functions are positive.



              In the part where one function is positive and one is negative, anything can happen. In this particular case, $fg$ is decreasing over $(-1,0)$ and increasing over $(0,1)$.



              You can try and look for examples where weird things happen. For instance, with $f(x)=x^3$ and $g(x)=x+1$, over the interval $(-1,0)$ the function has a minimum.






              share|cite|improve this answer
























                2












                2








                2






                Consider $f(x)=x-1$ and $g(x)=x+1$. Both are increasing; on the other hand,
                $$
                f(x)g(x)=x^2-1
                $$
                is decreasing over $(-infty,0]$ and increasing over $[0,infty)$.



                What's the problem? In the interval $(-infty,-1)$, both functions are negative, so their product is decreasing: if $a<b<-1$, we have $f(a)<f(b)<0$ and $g(a)<g(b)<0$, so
                $$
                f(a)g(a)>f(b)g(b)
                $$
                Similarly, the product is certainly increasing where both functions are positive.



                In the part where one function is positive and one is negative, anything can happen. In this particular case, $fg$ is decreasing over $(-1,0)$ and increasing over $(0,1)$.



                You can try and look for examples where weird things happen. For instance, with $f(x)=x^3$ and $g(x)=x+1$, over the interval $(-1,0)$ the function has a minimum.






                share|cite|improve this answer












                Consider $f(x)=x-1$ and $g(x)=x+1$. Both are increasing; on the other hand,
                $$
                f(x)g(x)=x^2-1
                $$
                is decreasing over $(-infty,0]$ and increasing over $[0,infty)$.



                What's the problem? In the interval $(-infty,-1)$, both functions are negative, so their product is decreasing: if $a<b<-1$, we have $f(a)<f(b)<0$ and $g(a)<g(b)<0$, so
                $$
                f(a)g(a)>f(b)g(b)
                $$
                Similarly, the product is certainly increasing where both functions are positive.



                In the part where one function is positive and one is negative, anything can happen. In this particular case, $fg$ is decreasing over $(-1,0)$ and increasing over $(0,1)$.



                You can try and look for examples where weird things happen. For instance, with $f(x)=x^3$ and $g(x)=x+1$, over the interval $(-1,0)$ the function has a minimum.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                egreg

                178k1484201




                178k1484201






























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