Pointwise product of uniformly continuous functions












2














True or false: Let $f(x)$ and $g(x)$ be uniformly continuous functions from $mathbb{R}$ to $mathbb{R}$.
Then their pointwise product $f(x)g(x)$ is uniformly continuous.



I think it will be true; take $f(x)= g(x) = sqrt x$.



Am I right or wrong?






real-analysis uniform-continuity







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  • 7




    You don't prove a statement true by finding an example for which the statement holds.
    – ab123
    Nov 11 at 12:52










  • It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
    – JonathanZ
    Nov 11 at 17:57


















2














True or false: Let $f(x)$ and $g(x)$ be uniformly continuous functions from $mathbb{R}$ to $mathbb{R}$.
Then their pointwise product $f(x)g(x)$ is uniformly continuous.



I think it will be true; take $f(x)= g(x) = sqrt x$.



Am I right or wrong?






real-analysis uniform-continuity







share|cite|improve this question




















  • 7




    You don't prove a statement true by finding an example for which the statement holds.
    – ab123
    Nov 11 at 12:52










  • It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
    – JonathanZ
    Nov 11 at 17:57
















2












2








2







True or false: Let $f(x)$ and $g(x)$ be uniformly continuous functions from $mathbb{R}$ to $mathbb{R}$.
Then their pointwise product $f(x)g(x)$ is uniformly continuous.



I think it will be true; take $f(x)= g(x) = sqrt x$.



Am I right or wrong?






real-analysis uniform-continuity







share|cite|improve this question















True or false: Let $f(x)$ and $g(x)$ be uniformly continuous functions from $mathbb{R}$ to $mathbb{R}$.
Then their pointwise product $f(x)g(x)$ is uniformly continuous.



I think it will be true; take $f(x)= g(x) = sqrt x$.



Am I right or wrong?






real-analysis uniform-continuity




real-analysis uniform-continuity






share|cite|improve this question















share|cite|improve this question













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edited Nov 11 at 19:04









jwodder

1,262918




1,262918










asked Nov 11 at 12:50









santosh

969




969








  • 7




    You don't prove a statement true by finding an example for which the statement holds.
    – ab123
    Nov 11 at 12:52










  • It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
    – JonathanZ
    Nov 11 at 17:57
















  • 7




    You don't prove a statement true by finding an example for which the statement holds.
    – ab123
    Nov 11 at 12:52










  • It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
    – JonathanZ
    Nov 11 at 17:57










7




7




You don't prove a statement true by finding an example for which the statement holds.
– ab123
Nov 11 at 12:52




You don't prove a statement true by finding an example for which the statement holds.
– ab123
Nov 11 at 12:52












It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
– JonathanZ
Nov 11 at 17:57






It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
– JonathanZ
Nov 11 at 17:57












2 Answers
2






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5














It is false. Just take $f(x)=g(x)=x$.






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    1














    The uniform continuity property refers to a spatially homogeneous behavior of continuity. Look at the graph of the following functions.




    1. $f(x)=x$

    2. $f(x)=sqrt{|x|}$

    3. $f(x)=x^2$

    4. $$ f(x)= begin{cases}
      xsin(frac{1}{x}) & xne 0 \
      0 & x=0
      end{cases}$$


    Note that (2) has a "vertiginous behavior" around $0$, since that $$f'(x)=frac{1}{2sqrt{x}}to infty$$ when $xdownarrow 0$.
    Also for (3) this behavior occurs when $|x|to infty$.
    (4) is continuous but is even "less uniform" than previous cases.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5














      It is false. Just take $f(x)=g(x)=x$.






      share|cite|improve this answer


























        5














        It is false. Just take $f(x)=g(x)=x$.






        share|cite|improve this answer
























          5












          5








          5






          It is false. Just take $f(x)=g(x)=x$.






          share|cite|improve this answer












          It is false. Just take $f(x)=g(x)=x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 11 at 12:53









          José Carlos Santos

          149k22117219




          149k22117219























              1














              The uniform continuity property refers to a spatially homogeneous behavior of continuity. Look at the graph of the following functions.




              1. $f(x)=x$

              2. $f(x)=sqrt{|x|}$

              3. $f(x)=x^2$

              4. $$ f(x)= begin{cases}
                xsin(frac{1}{x}) & xne 0 \
                0 & x=0
                end{cases}$$


              Note that (2) has a "vertiginous behavior" around $0$, since that $$f'(x)=frac{1}{2sqrt{x}}to infty$$ when $xdownarrow 0$.
              Also for (3) this behavior occurs when $|x|to infty$.
              (4) is continuous but is even "less uniform" than previous cases.






              share|cite|improve this answer


























                1














                The uniform continuity property refers to a spatially homogeneous behavior of continuity. Look at the graph of the following functions.




                1. $f(x)=x$

                2. $f(x)=sqrt{|x|}$

                3. $f(x)=x^2$

                4. $$ f(x)= begin{cases}
                  xsin(frac{1}{x}) & xne 0 \
                  0 & x=0
                  end{cases}$$


                Note that (2) has a "vertiginous behavior" around $0$, since that $$f'(x)=frac{1}{2sqrt{x}}to infty$$ when $xdownarrow 0$.
                Also for (3) this behavior occurs when $|x|to infty$.
                (4) is continuous but is even "less uniform" than previous cases.






                share|cite|improve this answer
























                  1












                  1








                  1






                  The uniform continuity property refers to a spatially homogeneous behavior of continuity. Look at the graph of the following functions.




                  1. $f(x)=x$

                  2. $f(x)=sqrt{|x|}$

                  3. $f(x)=x^2$

                  4. $$ f(x)= begin{cases}
                    xsin(frac{1}{x}) & xne 0 \
                    0 & x=0
                    end{cases}$$


                  Note that (2) has a "vertiginous behavior" around $0$, since that $$f'(x)=frac{1}{2sqrt{x}}to infty$$ when $xdownarrow 0$.
                  Also for (3) this behavior occurs when $|x|to infty$.
                  (4) is continuous but is even "less uniform" than previous cases.






                  share|cite|improve this answer












                  The uniform continuity property refers to a spatially homogeneous behavior of continuity. Look at the graph of the following functions.




                  1. $f(x)=x$

                  2. $f(x)=sqrt{|x|}$

                  3. $f(x)=x^2$

                  4. $$ f(x)= begin{cases}
                    xsin(frac{1}{x}) & xne 0 \
                    0 & x=0
                    end{cases}$$


                  Note that (2) has a "vertiginous behavior" around $0$, since that $$f'(x)=frac{1}{2sqrt{x}}to infty$$ when $xdownarrow 0$.
                  Also for (3) this behavior occurs when $|x|to infty$.
                  (4) is continuous but is even "less uniform" than previous cases.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 11 at 13:38









                  Daniel Camarena Perez

                  57538




                  57538






























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