Proving that 2 functions are bounded.












2














Let $f,g:mathbb R to mathbb R$. Let $h_1$ = $f-g$ and $h_2$ = $f+g$. Suppose $h_1$ and $h_2$ are bounded. Show that $f$ and $g$ are bounded.



I understand the concept of bounded but would've thought you would need to use limits or a derivative to do a proof. How would you accomplish this without one.






real-analysis







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  • 2




    Express $f,g $ in the form of $h_2, h_1$.
    – xbh
    Dec 18 at 4:57










  • $|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
    – Henno Brandsma
    Dec 18 at 5:11
















2














Let $f,g:mathbb R to mathbb R$. Let $h_1$ = $f-g$ and $h_2$ = $f+g$. Suppose $h_1$ and $h_2$ are bounded. Show that $f$ and $g$ are bounded.



I understand the concept of bounded but would've thought you would need to use limits or a derivative to do a proof. How would you accomplish this without one.






real-analysis







share|cite|improve this question




















  • 2




    Express $f,g $ in the form of $h_2, h_1$.
    – xbh
    Dec 18 at 4:57










  • $|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
    – Henno Brandsma
    Dec 18 at 5:11














2












2








2







Let $f,g:mathbb R to mathbb R$. Let $h_1$ = $f-g$ and $h_2$ = $f+g$. Suppose $h_1$ and $h_2$ are bounded. Show that $f$ and $g$ are bounded.



I understand the concept of bounded but would've thought you would need to use limits or a derivative to do a proof. How would you accomplish this without one.






real-analysis







share|cite|improve this question















Let $f,g:mathbb R to mathbb R$. Let $h_1$ = $f-g$ and $h_2$ = $f+g$. Suppose $h_1$ and $h_2$ are bounded. Show that $f$ and $g$ are bounded.



I understand the concept of bounded but would've thought you would need to use limits or a derivative to do a proof. How would you accomplish this without one.






real-analysis




real-analysis






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edited Dec 18 at 8:32









Carmeister

2,6742920




2,6742920










asked Dec 18 at 4:51









Pablo Tores

387




387








  • 2




    Express $f,g $ in the form of $h_2, h_1$.
    – xbh
    Dec 18 at 4:57










  • $|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
    – Henno Brandsma
    Dec 18 at 5:11














  • 2




    Express $f,g $ in the form of $h_2, h_1$.
    – xbh
    Dec 18 at 4:57










  • $|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
    – Henno Brandsma
    Dec 18 at 5:11








2




2




Express $f,g $ in the form of $h_2, h_1$.
– xbh
Dec 18 at 4:57




Express $f,g $ in the form of $h_2, h_1$.
– xbh
Dec 18 at 4:57












$|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
– Henno Brandsma
Dec 18 at 5:11




$|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
– Henno Brandsma
Dec 18 at 5:11










2 Answers
2






active

oldest

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5














The sum of bounded functions is bounded and so are scalar multiples. Note that $f= frac{h_1 + h_2}{2}$.



Likewise for differences of bounded functions and $g = frac{h_2 - h_1}{2}$.






share|cite|improve this answer





























    1














    Suppose $h_1, h_2$ are bounded. Let $D(f)$ denote the domain of $f$, where $f$ is a function. Then $exists M_1, M_2$ such that $$h_1(x) = f(x) - g(x) leq M_1 forall x in D(h_1), $$ $$h_2(x) = f(x) + g(x) leq M_2 forall x in D(h_2).$$ So, we have that for arbitrary $x in D(f) cap D(g)$, $$ f(x) - g(x) + f(x) + g(x) = 2f(x) leq M_1 + M_2 implies f(x) leq frac{M_1 + M_2}{2}, $$
    so $f$ is bounded, by definition. Similarly, notice that
    $$ g(x) - f(x) + g(x) + f(x) = 2g(x) leq M_2 - M_1 implies g(x) leq frac{M_2 - M_1}{2}, $$
    so $g$ is bounded, by definition. This completes the proof.






    share|cite|improve this answer



















    • 1




      Maybe it should be $xin D(h_1)cap D(h_2)$?
      – Shubham Johri
      Dec 18 at 5:54












    • Ah, you are right; good catch! I've edited my answer to include this.
      – alexsieusahai
      Dec 18 at 5:58











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    The sum of bounded functions is bounded and so are scalar multiples. Note that $f= frac{h_1 + h_2}{2}$.



    Likewise for differences of bounded functions and $g = frac{h_2 - h_1}{2}$.






    share|cite|improve this answer


























      5














      The sum of bounded functions is bounded and so are scalar multiples. Note that $f= frac{h_1 + h_2}{2}$.



      Likewise for differences of bounded functions and $g = frac{h_2 - h_1}{2}$.






      share|cite|improve this answer
























        5












        5








        5






        The sum of bounded functions is bounded and so are scalar multiples. Note that $f= frac{h_1 + h_2}{2}$.



        Likewise for differences of bounded functions and $g = frac{h_2 - h_1}{2}$.






        share|cite|improve this answer












        The sum of bounded functions is bounded and so are scalar multiples. Note that $f= frac{h_1 + h_2}{2}$.



        Likewise for differences of bounded functions and $g = frac{h_2 - h_1}{2}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 at 5:08









        Henno Brandsma

        105k346114




        105k346114























            1














            Suppose $h_1, h_2$ are bounded. Let $D(f)$ denote the domain of $f$, where $f$ is a function. Then $exists M_1, M_2$ such that $$h_1(x) = f(x) - g(x) leq M_1 forall x in D(h_1), $$ $$h_2(x) = f(x) + g(x) leq M_2 forall x in D(h_2).$$ So, we have that for arbitrary $x in D(f) cap D(g)$, $$ f(x) - g(x) + f(x) + g(x) = 2f(x) leq M_1 + M_2 implies f(x) leq frac{M_1 + M_2}{2}, $$
            so $f$ is bounded, by definition. Similarly, notice that
            $$ g(x) - f(x) + g(x) + f(x) = 2g(x) leq M_2 - M_1 implies g(x) leq frac{M_2 - M_1}{2}, $$
            so $g$ is bounded, by definition. This completes the proof.






            share|cite|improve this answer



















            • 1




              Maybe it should be $xin D(h_1)cap D(h_2)$?
              – Shubham Johri
              Dec 18 at 5:54












            • Ah, you are right; good catch! I've edited my answer to include this.
              – alexsieusahai
              Dec 18 at 5:58
















            1














            Suppose $h_1, h_2$ are bounded. Let $D(f)$ denote the domain of $f$, where $f$ is a function. Then $exists M_1, M_2$ such that $$h_1(x) = f(x) - g(x) leq M_1 forall x in D(h_1), $$ $$h_2(x) = f(x) + g(x) leq M_2 forall x in D(h_2).$$ So, we have that for arbitrary $x in D(f) cap D(g)$, $$ f(x) - g(x) + f(x) + g(x) = 2f(x) leq M_1 + M_2 implies f(x) leq frac{M_1 + M_2}{2}, $$
            so $f$ is bounded, by definition. Similarly, notice that
            $$ g(x) - f(x) + g(x) + f(x) = 2g(x) leq M_2 - M_1 implies g(x) leq frac{M_2 - M_1}{2}, $$
            so $g$ is bounded, by definition. This completes the proof.






            share|cite|improve this answer



















            • 1




              Maybe it should be $xin D(h_1)cap D(h_2)$?
              – Shubham Johri
              Dec 18 at 5:54












            • Ah, you are right; good catch! I've edited my answer to include this.
              – alexsieusahai
              Dec 18 at 5:58














            1












            1








            1






            Suppose $h_1, h_2$ are bounded. Let $D(f)$ denote the domain of $f$, where $f$ is a function. Then $exists M_1, M_2$ such that $$h_1(x) = f(x) - g(x) leq M_1 forall x in D(h_1), $$ $$h_2(x) = f(x) + g(x) leq M_2 forall x in D(h_2).$$ So, we have that for arbitrary $x in D(f) cap D(g)$, $$ f(x) - g(x) + f(x) + g(x) = 2f(x) leq M_1 + M_2 implies f(x) leq frac{M_1 + M_2}{2}, $$
            so $f$ is bounded, by definition. Similarly, notice that
            $$ g(x) - f(x) + g(x) + f(x) = 2g(x) leq M_2 - M_1 implies g(x) leq frac{M_2 - M_1}{2}, $$
            so $g$ is bounded, by definition. This completes the proof.






            share|cite|improve this answer














            Suppose $h_1, h_2$ are bounded. Let $D(f)$ denote the domain of $f$, where $f$ is a function. Then $exists M_1, M_2$ such that $$h_1(x) = f(x) - g(x) leq M_1 forall x in D(h_1), $$ $$h_2(x) = f(x) + g(x) leq M_2 forall x in D(h_2).$$ So, we have that for arbitrary $x in D(f) cap D(g)$, $$ f(x) - g(x) + f(x) + g(x) = 2f(x) leq M_1 + M_2 implies f(x) leq frac{M_1 + M_2}{2}, $$
            so $f$ is bounded, by definition. Similarly, notice that
            $$ g(x) - f(x) + g(x) + f(x) = 2g(x) leq M_2 - M_1 implies g(x) leq frac{M_2 - M_1}{2}, $$
            so $g$ is bounded, by definition. This completes the proof.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 18 at 5:58

























            answered Dec 18 at 5:37









            alexsieusahai

            638




            638








            • 1




              Maybe it should be $xin D(h_1)cap D(h_2)$?
              – Shubham Johri
              Dec 18 at 5:54












            • Ah, you are right; good catch! I've edited my answer to include this.
              – alexsieusahai
              Dec 18 at 5:58














            • 1




              Maybe it should be $xin D(h_1)cap D(h_2)$?
              – Shubham Johri
              Dec 18 at 5:54












            • Ah, you are right; good catch! I've edited my answer to include this.
              – alexsieusahai
              Dec 18 at 5:58








            1




            1




            Maybe it should be $xin D(h_1)cap D(h_2)$?
            – Shubham Johri
            Dec 18 at 5:54






            Maybe it should be $xin D(h_1)cap D(h_2)$?
            – Shubham Johri
            Dec 18 at 5:54














            Ah, you are right; good catch! I've edited my answer to include this.
            – alexsieusahai
            Dec 18 at 5:58




            Ah, you are right; good catch! I've edited my answer to include this.
            – alexsieusahai
            Dec 18 at 5:58


















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