Proving that 2 functions are bounded.











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Let $f,g:mathbb R -> mathbb R$. 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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  • 1




    Express $f,g $ in the form of $h_2, h_1$.
    – xbh
    1 hour ago










  • $|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
    – Henno Brandsma
    1 hour ago















up vote
2
down vote

favorite












Let $f,g:mathbb R -> mathbb R$. 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




















  • 1




    Express $f,g $ in the form of $h_2, h_1$.
    – xbh
    1 hour ago










  • $|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
    – Henno Brandsma
    1 hour ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $f,g:mathbb R -> mathbb R$. 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 $f,g:mathbb R -> mathbb R$. 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Henno Brandsma

103k346113




103k346113










asked 1 hour ago









Pablo Tores

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386








  • 1




    Express $f,g $ in the form of $h_2, h_1$.
    – xbh
    1 hour ago










  • $|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
    – Henno Brandsma
    1 hour ago














  • 1




    Express $f,g $ in the form of $h_2, h_1$.
    – xbh
    1 hour ago










  • $|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
    – Henno Brandsma
    1 hour ago








1




1




Express $f,g $ in the form of $h_2, h_1$.
– xbh
1 hour ago




Express $f,g $ in the form of $h_2, h_1$.
– xbh
1 hour ago












$|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
– Henno Brandsma
1 hour ago




$|f(x) pm g(x) | le |f(x)| + |g(x)|$ for all $f,g,x$ so the sum and difference of bounded functions is bounded.
– Henno Brandsma
1 hour ago










2 Answers
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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 $g = frac{h_2 - h_1}{2}$.






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    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
      48 mins ago












    • Ah, you are right; good catch! I've edited my answer to include this.
      – alexsieusahai
      44 mins ago











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    2 Answers
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    active

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    2 Answers
    2






    active

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    active

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    up vote
    5
    down vote













    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

























      up vote
      5
      down vote













      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























        up vote
        5
        down vote










        up vote
        5
        down vote









        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 1 hour ago









        Henno Brandsma

        103k346113




        103k346113






















            up vote
            1
            down vote













            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
              48 mins ago












            • Ah, you are right; good catch! I've edited my answer to include this.
              – alexsieusahai
              44 mins ago















            up vote
            1
            down vote













            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
              48 mins ago












            • Ah, you are right; good catch! I've edited my answer to include this.
              – alexsieusahai
              44 mins ago













            up vote
            1
            down vote










            up vote
            1
            down vote









            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 44 mins ago

























            answered 1 hour ago









            alexsieusahai

            636




            636








            • 1




              Maybe it should be $xin D(h_1)cap D(h_2)$?
              – Shubham Johri
              48 mins ago












            • Ah, you are right; good catch! I've edited my answer to include this.
              – alexsieusahai
              44 mins ago














            • 1




              Maybe it should be $xin D(h_1)cap D(h_2)$?
              – Shubham Johri
              48 mins ago












            • Ah, you are right; good catch! I've edited my answer to include this.
              – alexsieusahai
              44 mins ago








            1




            1




            Maybe it should be $xin D(h_1)cap D(h_2)$?
            – Shubham Johri
            48 mins ago






            Maybe it should be $xin D(h_1)cap D(h_2)$?
            – Shubham Johri
            48 mins ago














            Ah, you are right; good catch! I've edited my answer to include this.
            – alexsieusahai
            44 mins ago




            Ah, you are right; good catch! I've edited my answer to include this.
            – alexsieusahai
            44 mins ago


















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