Proving that 2 functions are bounded.











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












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

386




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
2






active

oldest

votes

















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
    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
      23 mins ago












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











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044784%2fproving-that-2-functions-are-bounded%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    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
              23 mins ago












            • Ah, you are right; good catch! I've edited my answer to include this.
              – alexsieusahai
              19 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
              23 mins ago












            • Ah, you are right; good catch! I've edited my answer to include this.
              – alexsieusahai
              19 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 18 mins ago

























            answered 40 mins ago









            alexsieusahai

            636




            636








            • 1




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












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














            • 1




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












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








            1




            1




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






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














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




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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044784%2fproving-that-2-functions-are-bounded%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
            Read more