Continuity in open intervals












1












$begingroup$


The problem:



Let $f$ and $g$ be continuous functions defined on an open interval $I$. Let $ain I$ such that $f(a)<g(a)$. Show that there exists an open interval $Jsubset I$ with $ain J$ such that $f(x)<g(x)$ for all $xin J$



I'm honestly struggling to even conceptualize what this problem is asking me to prove let alone to how start proving it. Can anyone guide me or help me through?






real-analysis calculus continuity







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$endgroup$












  • $begingroup$
    You are being told that $flt g$ for one value, i.e. $f(a)lt g(a)$. Given that both functions are continuous show that you can find an interval around $a$ such that $f(x)lt g(x)$ for all $x$ in the interval.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:51








  • 1




    $begingroup$
    That's the wrong question. Now that you know what is being asked you should spend some time trying to solve this problem for yourself.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:54










  • $begingroup$
    Do you know what continuity of $f$ at $a$ means in informal sense? If yes then the answer to your question is immediate.
    $endgroup$
    – Paramanand Singh
    Dec 20 '18 at 5:22
















1












$begingroup$


The problem:



Let $f$ and $g$ be continuous functions defined on an open interval $I$. Let $ain I$ such that $f(a)<g(a)$. Show that there exists an open interval $Jsubset I$ with $ain J$ such that $f(x)<g(x)$ for all $xin J$



I'm honestly struggling to even conceptualize what this problem is asking me to prove let alone to how start proving it. Can anyone guide me or help me through?






real-analysis calculus continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    You are being told that $flt g$ for one value, i.e. $f(a)lt g(a)$. Given that both functions are continuous show that you can find an interval around $a$ such that $f(x)lt g(x)$ for all $x$ in the interval.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:51








  • 1




    $begingroup$
    That's the wrong question. Now that you know what is being asked you should spend some time trying to solve this problem for yourself.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:54










  • $begingroup$
    Do you know what continuity of $f$ at $a$ means in informal sense? If yes then the answer to your question is immediate.
    $endgroup$
    – Paramanand Singh
    Dec 20 '18 at 5:22














1












1








1





$begingroup$


The problem:



Let $f$ and $g$ be continuous functions defined on an open interval $I$. Let $ain I$ such that $f(a)<g(a)$. Show that there exists an open interval $Jsubset I$ with $ain J$ such that $f(x)<g(x)$ for all $xin J$



I'm honestly struggling to even conceptualize what this problem is asking me to prove let alone to how start proving it. Can anyone guide me or help me through?






real-analysis calculus continuity







share|cite|improve this question









$endgroup$




The problem:



Let $f$ and $g$ be continuous functions defined on an open interval $I$. Let $ain I$ such that $f(a)<g(a)$. Show that there exists an open interval $Jsubset I$ with $ain J$ such that $f(x)<g(x)$ for all $xin J$



I'm honestly struggling to even conceptualize what this problem is asking me to prove let alone to how start proving it. Can anyone guide me or help me through?






real-analysis calculus continuity




real-analysis calculus continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 20 '18 at 3:42









Jess SavoieJess Savoie

597




597












  • $begingroup$
    You are being told that $flt g$ for one value, i.e. $f(a)lt g(a)$. Given that both functions are continuous show that you can find an interval around $a$ such that $f(x)lt g(x)$ for all $x$ in the interval.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:51








  • 1




    $begingroup$
    That's the wrong question. Now that you know what is being asked you should spend some time trying to solve this problem for yourself.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:54










  • $begingroup$
    Do you know what continuity of $f$ at $a$ means in informal sense? If yes then the answer to your question is immediate.
    $endgroup$
    – Paramanand Singh
    Dec 20 '18 at 5:22


















  • $begingroup$
    You are being told that $flt g$ for one value, i.e. $f(a)lt g(a)$. Given that both functions are continuous show that you can find an interval around $a$ such that $f(x)lt g(x)$ for all $x$ in the interval.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:51








  • 1




    $begingroup$
    That's the wrong question. Now that you know what is being asked you should spend some time trying to solve this problem for yourself.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:54










  • $begingroup$
    Do you know what continuity of $f$ at $a$ means in informal sense? If yes then the answer to your question is immediate.
    $endgroup$
    – Paramanand Singh
    Dec 20 '18 at 5:22
















$begingroup$
You are being told that $flt g$ for one value, i.e. $f(a)lt g(a)$. Given that both functions are continuous show that you can find an interval around $a$ such that $f(x)lt g(x)$ for all $x$ in the interval.
$endgroup$
– John Douma
Dec 20 '18 at 3:51






$begingroup$
You are being told that $flt g$ for one value, i.e. $f(a)lt g(a)$. Given that both functions are continuous show that you can find an interval around $a$ such that $f(x)lt g(x)$ for all $x$ in the interval.
$endgroup$
– John Douma
Dec 20 '18 at 3:51






1




1




$begingroup$
That's the wrong question. Now that you know what is being asked you should spend some time trying to solve this problem for yourself.
$endgroup$
– John Douma
Dec 20 '18 at 3:54




$begingroup$
That's the wrong question. Now that you know what is being asked you should spend some time trying to solve this problem for yourself.
$endgroup$
– John Douma
Dec 20 '18 at 3:54












$begingroup$
Do you know what continuity of $f$ at $a$ means in informal sense? If yes then the answer to your question is immediate.
$endgroup$
– Paramanand Singh
Dec 20 '18 at 5:22




$begingroup$
Do you know what continuity of $f$ at $a$ means in informal sense? If yes then the answer to your question is immediate.
$endgroup$
– Paramanand Singh
Dec 20 '18 at 5:22










2 Answers
2






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2












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Let $epsilon =frac {g(a)-f(a)} 2$. There exists $delta >0$ such that $|x-a| <delta$ implies $|f(x)-f(a)| <epsilon$ and $|g(x)-g(a)| <epsilon$. Verify that for $x in (a-delta, a+delta) $ we have $f(x) <g(x)$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $h(x)=g(x)-f(x)$. $h$ is continuous on $I$ and $h(a)>0$. Since $h$ is continuous, in particular at $a$, we have $$lim_{kto 0}h(a-k)=lim_{xto a^-}h(x)=h(a)=lim_{xto a^{+}}h(x)=lim_{kto 0}h(a+k)$$



    We see that for $k>0$, $$quad h(a-k)< h(a)< h(a+k)implies h(x)>0;text{for};xin(a,a+k)subset I$$






    share|cite|improve this answer











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      2 Answers
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      2 Answers
      2






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      active

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      2












      $begingroup$

      Let $epsilon =frac {g(a)-f(a)} 2$. There exists $delta >0$ such that $|x-a| <delta$ implies $|f(x)-f(a)| <epsilon$ and $|g(x)-g(a)| <epsilon$. Verify that for $x in (a-delta, a+delta) $ we have $f(x) <g(x)$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Let $epsilon =frac {g(a)-f(a)} 2$. There exists $delta >0$ such that $|x-a| <delta$ implies $|f(x)-f(a)| <epsilon$ and $|g(x)-g(a)| <epsilon$. Verify that for $x in (a-delta, a+delta) $ we have $f(x) <g(x)$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Let $epsilon =frac {g(a)-f(a)} 2$. There exists $delta >0$ such that $|x-a| <delta$ implies $|f(x)-f(a)| <epsilon$ and $|g(x)-g(a)| <epsilon$. Verify that for $x in (a-delta, a+delta) $ we have $f(x) <g(x)$.






          share|cite|improve this answer









          $endgroup$



          Let $epsilon =frac {g(a)-f(a)} 2$. There exists $delta >0$ such that $|x-a| <delta$ implies $|f(x)-f(a)| <epsilon$ and $|g(x)-g(a)| <epsilon$. Verify that for $x in (a-delta, a+delta) $ we have $f(x) <g(x)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 5:32









          Kavi Rama MurthyKavi Rama Murthy

          72.6k53170




          72.6k53170























              1












              $begingroup$

              $h(x)=g(x)-f(x)$. $h$ is continuous on $I$ and $h(a)>0$. Since $h$ is continuous, in particular at $a$, we have $$lim_{kto 0}h(a-k)=lim_{xto a^-}h(x)=h(a)=lim_{xto a^{+}}h(x)=lim_{kto 0}h(a+k)$$



              We see that for $k>0$, $$quad h(a-k)< h(a)< h(a+k)implies h(x)>0;text{for};xin(a,a+k)subset I$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                $h(x)=g(x)-f(x)$. $h$ is continuous on $I$ and $h(a)>0$. Since $h$ is continuous, in particular at $a$, we have $$lim_{kto 0}h(a-k)=lim_{xto a^-}h(x)=h(a)=lim_{xto a^{+}}h(x)=lim_{kto 0}h(a+k)$$



                We see that for $k>0$, $$quad h(a-k)< h(a)< h(a+k)implies h(x)>0;text{for};xin(a,a+k)subset I$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $h(x)=g(x)-f(x)$. $h$ is continuous on $I$ and $h(a)>0$. Since $h$ is continuous, in particular at $a$, we have $$lim_{kto 0}h(a-k)=lim_{xto a^-}h(x)=h(a)=lim_{xto a^{+}}h(x)=lim_{kto 0}h(a+k)$$



                  We see that for $k>0$, $$quad h(a-k)< h(a)< h(a+k)implies h(x)>0;text{for};xin(a,a+k)subset I$$






                  share|cite|improve this answer











                  $endgroup$



                  $h(x)=g(x)-f(x)$. $h$ is continuous on $I$ and $h(a)>0$. Since $h$ is continuous, in particular at $a$, we have $$lim_{kto 0}h(a-k)=lim_{xto a^-}h(x)=h(a)=lim_{xto a^{+}}h(x)=lim_{kto 0}h(a+k)$$



                  We see that for $k>0$, $$quad h(a-k)< h(a)< h(a+k)implies h(x)>0;text{for};xin(a,a+k)subset I$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 20 '18 at 6:39

























                  answered Dec 20 '18 at 3:53









                  Yadati KiranYadati Kiran

                  2,1161622




                  2,1161622






























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