Continuity of the following function












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Let $f:[a,b] to Bbb R$ be continuous . Define $g:[a,b] to Bbb R$ by $$g(x) = sup[f(t) | t in [a,x]].$$



Then is $g$ continuous? I tried to show the inequality $$|g(x)-g(y)| le sup(f(t) | t in [x,y])$$ but wasn’t able to. Any hints or suggestions will be appreciated.






real-analysis calculus continuity supremum-and-infimum







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  • 1




    $begingroup$
    Hint: $g$ is monotone increasing, so if there were a discontinuity at $x = x_0$, it would have to be the case that $lim_{x to x_0^-} g(x) < lim_{x to x_0^+} g(x)$. Apply the Intermediate Value Theorem to $f$.
    $endgroup$
    – Theo Bendit
    Oct 29 '18 at 13:47












  • $begingroup$
    Can you elaborate?
    $endgroup$
    – John Mitchell
    Oct 29 '18 at 16:53
















1












$begingroup$


Let $f:[a,b] to Bbb R$ be continuous . Define $g:[a,b] to Bbb R$ by $$g(x) = sup[f(t) | t in [a,x]].$$



Then is $g$ continuous? I tried to show the inequality $$|g(x)-g(y)| le sup(f(t) | t in [x,y])$$ but wasn’t able to. Any hints or suggestions will be appreciated.






real-analysis calculus continuity supremum-and-infimum







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: $g$ is monotone increasing, so if there were a discontinuity at $x = x_0$, it would have to be the case that $lim_{x to x_0^-} g(x) < lim_{x to x_0^+} g(x)$. Apply the Intermediate Value Theorem to $f$.
    $endgroup$
    – Theo Bendit
    Oct 29 '18 at 13:47












  • $begingroup$
    Can you elaborate?
    $endgroup$
    – John Mitchell
    Oct 29 '18 at 16:53














1












1








1





$begingroup$


Let $f:[a,b] to Bbb R$ be continuous . Define $g:[a,b] to Bbb R$ by $$g(x) = sup[f(t) | t in [a,x]].$$



Then is $g$ continuous? I tried to show the inequality $$|g(x)-g(y)| le sup(f(t) | t in [x,y])$$ but wasn’t able to. Any hints or suggestions will be appreciated.






real-analysis calculus continuity supremum-and-infimum







share|cite|improve this question











$endgroup$




Let $f:[a,b] to Bbb R$ be continuous . Define $g:[a,b] to Bbb R$ by $$g(x) = sup[f(t) | t in [a,x]].$$



Then is $g$ continuous? I tried to show the inequality $$|g(x)-g(y)| le sup(f(t) | t in [x,y])$$ but wasn’t able to. Any hints or suggestions will be appreciated.






real-analysis calculus continuity supremum-and-infimum




real-analysis calculus continuity supremum-and-infimum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 12:42









Davide Giraudo

127k16151264




127k16151264










asked Oct 29 '18 at 13:25









John MitchellJohn Mitchell

377210




377210








  • 1




    $begingroup$
    Hint: $g$ is monotone increasing, so if there were a discontinuity at $x = x_0$, it would have to be the case that $lim_{x to x_0^-} g(x) < lim_{x to x_0^+} g(x)$. Apply the Intermediate Value Theorem to $f$.
    $endgroup$
    – Theo Bendit
    Oct 29 '18 at 13:47












  • $begingroup$
    Can you elaborate?
    $endgroup$
    – John Mitchell
    Oct 29 '18 at 16:53














  • 1




    $begingroup$
    Hint: $g$ is monotone increasing, so if there were a discontinuity at $x = x_0$, it would have to be the case that $lim_{x to x_0^-} g(x) < lim_{x to x_0^+} g(x)$. Apply the Intermediate Value Theorem to $f$.
    $endgroup$
    – Theo Bendit
    Oct 29 '18 at 13:47












  • $begingroup$
    Can you elaborate?
    $endgroup$
    – John Mitchell
    Oct 29 '18 at 16:53








1




1




$begingroup$
Hint: $g$ is monotone increasing, so if there were a discontinuity at $x = x_0$, it would have to be the case that $lim_{x to x_0^-} g(x) < lim_{x to x_0^+} g(x)$. Apply the Intermediate Value Theorem to $f$.
$endgroup$
– Theo Bendit
Oct 29 '18 at 13:47






$begingroup$
Hint: $g$ is monotone increasing, so if there were a discontinuity at $x = x_0$, it would have to be the case that $lim_{x to x_0^-} g(x) < lim_{x to x_0^+} g(x)$. Apply the Intermediate Value Theorem to $f$.
$endgroup$
– Theo Bendit
Oct 29 '18 at 13:47














$begingroup$
Can you elaborate?
$endgroup$
– John Mitchell
Oct 29 '18 at 16:53




$begingroup$
Can you elaborate?
$endgroup$
– John Mitchell
Oct 29 '18 at 16:53










2 Answers
2






active

oldest

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0












$begingroup$

Let $aleqslant xlt y$. Then
$$
g(y)=maxleft{sup_{aleqslant tleqslant x}leftlvert f(t)rightrvert,sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}= maxleft{g(x),sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}.
$$
We use the bound
$$
sup_{xleqslant tleqslant y}leftlvert f(t)rightrvert leqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+ leftlvert f(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+g(x)
$$
in order to get
$$
g(y)leqslant maxleft{g(x),g(x)+sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}=g(x)+maxleft{0,sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}
$$
and since $g$ is non-decreasing, it follows that
$$
leftlvert g(y)-g(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert.
$$
Now the continuity of $g$ follows from the uniform continuity of $f$: fix a positive $varepsilon$ and choose $delta$ such that if $aleqslant t_1,t_2leqslant b$ are such that $leftlvert t_2-t_1rightrvertlt delta$, then $leftlvert f(t_2)-f(t_1)rightrvertlt varepsilon$.
If $aleqslant xlt yleqslant b$ and $y-xleqslant delta$, the previous inequality shows that $leftlvert g(y)-g(x)rightrvertleqslantvarepsilon$.






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$endgroup$





















    -1












    $begingroup$

    Your inequality is false. Suppose that $f:[0,1]rightarrow mathbf{R}$ is given by



    $$f(x) = 2x-1.$$



    then $g(x) = 2x-1$. What is $|g(1/4)-g(3/4)|$ and what is $sup{f(t)mid tin [1/4,3/4]}$?



    For a given $x>0$ consider that $exists xi_x$ such that $f(xi_x) = g(x)$ for every $x$ (why is that?). Suppose that $f(x)<f(xi_x)$ then $f(x+delta)<f(xi_x)$ for $|delta|<d$ where $d$ is sufficiently small (why is that?). Then $g$ is constant on $(x-d,x+d)$ and hence continuous.



    Suppose now that $f(x) = f(xi_x) = g(x)$, given $varepsilon>0$ pick $delta>0$ such that $|y-x|<delta$ implies that $|f(x)-f(y)|<varepsilon$. What can then be said about $sup{f(t)mid tin [0,x+delta]}$?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I think you’ve made a mistake. It should be g is continuous since g is constant.
      $endgroup$
      – John Mitchell
      Oct 30 '18 at 3:03










    • $begingroup$
      Yes it should be g
      $endgroup$
      – Olof Rubin
      Oct 30 '18 at 6:33











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    2 Answers
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    2 Answers
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    0












    $begingroup$

    Let $aleqslant xlt y$. Then
    $$
    g(y)=maxleft{sup_{aleqslant tleqslant x}leftlvert f(t)rightrvert,sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}= maxleft{g(x),sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}.
    $$
    We use the bound
    $$
    sup_{xleqslant tleqslant y}leftlvert f(t)rightrvert leqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+ leftlvert f(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+g(x)
    $$
    in order to get
    $$
    g(y)leqslant maxleft{g(x),g(x)+sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}=g(x)+maxleft{0,sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}
    $$
    and since $g$ is non-decreasing, it follows that
    $$
    leftlvert g(y)-g(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert.
    $$
    Now the continuity of $g$ follows from the uniform continuity of $f$: fix a positive $varepsilon$ and choose $delta$ such that if $aleqslant t_1,t_2leqslant b$ are such that $leftlvert t_2-t_1rightrvertlt delta$, then $leftlvert f(t_2)-f(t_1)rightrvertlt varepsilon$.
    If $aleqslant xlt yleqslant b$ and $y-xleqslant delta$, the previous inequality shows that $leftlvert g(y)-g(x)rightrvertleqslantvarepsilon$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let $aleqslant xlt y$. Then
      $$
      g(y)=maxleft{sup_{aleqslant tleqslant x}leftlvert f(t)rightrvert,sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}= maxleft{g(x),sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}.
      $$
      We use the bound
      $$
      sup_{xleqslant tleqslant y}leftlvert f(t)rightrvert leqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+ leftlvert f(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+g(x)
      $$
      in order to get
      $$
      g(y)leqslant maxleft{g(x),g(x)+sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}=g(x)+maxleft{0,sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}
      $$
      and since $g$ is non-decreasing, it follows that
      $$
      leftlvert g(y)-g(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert.
      $$
      Now the continuity of $g$ follows from the uniform continuity of $f$: fix a positive $varepsilon$ and choose $delta$ such that if $aleqslant t_1,t_2leqslant b$ are such that $leftlvert t_2-t_1rightrvertlt delta$, then $leftlvert f(t_2)-f(t_1)rightrvertlt varepsilon$.
      If $aleqslant xlt yleqslant b$ and $y-xleqslant delta$, the previous inequality shows that $leftlvert g(y)-g(x)rightrvertleqslantvarepsilon$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $aleqslant xlt y$. Then
        $$
        g(y)=maxleft{sup_{aleqslant tleqslant x}leftlvert f(t)rightrvert,sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}= maxleft{g(x),sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}.
        $$
        We use the bound
        $$
        sup_{xleqslant tleqslant y}leftlvert f(t)rightrvert leqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+ leftlvert f(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+g(x)
        $$
        in order to get
        $$
        g(y)leqslant maxleft{g(x),g(x)+sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}=g(x)+maxleft{0,sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}
        $$
        and since $g$ is non-decreasing, it follows that
        $$
        leftlvert g(y)-g(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert.
        $$
        Now the continuity of $g$ follows from the uniform continuity of $f$: fix a positive $varepsilon$ and choose $delta$ such that if $aleqslant t_1,t_2leqslant b$ are such that $leftlvert t_2-t_1rightrvertlt delta$, then $leftlvert f(t_2)-f(t_1)rightrvertlt varepsilon$.
        If $aleqslant xlt yleqslant b$ and $y-xleqslant delta$, the previous inequality shows that $leftlvert g(y)-g(x)rightrvertleqslantvarepsilon$.






        share|cite|improve this answer









        $endgroup$



        Let $aleqslant xlt y$. Then
        $$
        g(y)=maxleft{sup_{aleqslant tleqslant x}leftlvert f(t)rightrvert,sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}= maxleft{g(x),sup_{xleqslant tleqslant y}leftlvert f(t)rightrvertright}.
        $$
        We use the bound
        $$
        sup_{xleqslant tleqslant y}leftlvert f(t)rightrvert leqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+ leftlvert f(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert+g(x)
        $$
        in order to get
        $$
        g(y)leqslant maxleft{g(x),g(x)+sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}=g(x)+maxleft{0,sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvertright}
        $$
        and since $g$ is non-decreasing, it follows that
        $$
        leftlvert g(y)-g(x)rightrvertleqslant sup_{xleqslant tleqslant y}leftlvert f(t)-f(x)rightrvert.
        $$
        Now the continuity of $g$ follows from the uniform continuity of $f$: fix a positive $varepsilon$ and choose $delta$ such that if $aleqslant t_1,t_2leqslant b$ are such that $leftlvert t_2-t_1rightrvertlt delta$, then $leftlvert f(t_2)-f(t_1)rightrvertlt varepsilon$.
        If $aleqslant xlt yleqslant b$ and $y-xleqslant delta$, the previous inequality shows that $leftlvert g(y)-g(x)rightrvertleqslantvarepsilon$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 11:35









        Davide GiraudoDavide Giraudo

        127k16151264




        127k16151264























            -1












            $begingroup$

            Your inequality is false. Suppose that $f:[0,1]rightarrow mathbf{R}$ is given by



            $$f(x) = 2x-1.$$



            then $g(x) = 2x-1$. What is $|g(1/4)-g(3/4)|$ and what is $sup{f(t)mid tin [1/4,3/4]}$?



            For a given $x>0$ consider that $exists xi_x$ such that $f(xi_x) = g(x)$ for every $x$ (why is that?). Suppose that $f(x)<f(xi_x)$ then $f(x+delta)<f(xi_x)$ for $|delta|<d$ where $d$ is sufficiently small (why is that?). Then $g$ is constant on $(x-d,x+d)$ and hence continuous.



            Suppose now that $f(x) = f(xi_x) = g(x)$, given $varepsilon>0$ pick $delta>0$ such that $|y-x|<delta$ implies that $|f(x)-f(y)|<varepsilon$. What can then be said about $sup{f(t)mid tin [0,x+delta]}$?






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I think you’ve made a mistake. It should be g is continuous since g is constant.
              $endgroup$
              – John Mitchell
              Oct 30 '18 at 3:03










            • $begingroup$
              Yes it should be g
              $endgroup$
              – Olof Rubin
              Oct 30 '18 at 6:33
















            -1












            $begingroup$

            Your inequality is false. Suppose that $f:[0,1]rightarrow mathbf{R}$ is given by



            $$f(x) = 2x-1.$$



            then $g(x) = 2x-1$. What is $|g(1/4)-g(3/4)|$ and what is $sup{f(t)mid tin [1/4,3/4]}$?



            For a given $x>0$ consider that $exists xi_x$ such that $f(xi_x) = g(x)$ for every $x$ (why is that?). Suppose that $f(x)<f(xi_x)$ then $f(x+delta)<f(xi_x)$ for $|delta|<d$ where $d$ is sufficiently small (why is that?). Then $g$ is constant on $(x-d,x+d)$ and hence continuous.



            Suppose now that $f(x) = f(xi_x) = g(x)$, given $varepsilon>0$ pick $delta>0$ such that $|y-x|<delta$ implies that $|f(x)-f(y)|<varepsilon$. What can then be said about $sup{f(t)mid tin [0,x+delta]}$?






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I think you’ve made a mistake. It should be g is continuous since g is constant.
              $endgroup$
              – John Mitchell
              Oct 30 '18 at 3:03










            • $begingroup$
              Yes it should be g
              $endgroup$
              – Olof Rubin
              Oct 30 '18 at 6:33














            -1












            -1








            -1





            $begingroup$

            Your inequality is false. Suppose that $f:[0,1]rightarrow mathbf{R}$ is given by



            $$f(x) = 2x-1.$$



            then $g(x) = 2x-1$. What is $|g(1/4)-g(3/4)|$ and what is $sup{f(t)mid tin [1/4,3/4]}$?



            For a given $x>0$ consider that $exists xi_x$ such that $f(xi_x) = g(x)$ for every $x$ (why is that?). Suppose that $f(x)<f(xi_x)$ then $f(x+delta)<f(xi_x)$ for $|delta|<d$ where $d$ is sufficiently small (why is that?). Then $g$ is constant on $(x-d,x+d)$ and hence continuous.



            Suppose now that $f(x) = f(xi_x) = g(x)$, given $varepsilon>0$ pick $delta>0$ such that $|y-x|<delta$ implies that $|f(x)-f(y)|<varepsilon$. What can then be said about $sup{f(t)mid tin [0,x+delta]}$?






            share|cite|improve this answer











            $endgroup$



            Your inequality is false. Suppose that $f:[0,1]rightarrow mathbf{R}$ is given by



            $$f(x) = 2x-1.$$



            then $g(x) = 2x-1$. What is $|g(1/4)-g(3/4)|$ and what is $sup{f(t)mid tin [1/4,3/4]}$?



            For a given $x>0$ consider that $exists xi_x$ such that $f(xi_x) = g(x)$ for every $x$ (why is that?). Suppose that $f(x)<f(xi_x)$ then $f(x+delta)<f(xi_x)$ for $|delta|<d$ where $d$ is sufficiently small (why is that?). Then $g$ is constant on $(x-d,x+d)$ and hence continuous.



            Suppose now that $f(x) = f(xi_x) = g(x)$, given $varepsilon>0$ pick $delta>0$ such that $|y-x|<delta$ implies that $|f(x)-f(y)|<varepsilon$. What can then be said about $sup{f(t)mid tin [0,x+delta]}$?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Oct 30 '18 at 8:02

























            answered Oct 29 '18 at 19:35









            Olof RubinOlof Rubin

            1,156316




            1,156316












            • $begingroup$
              I think you’ve made a mistake. It should be g is continuous since g is constant.
              $endgroup$
              – John Mitchell
              Oct 30 '18 at 3:03










            • $begingroup$
              Yes it should be g
              $endgroup$
              – Olof Rubin
              Oct 30 '18 at 6:33


















            • $begingroup$
              I think you’ve made a mistake. It should be g is continuous since g is constant.
              $endgroup$
              – John Mitchell
              Oct 30 '18 at 3:03










            • $begingroup$
              Yes it should be g
              $endgroup$
              – Olof Rubin
              Oct 30 '18 at 6:33
















            $begingroup$
            I think you’ve made a mistake. It should be g is continuous since g is constant.
            $endgroup$
            – John Mitchell
            Oct 30 '18 at 3:03




            $begingroup$
            I think you’ve made a mistake. It should be g is continuous since g is constant.
            $endgroup$
            – John Mitchell
            Oct 30 '18 at 3:03












            $begingroup$
            Yes it should be g
            $endgroup$
            – Olof Rubin
            Oct 30 '18 at 6:33




            $begingroup$
            Yes it should be g
            $endgroup$
            – Olof Rubin
            Oct 30 '18 at 6:33


















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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2
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