Continuous function:











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Prove: If $f$ is continuous on $I=(0,1]$ and $lim_{x to 0^+}f(x)=L$ is finite then $f$ is bounded on $I$




I'm not too sure how to start this problem, I know $f(x)$ doesn't attain its least upper bound of $0$.



And because $f(x)$ is continuous on $I=(0,1]$ then it is bounded everywhere in this interval but not at $0$.






real-analysis analysis







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    Prove: If $f$ is continuous on $I=(0,1]$ and $lim_{x to 0^+}f(x)=L$ is finite then $f$ is bounded on $I$




    I'm not too sure how to start this problem, I know $f(x)$ doesn't attain its least upper bound of $0$.



    And because $f(x)$ is continuous on $I=(0,1]$ then it is bounded everywhere in this interval but not at $0$.






    real-analysis analysis







    share|cite|improve this question


























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      down vote

      favorite









      up vote
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      down vote

      favorite












      Prove: If $f$ is continuous on $I=(0,1]$ and $lim_{x to 0^+}f(x)=L$ is finite then $f$ is bounded on $I$




      I'm not too sure how to start this problem, I know $f(x)$ doesn't attain its least upper bound of $0$.



      And because $f(x)$ is continuous on $I=(0,1]$ then it is bounded everywhere in this interval but not at $0$.






      real-analysis analysis







      share|cite|improve this question
















      Prove: If $f$ is continuous on $I=(0,1]$ and $lim_{x to 0^+}f(x)=L$ is finite then $f$ is bounded on $I$




      I'm not too sure how to start this problem, I know $f(x)$ doesn't attain its least upper bound of $0$.



      And because $f(x)$ is continuous on $I=(0,1]$ then it is bounded everywhere in this interval but not at $0$.






      real-analysis analysis




      real-analysis analysis






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      share|cite|improve this question













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      edited yesterday









      idea

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      2,07121023










      asked yesterday









      Guy

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          A quick way to show this is by extending f to $[0,1]$: define $f(0)=L$ and note that f is now continuous on a compact set. We know that continuous functions on compacts are bounded, whence the result.






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            HINT: Your last sentence correctly identifies why $f$ is bounded everywhere except near $0$. Near $0$, $f$ is bounded for a different reason: the values of $f$ are near $L$. So split your interval into $(0,delta)$ and $[delta, 1]$, then prove $f$ is bounded on each of these intervals for different reasons.






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              Suppose that $f$ is not bounded. Then, for each $ninmathbb N$, there is a $a_nin(0,1]$ such that $bigllvert f(a_n)bigrrvertgeqslant n$. The sequence $(a_n)_{ninmathbb N}$ is bounded and therefore it has a convergent subsequence. Without loss of generality, we can assume that $(a_n)_{ninmathbb N}$ converges. Let $l$ be its limit. Then $lin[0,1]$ and there are two possibilities:





              1. $lin(0,1]$. Then, since $f$ is continuous at $l$, $lim_{ntoinfty}f(a_n)=f(l)$, which is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.


              2. $l=0$. Then $lim_{ntoinfty}f(a_n)=L$. Again, this is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.






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                3 Answers
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                3 Answers
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                active

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                active

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                up vote
                3
                down vote













                A quick way to show this is by extending f to $[0,1]$: define $f(0)=L$ and note that f is now continuous on a compact set. We know that continuous functions on compacts are bounded, whence the result.






                share|cite|improve this answer

























                  up vote
                  3
                  down vote













                  A quick way to show this is by extending f to $[0,1]$: define $f(0)=L$ and note that f is now continuous on a compact set. We know that continuous functions on compacts are bounded, whence the result.






                  share|cite|improve this answer























                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    A quick way to show this is by extending f to $[0,1]$: define $f(0)=L$ and note that f is now continuous on a compact set. We know that continuous functions on compacts are bounded, whence the result.






                    share|cite|improve this answer












                    A quick way to show this is by extending f to $[0,1]$: define $f(0)=L$ and note that f is now continuous on a compact set. We know that continuous functions on compacts are bounded, whence the result.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered yesterday









                    Sorin Tirc

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                        HINT: Your last sentence correctly identifies why $f$ is bounded everywhere except near $0$. Near $0$, $f$ is bounded for a different reason: the values of $f$ are near $L$. So split your interval into $(0,delta)$ and $[delta, 1]$, then prove $f$ is bounded on each of these intervals for different reasons.






                        share|cite|improve this answer

























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                          0
                          down vote













                          HINT: Your last sentence correctly identifies why $f$ is bounded everywhere except near $0$. Near $0$, $f$ is bounded for a different reason: the values of $f$ are near $L$. So split your interval into $(0,delta)$ and $[delta, 1]$, then prove $f$ is bounded on each of these intervals for different reasons.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            HINT: Your last sentence correctly identifies why $f$ is bounded everywhere except near $0$. Near $0$, $f$ is bounded for a different reason: the values of $f$ are near $L$. So split your interval into $(0,delta)$ and $[delta, 1]$, then prove $f$ is bounded on each of these intervals for different reasons.






                            share|cite|improve this answer












                            HINT: Your last sentence correctly identifies why $f$ is bounded everywhere except near $0$. Near $0$, $f$ is bounded for a different reason: the values of $f$ are near $L$. So split your interval into $(0,delta)$ and $[delta, 1]$, then prove $f$ is bounded on each of these intervals for different reasons.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            Y. Forman

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                            11.3k423






















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                                Suppose that $f$ is not bounded. Then, for each $ninmathbb N$, there is a $a_nin(0,1]$ such that $bigllvert f(a_n)bigrrvertgeqslant n$. The sequence $(a_n)_{ninmathbb N}$ is bounded and therefore it has a convergent subsequence. Without loss of generality, we can assume that $(a_n)_{ninmathbb N}$ converges. Let $l$ be its limit. Then $lin[0,1]$ and there are two possibilities:





                                1. $lin(0,1]$. Then, since $f$ is continuous at $l$, $lim_{ntoinfty}f(a_n)=f(l)$, which is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.


                                2. $l=0$. Then $lim_{ntoinfty}f(a_n)=L$. Again, this is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.






                                share|cite|improve this answer

























                                  up vote
                                  0
                                  down vote













                                  Suppose that $f$ is not bounded. Then, for each $ninmathbb N$, there is a $a_nin(0,1]$ such that $bigllvert f(a_n)bigrrvertgeqslant n$. The sequence $(a_n)_{ninmathbb N}$ is bounded and therefore it has a convergent subsequence. Without loss of generality, we can assume that $(a_n)_{ninmathbb N}$ converges. Let $l$ be its limit. Then $lin[0,1]$ and there are two possibilities:





                                  1. $lin(0,1]$. Then, since $f$ is continuous at $l$, $lim_{ntoinfty}f(a_n)=f(l)$, which is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.


                                  2. $l=0$. Then $lim_{ntoinfty}f(a_n)=L$. Again, this is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.






                                  share|cite|improve this answer























                                    up vote
                                    0
                                    down vote










                                    up vote
                                    0
                                    down vote









                                    Suppose that $f$ is not bounded. Then, for each $ninmathbb N$, there is a $a_nin(0,1]$ such that $bigllvert f(a_n)bigrrvertgeqslant n$. The sequence $(a_n)_{ninmathbb N}$ is bounded and therefore it has a convergent subsequence. Without loss of generality, we can assume that $(a_n)_{ninmathbb N}$ converges. Let $l$ be its limit. Then $lin[0,1]$ and there are two possibilities:





                                    1. $lin(0,1]$. Then, since $f$ is continuous at $l$, $lim_{ntoinfty}f(a_n)=f(l)$, which is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.


                                    2. $l=0$. Then $lim_{ntoinfty}f(a_n)=L$. Again, this is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.






                                    share|cite|improve this answer












                                    Suppose that $f$ is not bounded. Then, for each $ninmathbb N$, there is a $a_nin(0,1]$ such that $bigllvert f(a_n)bigrrvertgeqslant n$. The sequence $(a_n)_{ninmathbb N}$ is bounded and therefore it has a convergent subsequence. Without loss of generality, we can assume that $(a_n)_{ninmathbb N}$ converges. Let $l$ be its limit. Then $lin[0,1]$ and there are two possibilities:





                                    1. $lin(0,1]$. Then, since $f$ is continuous at $l$, $lim_{ntoinfty}f(a_n)=f(l)$, which is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.


                                    2. $l=0$. Then $lim_{ntoinfty}f(a_n)=L$. Again, this is impossible, since $bigl(f(a_n)bigr)_{ninmathbb N}$ is unbounded.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered yesterday









                                    José Carlos Santos

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