${ x in X : |f(x)| geq epsilon }$ is compact then $f$ is uniformly continuous on $X$.












4














Past year paper question.



Let $(X, d)$ be a metric space and let $f: X to mathbb{R}$ be a continuous function, where $mathbb{R}$ is given the standard metric. Assume that for any $epsilon > 0$, the set ${ x in X : |f(x)| geq epsilon }$ is a compact metric subspace of $X$. Show that $f$ is uniformly continuous on $X$.



Attempt:



Let $epsilon >0$ be given. Let $K := { x in X : |f(x)| geq frac{epsilon}{2} }$. $f$ is continuous on $K$ $implies f$ is uniform continuous on $K$. Then there exists $delta$ such that $d(x,y)<delta implies |f(x)-f(y)| < frac{epsilon}{2}$.



Take any $2$ points $x,y in X$, such that $d(x,y)<delta$. If $x,y in K$, then $|f(x)-f(y)| < frac{epsilon}{2} < epsilon$. If $x,y notin K$, then $|f(x)-f(y)| leq |f(x)|+|f(y)| leq frac{epsilon}{2} + frac{epsilon}{2} leq epsilon$.



But I am stuck here, because what if $x in K$ and $y notin K$.






analysis metric-spaces







share|cite|improve this question



























    4














    Past year paper question.



    Let $(X, d)$ be a metric space and let $f: X to mathbb{R}$ be a continuous function, where $mathbb{R}$ is given the standard metric. Assume that for any $epsilon > 0$, the set ${ x in X : |f(x)| geq epsilon }$ is a compact metric subspace of $X$. Show that $f$ is uniformly continuous on $X$.



    Attempt:



    Let $epsilon >0$ be given. Let $K := { x in X : |f(x)| geq frac{epsilon}{2} }$. $f$ is continuous on $K$ $implies f$ is uniform continuous on $K$. Then there exists $delta$ such that $d(x,y)<delta implies |f(x)-f(y)| < frac{epsilon}{2}$.



    Take any $2$ points $x,y in X$, such that $d(x,y)<delta$. If $x,y in K$, then $|f(x)-f(y)| < frac{epsilon}{2} < epsilon$. If $x,y notin K$, then $|f(x)-f(y)| leq |f(x)|+|f(y)| leq frac{epsilon}{2} + frac{epsilon}{2} leq epsilon$.



    But I am stuck here, because what if $x in K$ and $y notin K$.






    analysis metric-spaces







    share|cite|improve this question

























      4












      4








      4


      2





      Past year paper question.



      Let $(X, d)$ be a metric space and let $f: X to mathbb{R}$ be a continuous function, where $mathbb{R}$ is given the standard metric. Assume that for any $epsilon > 0$, the set ${ x in X : |f(x)| geq epsilon }$ is a compact metric subspace of $X$. Show that $f$ is uniformly continuous on $X$.



      Attempt:



      Let $epsilon >0$ be given. Let $K := { x in X : |f(x)| geq frac{epsilon}{2} }$. $f$ is continuous on $K$ $implies f$ is uniform continuous on $K$. Then there exists $delta$ such that $d(x,y)<delta implies |f(x)-f(y)| < frac{epsilon}{2}$.



      Take any $2$ points $x,y in X$, such that $d(x,y)<delta$. If $x,y in K$, then $|f(x)-f(y)| < frac{epsilon}{2} < epsilon$. If $x,y notin K$, then $|f(x)-f(y)| leq |f(x)|+|f(y)| leq frac{epsilon}{2} + frac{epsilon}{2} leq epsilon$.



      But I am stuck here, because what if $x in K$ and $y notin K$.






      analysis metric-spaces







      share|cite|improve this question













      Past year paper question.



      Let $(X, d)$ be a metric space and let $f: X to mathbb{R}$ be a continuous function, where $mathbb{R}$ is given the standard metric. Assume that for any $epsilon > 0$, the set ${ x in X : |f(x)| geq epsilon }$ is a compact metric subspace of $X$. Show that $f$ is uniformly continuous on $X$.



      Attempt:



      Let $epsilon >0$ be given. Let $K := { x in X : |f(x)| geq frac{epsilon}{2} }$. $f$ is continuous on $K$ $implies f$ is uniform continuous on $K$. Then there exists $delta$ such that $d(x,y)<delta implies |f(x)-f(y)| < frac{epsilon}{2}$.



      Take any $2$ points $x,y in X$, such that $d(x,y)<delta$. If $x,y in K$, then $|f(x)-f(y)| < frac{epsilon}{2} < epsilon$. If $x,y notin K$, then $|f(x)-f(y)| leq |f(x)|+|f(y)| leq frac{epsilon}{2} + frac{epsilon}{2} leq epsilon$.



      But I am stuck here, because what if $x in K$ and $y notin K$.






      analysis metric-spaces




      analysis metric-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 20 at 3:05









      eatfood

      1827




      1827






















          1 Answer
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          Suppose $f$ is not uniformly continuous. Then there exists $delta >0$ and sequences ${x_n},{y_n}$ such that $d(x_n,y_n) to 0$ but $|f(x_n)-f(y_n)| geq delta$ for each $n$. For each $n$ either $|f(x_n)| geq delta /2$ or $|f(y_n)| geq delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| geq delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences ${x_n},{y_n}$ converge to the same limit $x$ because $d(x_n,y_n) to 0$. This contradicts continuity of $f$ at $x$.






          share|cite|improve this answer





















          • Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
            – eatfood
            Nov 20 at 7:05












          • This has solved my problem, thank you!
            – eatfood
            Nov 20 at 7:06











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          Suppose $f$ is not uniformly continuous. Then there exists $delta >0$ and sequences ${x_n},{y_n}$ such that $d(x_n,y_n) to 0$ but $|f(x_n)-f(y_n)| geq delta$ for each $n$. For each $n$ either $|f(x_n)| geq delta /2$ or $|f(y_n)| geq delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| geq delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences ${x_n},{y_n}$ converge to the same limit $x$ because $d(x_n,y_n) to 0$. This contradicts continuity of $f$ at $x$.






          share|cite|improve this answer





















          • Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
            – eatfood
            Nov 20 at 7:05












          • This has solved my problem, thank you!
            – eatfood
            Nov 20 at 7:06
















          3














          Suppose $f$ is not uniformly continuous. Then there exists $delta >0$ and sequences ${x_n},{y_n}$ such that $d(x_n,y_n) to 0$ but $|f(x_n)-f(y_n)| geq delta$ for each $n$. For each $n$ either $|f(x_n)| geq delta /2$ or $|f(y_n)| geq delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| geq delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences ${x_n},{y_n}$ converge to the same limit $x$ because $d(x_n,y_n) to 0$. This contradicts continuity of $f$ at $x$.






          share|cite|improve this answer





















          • Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
            – eatfood
            Nov 20 at 7:05












          • This has solved my problem, thank you!
            – eatfood
            Nov 20 at 7:06














          3












          3








          3






          Suppose $f$ is not uniformly continuous. Then there exists $delta >0$ and sequences ${x_n},{y_n}$ such that $d(x_n,y_n) to 0$ but $|f(x_n)-f(y_n)| geq delta$ for each $n$. For each $n$ either $|f(x_n)| geq delta /2$ or $|f(y_n)| geq delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| geq delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences ${x_n},{y_n}$ converge to the same limit $x$ because $d(x_n,y_n) to 0$. This contradicts continuity of $f$ at $x$.






          share|cite|improve this answer












          Suppose $f$ is not uniformly continuous. Then there exists $delta >0$ and sequences ${x_n},{y_n}$ such that $d(x_n,y_n) to 0$ but $|f(x_n)-f(y_n)| geq delta$ for each $n$. For each $n$ either $|f(x_n)| geq delta /2$ or $|f(y_n)| geq delta /2$. One of these holds for infinitely many $n$. Suppose $|f(x_n)| geq delta/2 $ along a subsequence. The hypothesis tells you that the subsequence lies in a compact set, so it has a convergent subsequence. Along this subsequence both sequences ${x_n},{y_n}$ converge to the same limit $x$ because $d(x_n,y_n) to 0$. This contradicts continuity of $f$ at $x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 6:37









          Kavi Rama Murthy

          49.4k31854




          49.4k31854












          • Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
            – eatfood
            Nov 20 at 7:05












          • This has solved my problem, thank you!
            – eatfood
            Nov 20 at 7:06


















          • Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
            – eatfood
            Nov 20 at 7:05












          • This has solved my problem, thank you!
            – eatfood
            Nov 20 at 7:06
















          Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
          – eatfood
          Nov 20 at 7:05






          Writing out formally, So $(x_{n_k}) to x$ for some subsequence. Then $(y_{n_k}) to x$ too, since $|x_{n_k} - y_{n_k}| to 0$. But $lim_{kto infty}f(x_{n_k}) neq lim_{kto infty}f(y_{n_k})$, since $|f(x_{n_k})-f(y_{n_k})| geq delta$. Contradicting continuity of $f$ at $x$.
          – eatfood
          Nov 20 at 7:05














          This has solved my problem, thank you!
          – eatfood
          Nov 20 at 7:06




          This has solved my problem, thank you!
          – eatfood
          Nov 20 at 7:06


















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