Proof lower level sets of continuous strictly convex function are closed












0












$begingroup$


I need to prove that the following set:
$$mathcal L:={xin mathbb{R}^n , : , f(x) leq f(x_0) }$$
which apparently is called lower level set, is compact. I've already proved boundedness and I just need to show closeness.



The function $f$ is twice continuously differentiable on all of $mathbb{R}^n$ and so it is continuous everywhere.



I am looking for a "simple" proof, i.e. one that uses the notion of continuity and the notion of closeness/openness, NOT the notion of semi-continuity, which is used in every single proof I've found.



My Attempt



$f$ is continuous on all of $mathbb{R}^n$ means that $forall xinmathbb{R}^n$:
$$forall epsilon >0, , exists delta > 0 , forall yin mathbb{R}^n: ||x - y|| < delta Longrightarrow ||f(x)-f(y)||< epsilon$$
On the other hand, $mathcal L$ is closed means that its complement wrt $mathbb{R}^n$ is open, that is:
$$forall xinbar{mathcal{L}}, exists epsilon > 0 , forall yin mathbb{R}^n , : , ||x - y||<epsilon Longrightarrow yin bar{mathcal{L}}$$



Now using continuity we know:
$$||x-y|| < delta Longrightarrow ||f(x)-f(y)||<epsilon Longrightarrow -epsilon < f(x) - f(y) < epsilon$$
And so $Longrightarrow f(y)-epsilon < f(x) < epsilon+f(y)$. How do I continue? Is this correct?






real-analysis calculus optimization







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I need to prove that the following set:
    $$mathcal L:={xin mathbb{R}^n , : , f(x) leq f(x_0) }$$
    which apparently is called lower level set, is compact. I've already proved boundedness and I just need to show closeness.



    The function $f$ is twice continuously differentiable on all of $mathbb{R}^n$ and so it is continuous everywhere.



    I am looking for a "simple" proof, i.e. one that uses the notion of continuity and the notion of closeness/openness, NOT the notion of semi-continuity, which is used in every single proof I've found.



    My Attempt



    $f$ is continuous on all of $mathbb{R}^n$ means that $forall xinmathbb{R}^n$:
    $$forall epsilon >0, , exists delta > 0 , forall yin mathbb{R}^n: ||x - y|| < delta Longrightarrow ||f(x)-f(y)||< epsilon$$
    On the other hand, $mathcal L$ is closed means that its complement wrt $mathbb{R}^n$ is open, that is:
    $$forall xinbar{mathcal{L}}, exists epsilon > 0 , forall yin mathbb{R}^n , : , ||x - y||<epsilon Longrightarrow yin bar{mathcal{L}}$$



    Now using continuity we know:
    $$||x-y|| < delta Longrightarrow ||f(x)-f(y)||<epsilon Longrightarrow -epsilon < f(x) - f(y) < epsilon$$
    And so $Longrightarrow f(y)-epsilon < f(x) < epsilon+f(y)$. How do I continue? Is this correct?






    real-analysis calculus optimization







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      I need to prove that the following set:
      $$mathcal L:={xin mathbb{R}^n , : , f(x) leq f(x_0) }$$
      which apparently is called lower level set, is compact. I've already proved boundedness and I just need to show closeness.



      The function $f$ is twice continuously differentiable on all of $mathbb{R}^n$ and so it is continuous everywhere.



      I am looking for a "simple" proof, i.e. one that uses the notion of continuity and the notion of closeness/openness, NOT the notion of semi-continuity, which is used in every single proof I've found.



      My Attempt



      $f$ is continuous on all of $mathbb{R}^n$ means that $forall xinmathbb{R}^n$:
      $$forall epsilon >0, , exists delta > 0 , forall yin mathbb{R}^n: ||x - y|| < delta Longrightarrow ||f(x)-f(y)||< epsilon$$
      On the other hand, $mathcal L$ is closed means that its complement wrt $mathbb{R}^n$ is open, that is:
      $$forall xinbar{mathcal{L}}, exists epsilon > 0 , forall yin mathbb{R}^n , : , ||x - y||<epsilon Longrightarrow yin bar{mathcal{L}}$$



      Now using continuity we know:
      $$||x-y|| < delta Longrightarrow ||f(x)-f(y)||<epsilon Longrightarrow -epsilon < f(x) - f(y) < epsilon$$
      And so $Longrightarrow f(y)-epsilon < f(x) < epsilon+f(y)$. How do I continue? Is this correct?






      real-analysis calculus optimization







      share|cite|improve this question









      $endgroup$




      I need to prove that the following set:
      $$mathcal L:={xin mathbb{R}^n , : , f(x) leq f(x_0) }$$
      which apparently is called lower level set, is compact. I've already proved boundedness and I just need to show closeness.



      The function $f$ is twice continuously differentiable on all of $mathbb{R}^n$ and so it is continuous everywhere.



      I am looking for a "simple" proof, i.e. one that uses the notion of continuity and the notion of closeness/openness, NOT the notion of semi-continuity, which is used in every single proof I've found.



      My Attempt



      $f$ is continuous on all of $mathbb{R}^n$ means that $forall xinmathbb{R}^n$:
      $$forall epsilon >0, , exists delta > 0 , forall yin mathbb{R}^n: ||x - y|| < delta Longrightarrow ||f(x)-f(y)||< epsilon$$
      On the other hand, $mathcal L$ is closed means that its complement wrt $mathbb{R}^n$ is open, that is:
      $$forall xinbar{mathcal{L}}, exists epsilon > 0 , forall yin mathbb{R}^n , : , ||x - y||<epsilon Longrightarrow yin bar{mathcal{L}}$$



      Now using continuity we know:
      $$||x-y|| < delta Longrightarrow ||f(x)-f(y)||<epsilon Longrightarrow -epsilon < f(x) - f(y) < epsilon$$
      And so $Longrightarrow f(y)-epsilon < f(x) < epsilon+f(y)$. How do I continue? Is this correct?






      real-analysis calculus optimization




      real-analysis calculus optimization






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 4 '18 at 18:57









      Euler_SalterEuler_Salter

      2,0671336




      2,0671336






















          1 Answer
          1






          active

          oldest

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          2












          $begingroup$

          If we define
          $$
          g(x)=f(x)-f(x_0)
          $$
          a continuous function, the set is
          $$
          g^{-1}((-infty,0])
          $$
          the preimage of a closed set under a continuous map.



          If you are wedded to epsilon's and delta's, then pick $xin mathcal{L}^c$, then
          $$
          g(x)>0
          $$
          and for $epsilon=frac{g(x)}{2}$, we have for some $delta>0$ corresponding to this $epsilon$,
          $$
          g(y)>0
          $$
          for all $yin (x-delta,x+delta)$. So $(x-delta,x+delta)subset mathcal{L}^c$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you please expand a little bit more? It seems interesting but I’m struggling to understand
            $endgroup$
            – Euler_Salter
            Dec 4 '18 at 19:06










          • $begingroup$
            I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
            $endgroup$
            – qbert
            Dec 4 '18 at 19:08











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          If we define
          $$
          g(x)=f(x)-f(x_0)
          $$
          a continuous function, the set is
          $$
          g^{-1}((-infty,0])
          $$
          the preimage of a closed set under a continuous map.



          If you are wedded to epsilon's and delta's, then pick $xin mathcal{L}^c$, then
          $$
          g(x)>0
          $$
          and for $epsilon=frac{g(x)}{2}$, we have for some $delta>0$ corresponding to this $epsilon$,
          $$
          g(y)>0
          $$
          for all $yin (x-delta,x+delta)$. So $(x-delta,x+delta)subset mathcal{L}^c$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you please expand a little bit more? It seems interesting but I’m struggling to understand
            $endgroup$
            – Euler_Salter
            Dec 4 '18 at 19:06










          • $begingroup$
            I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
            $endgroup$
            – qbert
            Dec 4 '18 at 19:08
















          2












          $begingroup$

          If we define
          $$
          g(x)=f(x)-f(x_0)
          $$
          a continuous function, the set is
          $$
          g^{-1}((-infty,0])
          $$
          the preimage of a closed set under a continuous map.



          If you are wedded to epsilon's and delta's, then pick $xin mathcal{L}^c$, then
          $$
          g(x)>0
          $$
          and for $epsilon=frac{g(x)}{2}$, we have for some $delta>0$ corresponding to this $epsilon$,
          $$
          g(y)>0
          $$
          for all $yin (x-delta,x+delta)$. So $(x-delta,x+delta)subset mathcal{L}^c$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you please expand a little bit more? It seems interesting but I’m struggling to understand
            $endgroup$
            – Euler_Salter
            Dec 4 '18 at 19:06










          • $begingroup$
            I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
            $endgroup$
            – qbert
            Dec 4 '18 at 19:08














          2












          2








          2





          $begingroup$

          If we define
          $$
          g(x)=f(x)-f(x_0)
          $$
          a continuous function, the set is
          $$
          g^{-1}((-infty,0])
          $$
          the preimage of a closed set under a continuous map.



          If you are wedded to epsilon's and delta's, then pick $xin mathcal{L}^c$, then
          $$
          g(x)>0
          $$
          and for $epsilon=frac{g(x)}{2}$, we have for some $delta>0$ corresponding to this $epsilon$,
          $$
          g(y)>0
          $$
          for all $yin (x-delta,x+delta)$. So $(x-delta,x+delta)subset mathcal{L}^c$.






          share|cite|improve this answer











          $endgroup$



          If we define
          $$
          g(x)=f(x)-f(x_0)
          $$
          a continuous function, the set is
          $$
          g^{-1}((-infty,0])
          $$
          the preimage of a closed set under a continuous map.



          If you are wedded to epsilon's and delta's, then pick $xin mathcal{L}^c$, then
          $$
          g(x)>0
          $$
          and for $epsilon=frac{g(x)}{2}$, we have for some $delta>0$ corresponding to this $epsilon$,
          $$
          g(y)>0
          $$
          for all $yin (x-delta,x+delta)$. So $(x-delta,x+delta)subset mathcal{L}^c$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '18 at 19:07

























          answered Dec 4 '18 at 19:03









          qbertqbert

          22.1k32561




          22.1k32561












          • $begingroup$
            Can you please expand a little bit more? It seems interesting but I’m struggling to understand
            $endgroup$
            – Euler_Salter
            Dec 4 '18 at 19:06










          • $begingroup$
            I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
            $endgroup$
            – qbert
            Dec 4 '18 at 19:08


















          • $begingroup$
            Can you please expand a little bit more? It seems interesting but I’m struggling to understand
            $endgroup$
            – Euler_Salter
            Dec 4 '18 at 19:06










          • $begingroup$
            I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
            $endgroup$
            – qbert
            Dec 4 '18 at 19:08
















          $begingroup$
          Can you please expand a little bit more? It seems interesting but I’m struggling to understand
          $endgroup$
          – Euler_Salter
          Dec 4 '18 at 19:06




          $begingroup$
          Can you please expand a little bit more? It seems interesting but I’m struggling to understand
          $endgroup$
          – Euler_Salter
          Dec 4 '18 at 19:06












          $begingroup$
          I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
          $endgroup$
          – qbert
          Dec 4 '18 at 19:08




          $begingroup$
          I added a solution using the epsilon-delta definition of continuity. In general in topology, a continuous function is one which has the property that preimages of open sets are open. This then implies that preimages of closed sets are closed.
          $endgroup$
          – qbert
          Dec 4 '18 at 19:08


















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