Proving a function is always nonnegative












3












$begingroup$



Suppose $f : mathbb{R}^{n} rightarrow mathbb{R}$ is continuous and
$f(u) > 0$ if the point $u in mathbb{R}^{n}$ has at least one
rational component. Prove that $f(u) geq 0 $ for all $u in
mathbb{R}^{n}$.




Partial Solution:



What remains to be proven is that $f(u) geq 0$ for every $u in mathbb{R}^{n}$ with at all irrational components. Let $v$ be such a vector. Since every real number is the limit point of a rational sequence, we can define sequences ${r_{1}}, {r_{2}}, ldots, {r_{n}},$ where ${r_{k}}$ converges to the $k^{text{th}}$ component of $v$.




I don't really know how to finish from here. Can someone please help me?






real-analysis general-topology limits functions convergence







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stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    3












    $begingroup$



    Suppose $f : mathbb{R}^{n} rightarrow mathbb{R}$ is continuous and
    $f(u) > 0$ if the point $u in mathbb{R}^{n}$ has at least one
    rational component. Prove that $f(u) geq 0 $ for all $u in
    mathbb{R}^{n}$    .




    Partial Solution:



    What remains to be proven is that $f(u) geq 0$ for every $u in mathbb{R}^{n}$ with at all irrational components. Let $v$ be such a vector. Since every real number is the limit point of a rational sequence, we can define sequences ${r_{1}}, {r_{2}}, ldots, {r_{n}},$ where ${r_{k}}$ converges to the $k^{text{th}}$ component of $v$.




    I don't really know how to finish from here. Can someone please help me?






    real-analysis general-topology limits functions convergence







    share|cite|improve this question









    New contributor




    stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3





      $begingroup$



      Suppose $f : mathbb{R}^{n} rightarrow mathbb{R}$ is continuous and
      $f(u) > 0$ if the point $u in mathbb{R}^{n}$ has at least one
      rational component. Prove that $f(u) geq 0 $ for all $u in
      mathbb{R}^{n}$      .




      Partial Solution:



      What remains to be proven is that $f(u) geq 0$ for every $u in mathbb{R}^{n}$ with at all irrational components. Let $v$ be such a vector. Since every real number is the limit point of a rational sequence, we can define sequences ${r_{1}}, {r_{2}}, ldots, {r_{n}},$ where ${r_{k}}$ converges to the $k^{text{th}}$ component of $v$.




      I don't really know how to finish from here. Can someone please help me?






      real-analysis general-topology limits functions convergence







      share|cite|improve this question









      New contributor




      stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$





      Suppose $f : mathbb{R}^{n} rightarrow mathbb{R}$ is continuous and
      $f(u) > 0$ if the point $u in mathbb{R}^{n}$ has at least one
      rational component. Prove that $f(u) geq 0 $ for all $u in
      mathbb{R}^{n}$      .




      Partial Solution:



      What remains to be proven is that $f(u) geq 0$ for every $u in mathbb{R}^{n}$ with at all irrational components. Let $v$ be such a vector. Since every real number is the limit point of a rational sequence, we can define sequences ${r_{1}}, {r_{2}}, ldots, {r_{n}},$ where ${r_{k}}$ converges to the $k^{text{th}}$ component of $v$.




      I don't really know how to finish from here. Can someone please help me?






      real-analysis general-topology limits functions convergence




      real-analysis general-topology limits functions convergence






      share|cite|improve this question









      New contributor




      stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 14 hours ago









      rash

      22411




      22411






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      asked 15 hours ago









      stackofhay42stackofhay42

      1653




      1653




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      New contributor





      stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          For fixed $v in mathbb{R}^n$ write $v=(v^{(1)},ldots,v^{(n)})$, i.e. $v^{(k)}$ is the $k$-th component of $v$ for $k in {1,ldots,n}$.



          As you suggest, we choose for some some $k in {1,ldots,n}$ a rational sequence $(r_i)_{i in mathbb{N}}$ such that $r_i$ converges to $v^{(k)}$ as $i to infty$. Now define $$u_i := (v^{(1)},ldots,v^{(k-1)},r_i, v^{(k+1)},ldots,v^{(n)})$$ for $i geq 1$. By construction, we have $f(u_i) geq 0$ for all $i$ and $u_i to v$ as $i to infty$.



          Hint: Use the continuity of $f$ to conclude that $f(v) geq 0$. (How can you express $f(v)$ in terms of $f(u_i)$?)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Does $v^{(k)}$ mean the $k^{text{th}}$ component of $v$?
            $endgroup$
            – stackofhay42
            15 hours ago










          • $begingroup$
            I am still unsure on how to finish
            $endgroup$
            – stackofhay42
            15 hours ago










          • $begingroup$
            @stackofhay42 Maybe the second part of my answer could help.
            $endgroup$
            – Holding Arthur
            15 hours ago










          • $begingroup$
            @stackofhay42 Yes, it does denote the $k$-th component. Regarding your 2nd question: $f$ is (sequentially) continuous, so $x_k to x$ implies $f(x_k) to f(x)$, right?
            $endgroup$
            – saz
            14 hours ago










          • $begingroup$
            Yes, I got this: $f(u_{i}) rightarrow f(v)$, but I do not see how to conclude ?
            $endgroup$
            – stackofhay42
            14 hours ago





















          3












          $begingroup$

          Let's try to prove a stronger result:




          Theorem. Suppose $f : mathbb{R}^{n} rightarrow mathbb{R}$ is continuous and
          $f(u) > 0$ if the point $u in mathbb{R}^{n}$ has all its components rational. Then $f(u) geq 0 $ for all $u in
          mathbb{R}^{n}$          .




          Proof. Let $v=(v_1,v_2,...,v_n)inmathbb{R}^n$. Choose rational sequences ${x_{mk}}$ such that $lim_{kto infty}x_{mk}=v_m$, for all $m=1,2,...,n$. The sequence of points $x_k=(x_{1k},...,x_{nk})$. Please tell me if you want to know in details why such choice is possible.



          Then $x_k to v$. Since $f$ is continuous, $f(x_k) to f(v)$. We know that $f(x_k)>0$ for all $k$. To show that $f(v)geq0$, suppose that $f(v)<0$. Then for all $k$, $|f(x_k)-f(v)|geq|f(v)|$. This is a contradiction, because for $f(x_k) to f(v)$, there should exist $k_0$ such that for any $kgeq k_0$, $|f(x_k)-f(v)|leq|f(v)|$. Here, we take $|f(v)|>0$ as the arbitrary number $epsilon>0$.



          Note. The strict inequality becomes NOT strict after taking the limit. We can actually find examples to show that equality can hold in $f(u) geq 0 $.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Hint: Every $uinBbb{R}^n$ is the limit of a sequence $(u_k)_k$ where each of the $u_k$ has at least one rational component, and hence $f(u)=lim_{ktoinfty}f(u_k)geq0$.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              For fixed $v in mathbb{R}^n$ write $v=(v^{(1)},ldots,v^{(n)})$, i.e. $v^{(k)}$ is the $k$-th component of $v$ for $k in {1,ldots,n}$.



              As you suggest, we choose for some some $k in {1,ldots,n}$ a rational sequence $(r_i)_{i in mathbb{N}}$ such that $r_i$ converges to $v^{(k)}$ as $i to infty$. Now define $$u_i := (v^{(1)},ldots,v^{(k-1)},r_i, v^{(k+1)},ldots,v^{(n)})$$ for $i geq 1$. By construction, we have $f(u_i) geq 0$ for all $i$ and $u_i to v$ as $i to infty$.



              Hint: Use the continuity of $f$ to conclude that $f(v) geq 0$. (How can you express $f(v)$ in terms of $f(u_i)$?)






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Does $v^{(k)}$ mean the $k^{text{th}}$ component of $v$?
                $endgroup$
                – stackofhay42
                15 hours ago










              • $begingroup$
                I am still unsure on how to finish
                $endgroup$
                – stackofhay42
                15 hours ago










              • $begingroup$
                @stackofhay42 Maybe the second part of my answer could help.
                $endgroup$
                – Holding Arthur
                15 hours ago










              • $begingroup$
                @stackofhay42 Yes, it does denote the $k$-th component. Regarding your 2nd question: $f$ is (sequentially) continuous, so $x_k to x$ implies $f(x_k) to f(x)$, right?
                $endgroup$
                – saz
                14 hours ago










              • $begingroup$
                Yes, I got this: $f(u_{i}) rightarrow f(v)$, but I do not see how to conclude ?
                $endgroup$
                – stackofhay42
                14 hours ago


















              3












              $begingroup$

              For fixed $v in mathbb{R}^n$ write $v=(v^{(1)},ldots,v^{(n)})$, i.e. $v^{(k)}$ is the $k$-th component of $v$ for $k in {1,ldots,n}$.



              As you suggest, we choose for some some $k in {1,ldots,n}$ a rational sequence $(r_i)_{i in mathbb{N}}$ such that $r_i$ converges to $v^{(k)}$ as $i to infty$. Now define $$u_i := (v^{(1)},ldots,v^{(k-1)},r_i, v^{(k+1)},ldots,v^{(n)})$$ for $i geq 1$. By construction, we have $f(u_i) geq 0$ for all $i$ and $u_i to v$ as $i to infty$.



              Hint: Use the continuity of $f$ to conclude that $f(v) geq 0$. (How can you express $f(v)$ in terms of $f(u_i)$?)






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Does $v^{(k)}$ mean the $k^{text{th}}$ component of $v$?
                $endgroup$
                – stackofhay42
                15 hours ago










              • $begingroup$
                I am still unsure on how to finish
                $endgroup$
                – stackofhay42
                15 hours ago










              • $begingroup$
                @stackofhay42 Maybe the second part of my answer could help.
                $endgroup$
                – Holding Arthur
                15 hours ago










              • $begingroup$
                @stackofhay42 Yes, it does denote the $k$-th component. Regarding your 2nd question: $f$ is (sequentially) continuous, so $x_k to x$ implies $f(x_k) to f(x)$, right?
                $endgroup$
                – saz
                14 hours ago










              • $begingroup$
                Yes, I got this: $f(u_{i}) rightarrow f(v)$, but I do not see how to conclude ?
                $endgroup$
                – stackofhay42
                14 hours ago
















              3












              3








              3





              $begingroup$

              For fixed $v in mathbb{R}^n$ write $v=(v^{(1)},ldots,v^{(n)})$, i.e. $v^{(k)}$ is the $k$-th component of $v$ for $k in {1,ldots,n}$.



              As you suggest, we choose for some some $k in {1,ldots,n}$ a rational sequence $(r_i)_{i in mathbb{N}}$ such that $r_i$ converges to $v^{(k)}$ as $i to infty$. Now define $$u_i := (v^{(1)},ldots,v^{(k-1)},r_i, v^{(k+1)},ldots,v^{(n)})$$ for $i geq 1$. By construction, we have $f(u_i) geq 0$ for all $i$ and $u_i to v$ as $i to infty$.



              Hint: Use the continuity of $f$ to conclude that $f(v) geq 0$. (How can you express $f(v)$ in terms of $f(u_i)$?)






              share|cite|improve this answer











              $endgroup$



              For fixed $v in mathbb{R}^n$ write $v=(v^{(1)},ldots,v^{(n)})$, i.e. $v^{(k)}$ is the $k$-th component of $v$ for $k in {1,ldots,n}$.



              As you suggest, we choose for some some $k in {1,ldots,n}$ a rational sequence $(r_i)_{i in mathbb{N}}$ such that $r_i$ converges to $v^{(k)}$ as $i to infty$. Now define $$u_i := (v^{(1)},ldots,v^{(k-1)},r_i, v^{(k+1)},ldots,v^{(n)})$$ for $i geq 1$. By construction, we have $f(u_i) geq 0$ for all $i$ and $u_i to v$ as $i to infty$.



              Hint: Use the continuity of $f$ to conclude that $f(v) geq 0$. (How can you express $f(v)$ in terms of $f(u_i)$?)







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 14 hours ago

























              answered 15 hours ago









              sazsaz

              81.6k861127




              81.6k861127












              • $begingroup$
                Does $v^{(k)}$ mean the $k^{text{th}}$ component of $v$?
                $endgroup$
                – stackofhay42
                15 hours ago










              • $begingroup$
                I am still unsure on how to finish
                $endgroup$
                – stackofhay42
                15 hours ago










              • $begingroup$
                @stackofhay42 Maybe the second part of my answer could help.
                $endgroup$
                – Holding Arthur
                15 hours ago










              • $begingroup$
                @stackofhay42 Yes, it does denote the $k$-th component. Regarding your 2nd question: $f$ is (sequentially) continuous, so $x_k to x$ implies $f(x_k) to f(x)$, right?
                $endgroup$
                – saz
                14 hours ago










              • $begingroup$
                Yes, I got this: $f(u_{i}) rightarrow f(v)$, but I do not see how to conclude ?
                $endgroup$
                – stackofhay42
                14 hours ago




















              • $begingroup$
                Does $v^{(k)}$ mean the $k^{text{th}}$ component of $v$?
                $endgroup$
                – stackofhay42
                15 hours ago










              • $begingroup$
                I am still unsure on how to finish
                $endgroup$
                – stackofhay42
                15 hours ago










              • $begingroup$
                @stackofhay42 Maybe the second part of my answer could help.
                $endgroup$
                – Holding Arthur
                15 hours ago










              • $begingroup$
                @stackofhay42 Yes, it does denote the $k$-th component. Regarding your 2nd question: $f$ is (sequentially) continuous, so $x_k to x$ implies $f(x_k) to f(x)$, right?
                $endgroup$
                – saz
                14 hours ago










              • $begingroup$
                Yes, I got this: $f(u_{i}) rightarrow f(v)$, but I do not see how to conclude ?
                $endgroup$
                – stackofhay42
                14 hours ago


















              $begingroup$
              Does $v^{(k)}$ mean the $k^{text{th}}$ component of $v$?
              $endgroup$
              – stackofhay42
              15 hours ago




              $begingroup$
              Does $v^{(k)}$ mean the $k^{text{th}}$ component of $v$?
              $endgroup$
              – stackofhay42
              15 hours ago












              $begingroup$
              I am still unsure on how to finish
              $endgroup$
              – stackofhay42
              15 hours ago




              $begingroup$
              I am still unsure on how to finish
              $endgroup$
              – stackofhay42
              15 hours ago












              $begingroup$
              @stackofhay42 Maybe the second part of my answer could help.
              $endgroup$
              – Holding Arthur
              15 hours ago




              $begingroup$
              @stackofhay42 Maybe the second part of my answer could help.
              $endgroup$
              – Holding Arthur
              15 hours ago












              $begingroup$
              @stackofhay42 Yes, it does denote the $k$-th component. Regarding your 2nd question: $f$ is (sequentially) continuous, so $x_k to x$ implies $f(x_k) to f(x)$, right?
              $endgroup$
              – saz
              14 hours ago




              $begingroup$
              @stackofhay42 Yes, it does denote the $k$-th component. Regarding your 2nd question: $f$ is (sequentially) continuous, so $x_k to x$ implies $f(x_k) to f(x)$, right?
              $endgroup$
              – saz
              14 hours ago












              $begingroup$
              Yes, I got this: $f(u_{i}) rightarrow f(v)$, but I do not see how to conclude ?
              $endgroup$
              – stackofhay42
              14 hours ago






              $begingroup$
              Yes, I got this: $f(u_{i}) rightarrow f(v)$, but I do not see how to conclude ?
              $endgroup$
              – stackofhay42
              14 hours ago













              3












              $begingroup$

              Let's try to prove a stronger result:




              Theorem. Suppose $f : mathbb{R}^{n} rightarrow mathbb{R}$ is continuous and
              $f(u) > 0$ if the point $u in mathbb{R}^{n}$ has all its components rational. Then $f(u) geq 0 $ for all $u in
              mathbb{R}^{n}$              .




              Proof. Let $v=(v_1,v_2,...,v_n)inmathbb{R}^n$. Choose rational sequences ${x_{mk}}$ such that $lim_{kto infty}x_{mk}=v_m$, for all $m=1,2,...,n$. The sequence of points $x_k=(x_{1k},...,x_{nk})$. Please tell me if you want to know in details why such choice is possible.



              Then $x_k to v$. Since $f$ is continuous, $f(x_k) to f(v)$. We know that $f(x_k)>0$ for all $k$. To show that $f(v)geq0$, suppose that $f(v)<0$. Then for all $k$, $|f(x_k)-f(v)|geq|f(v)|$. This is a contradiction, because for $f(x_k) to f(v)$, there should exist $k_0$ such that for any $kgeq k_0$, $|f(x_k)-f(v)|leq|f(v)|$. Here, we take $|f(v)|>0$ as the arbitrary number $epsilon>0$.



              Note. The strict inequality becomes NOT strict after taking the limit. We can actually find examples to show that equality can hold in $f(u) geq 0 $.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Let's try to prove a stronger result:




                Theorem. Suppose $f : mathbb{R}^{n} rightarrow mathbb{R}$ is continuous and
                $f(u) > 0$ if the point $u in mathbb{R}^{n}$ has all its components rational. Then $f(u) geq 0 $ for all $u in
                mathbb{R}^{n}$                .




                Proof. Let $v=(v_1,v_2,...,v_n)inmathbb{R}^n$. Choose rational sequences ${x_{mk}}$ such that $lim_{kto infty}x_{mk}=v_m$, for all $m=1,2,...,n$. The sequence of points $x_k=(x_{1k},...,x_{nk})$. Please tell me if you want to know in details why such choice is possible.



                Then $x_k to v$. Since $f$ is continuous, $f(x_k) to f(v)$. We know that $f(x_k)>0$ for all $k$. To show that $f(v)geq0$, suppose that $f(v)<0$. Then for all $k$, $|f(x_k)-f(v)|geq|f(v)|$. This is a contradiction, because for $f(x_k) to f(v)$, there should exist $k_0$ such that for any $kgeq k_0$, $|f(x_k)-f(v)|leq|f(v)|$. Here, we take $|f(v)|>0$ as the arbitrary number $epsilon>0$.



                Note. The strict inequality becomes NOT strict after taking the limit. We can actually find examples to show that equality can hold in $f(u) geq 0 $.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Let's try to prove a stronger result:




                  Theorem. Suppose $f : mathbb{R}^{n} rightarrow mathbb{R}$ is continuous and
                  $f(u) > 0$ if the point $u in mathbb{R}^{n}$ has all its components rational. Then $f(u) geq 0 $ for all $u in
                  mathbb{R}^{n}$                  .




                  Proof. Let $v=(v_1,v_2,...,v_n)inmathbb{R}^n$. Choose rational sequences ${x_{mk}}$ such that $lim_{kto infty}x_{mk}=v_m$, for all $m=1,2,...,n$. The sequence of points $x_k=(x_{1k},...,x_{nk})$. Please tell me if you want to know in details why such choice is possible.



                  Then $x_k to v$. Since $f$ is continuous, $f(x_k) to f(v)$. We know that $f(x_k)>0$ for all $k$. To show that $f(v)geq0$, suppose that $f(v)<0$. Then for all $k$, $|f(x_k)-f(v)|geq|f(v)|$. This is a contradiction, because for $f(x_k) to f(v)$, there should exist $k_0$ such that for any $kgeq k_0$, $|f(x_k)-f(v)|leq|f(v)|$. Here, we take $|f(v)|>0$ as the arbitrary number $epsilon>0$.



                  Note. The strict inequality becomes NOT strict after taking the limit. We can actually find examples to show that equality can hold in $f(u) geq 0 $.






                  share|cite|improve this answer









                  $endgroup$



                  Let's try to prove a stronger result:




                  Theorem. Suppose $f : mathbb{R}^{n} rightarrow mathbb{R}$ is continuous and
                  $f(u) > 0$ if the point $u in mathbb{R}^{n}$ has all its components rational. Then $f(u) geq 0 $ for all $u in
                  mathbb{R}^{n}$                  .




                  Proof. Let $v=(v_1,v_2,...,v_n)inmathbb{R}^n$. Choose rational sequences ${x_{mk}}$ such that $lim_{kto infty}x_{mk}=v_m$, for all $m=1,2,...,n$. The sequence of points $x_k=(x_{1k},...,x_{nk})$. Please tell me if you want to know in details why such choice is possible.



                  Then $x_k to v$. Since $f$ is continuous, $f(x_k) to f(v)$. We know that $f(x_k)>0$ for all $k$. To show that $f(v)geq0$, suppose that $f(v)<0$. Then for all $k$, $|f(x_k)-f(v)|geq|f(v)|$. This is a contradiction, because for $f(x_k) to f(v)$, there should exist $k_0$ such that for any $kgeq k_0$, $|f(x_k)-f(v)|leq|f(v)|$. Here, we take $|f(v)|>0$ as the arbitrary number $epsilon>0$.



                  Note. The strict inequality becomes NOT strict after taking the limit. We can actually find examples to show that equality can hold in $f(u) geq 0 $.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 15 hours ago









                  Holding ArthurHolding Arthur

                  1,197417




                  1,197417























                      0












                      $begingroup$

                      Hint: Every $uinBbb{R}^n$ is the limit of a sequence $(u_k)_k$ where each of the $u_k$ has at least one rational component, and hence $f(u)=lim_{ktoinfty}f(u_k)geq0$.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Hint: Every $uinBbb{R}^n$ is the limit of a sequence $(u_k)_k$ where each of the $u_k$ has at least one rational component, and hence $f(u)=lim_{ktoinfty}f(u_k)geq0$.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Hint: Every $uinBbb{R}^n$ is the limit of a sequence $(u_k)_k$ where each of the $u_k$ has at least one rational component, and hence $f(u)=lim_{ktoinfty}f(u_k)geq0$.






                          share|cite|improve this answer









                          $endgroup$



                          Hint: Every $uinBbb{R}^n$ is the limit of a sequence $(u_k)_k$ where each of the $u_k$ has at least one rational component, and hence $f(u)=lim_{ktoinfty}f(u_k)geq0$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 12 hours ago









                          ServaesServaes

                          27.9k34099




                          27.9k34099






















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