Question about the proof of Stone-Weierstrass theorem (Weierstrass approximation theorem) in Rudin












2












$begingroup$


In Rudin's Principles of Mathematical Analysis, a proof of the Stone-Weierstrass theorem in its original statement is included (3ed, p159):



My question is about the step after (51), $P_n(x)=int_{-1}^1f(x+t)Q_n(t)operatorname{d}t$.
How does one proceed from this, by a change of variable, to the next step, namely $P_n(x)=int_{-x}^{1-x}f(x+t)Q_n(t)operatorname{d}t$?



And another question is why $P_n(x)=int_0^1f(t)Q_n(t-x)operatorname{d}t$ is a polynomial.






real-analysis complex-analysis proof-explanation







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    In Rudin's Principles of Mathematical Analysis, a proof of the Stone-Weierstrass theorem in its original statement is included (3ed, p159):



    My question is about the step after (51), $P_n(x)=int_{-1}^1f(x+t)Q_n(t)operatorname{d}t$.
    How does one proceed from this, by a change of variable, to the next step, namely $P_n(x)=int_{-x}^{1-x}f(x+t)Q_n(t)operatorname{d}t$?



    And another question is why $P_n(x)=int_0^1f(t)Q_n(t-x)operatorname{d}t$ is a polynomial.






    real-analysis complex-analysis proof-explanation







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      In Rudin's Principles of Mathematical Analysis, a proof of the Stone-Weierstrass theorem in its original statement is included (3ed, p159):



      My question is about the step after (51), $P_n(x)=int_{-1}^1f(x+t)Q_n(t)operatorname{d}t$.
      How does one proceed from this, by a change of variable, to the next step, namely $P_n(x)=int_{-x}^{1-x}f(x+t)Q_n(t)operatorname{d}t$?



      And another question is why $P_n(x)=int_0^1f(t)Q_n(t-x)operatorname{d}t$ is a polynomial.






      real-analysis complex-analysis proof-explanation







      share|cite|improve this question











      $endgroup$




      In Rudin's Principles of Mathematical Analysis, a proof of the Stone-Weierstrass theorem in its original statement is included (3ed, p159):



      My question is about the step after (51), $P_n(x)=int_{-1}^1f(x+t)Q_n(t)operatorname{d}t$.
      How does one proceed from this, by a change of variable, to the next step, namely $P_n(x)=int_{-x}^{1-x}f(x+t)Q_n(t)operatorname{d}t$?



      And another question is why $P_n(x)=int_0^1f(t)Q_n(t-x)operatorname{d}t$ is a polynomial.






      real-analysis complex-analysis proof-explanation




      real-analysis complex-analysis proof-explanation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 14 '18 at 12:08









      Scientifica

      6,81141335




      6,81141335










      asked Dec 14 '18 at 12:05









      Yutong ZhangYutong Zhang

      417




      417






















          1 Answer
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          3












          $begingroup$

          Well the first equality, namely $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt $ follows just from the fact that f is $0$ outside $[0,1]$ which is one of the simplificating assumptions Rudin makes.



          Now $int_{-x}^{1-x}f(x+t)Q_n(t)dt = int_{0}^{1}f(t)Q_n(t-x)dt $ follows by the substitution t = t-x.



          The fact that $int_{0}^{1}f(t)Q_n(t-x)dt $ is a poly in $x $ follows from writing $Q_n(t+x) = sum_{k=0}^{n}a_i(t+x)^k=sum_{k=0}^{n}b_i(t)x^k$ and now $int_{0}^{1}f(t)Q_n(t-x)dt = sum_{k=0}^{n}(int_{0}^{1}b_i(t)dt)x^k$, where $b_i(t)$ are just the functions(polys) obtained by expanding each $(t+x)^k$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay, now I see why that is a polynomial: apply binomial expansion multiple times, first on $(1-(t-x)^2)^n $ and on those $(t-x)^{2i}$, then the integral will become $$sum_{i=0}^{2n}c_nk(i)left(int_0^1f(t)cdot t^{2n-i}operatorname{d}tright)x^i$$ where $k(i)$ are the merged binomial coefficients.
            $endgroup$
            – Yutong Zhang
            Dec 16 '18 at 11:46












          • $begingroup$
            But I am still wondering about the $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt$ part, would you explain this part with a little bit more details? Thanks.
            $endgroup$
            – Yutong Zhang
            Dec 16 '18 at 11:50






          • 1




            $begingroup$
            Ok, so the integral has t varying from -1 to 1, but in fact f(x+t) is $0$ for t < -x so the integral from -1 to -x of f(x+t)$Q_n(t)$ will be 0. Is this clear?
            $endgroup$
            – Sorin Tirc
            Dec 16 '18 at 16:04












          • $begingroup$
            Yes, now I get it. Thank you.
            $endgroup$
            – Yutong Zhang
            Dec 17 '18 at 2:18











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          3












          $begingroup$

          Well the first equality, namely $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt $ follows just from the fact that f is $0$ outside $[0,1]$ which is one of the simplificating assumptions Rudin makes.



          Now $int_{-x}^{1-x}f(x+t)Q_n(t)dt = int_{0}^{1}f(t)Q_n(t-x)dt $ follows by the substitution t = t-x.



          The fact that $int_{0}^{1}f(t)Q_n(t-x)dt $ is a poly in $x $ follows from writing $Q_n(t+x) = sum_{k=0}^{n}a_i(t+x)^k=sum_{k=0}^{n}b_i(t)x^k$ and now $int_{0}^{1}f(t)Q_n(t-x)dt = sum_{k=0}^{n}(int_{0}^{1}b_i(t)dt)x^k$, where $b_i(t)$ are just the functions(polys) obtained by expanding each $(t+x)^k$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay, now I see why that is a polynomial: apply binomial expansion multiple times, first on $(1-(t-x)^2)^n $ and on those $(t-x)^{2i}$, then the integral will become $$sum_{i=0}^{2n}c_nk(i)left(int_0^1f(t)cdot t^{2n-i}operatorname{d}tright)x^i$$ where $k(i)$ are the merged binomial coefficients.
            $endgroup$
            – Yutong Zhang
            Dec 16 '18 at 11:46












          • $begingroup$
            But I am still wondering about the $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt$ part, would you explain this part with a little bit more details? Thanks.
            $endgroup$
            – Yutong Zhang
            Dec 16 '18 at 11:50






          • 1




            $begingroup$
            Ok, so the integral has t varying from -1 to 1, but in fact f(x+t) is $0$ for t < -x so the integral from -1 to -x of f(x+t)$Q_n(t)$ will be 0. Is this clear?
            $endgroup$
            – Sorin Tirc
            Dec 16 '18 at 16:04












          • $begingroup$
            Yes, now I get it. Thank you.
            $endgroup$
            – Yutong Zhang
            Dec 17 '18 at 2:18
















          3












          $begingroup$

          Well the first equality, namely $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt $ follows just from the fact that f is $0$ outside $[0,1]$ which is one of the simplificating assumptions Rudin makes.



          Now $int_{-x}^{1-x}f(x+t)Q_n(t)dt = int_{0}^{1}f(t)Q_n(t-x)dt $ follows by the substitution t = t-x.



          The fact that $int_{0}^{1}f(t)Q_n(t-x)dt $ is a poly in $x $ follows from writing $Q_n(t+x) = sum_{k=0}^{n}a_i(t+x)^k=sum_{k=0}^{n}b_i(t)x^k$ and now $int_{0}^{1}f(t)Q_n(t-x)dt = sum_{k=0}^{n}(int_{0}^{1}b_i(t)dt)x^k$, where $b_i(t)$ are just the functions(polys) obtained by expanding each $(t+x)^k$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay, now I see why that is a polynomial: apply binomial expansion multiple times, first on $(1-(t-x)^2)^n $ and on those $(t-x)^{2i}$, then the integral will become $$sum_{i=0}^{2n}c_nk(i)left(int_0^1f(t)cdot t^{2n-i}operatorname{d}tright)x^i$$ where $k(i)$ are the merged binomial coefficients.
            $endgroup$
            – Yutong Zhang
            Dec 16 '18 at 11:46












          • $begingroup$
            But I am still wondering about the $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt$ part, would you explain this part with a little bit more details? Thanks.
            $endgroup$
            – Yutong Zhang
            Dec 16 '18 at 11:50






          • 1




            $begingroup$
            Ok, so the integral has t varying from -1 to 1, but in fact f(x+t) is $0$ for t < -x so the integral from -1 to -x of f(x+t)$Q_n(t)$ will be 0. Is this clear?
            $endgroup$
            – Sorin Tirc
            Dec 16 '18 at 16:04












          • $begingroup$
            Yes, now I get it. Thank you.
            $endgroup$
            – Yutong Zhang
            Dec 17 '18 at 2:18














          3












          3








          3





          $begingroup$

          Well the first equality, namely $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt $ follows just from the fact that f is $0$ outside $[0,1]$ which is one of the simplificating assumptions Rudin makes.



          Now $int_{-x}^{1-x}f(x+t)Q_n(t)dt = int_{0}^{1}f(t)Q_n(t-x)dt $ follows by the substitution t = t-x.



          The fact that $int_{0}^{1}f(t)Q_n(t-x)dt $ is a poly in $x $ follows from writing $Q_n(t+x) = sum_{k=0}^{n}a_i(t+x)^k=sum_{k=0}^{n}b_i(t)x^k$ and now $int_{0}^{1}f(t)Q_n(t-x)dt = sum_{k=0}^{n}(int_{0}^{1}b_i(t)dt)x^k$, where $b_i(t)$ are just the functions(polys) obtained by expanding each $(t+x)^k$.






          share|cite|improve this answer









          $endgroup$



          Well the first equality, namely $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt $ follows just from the fact that f is $0$ outside $[0,1]$ which is one of the simplificating assumptions Rudin makes.



          Now $int_{-x}^{1-x}f(x+t)Q_n(t)dt = int_{0}^{1}f(t)Q_n(t-x)dt $ follows by the substitution t = t-x.



          The fact that $int_{0}^{1}f(t)Q_n(t-x)dt $ is a poly in $x $ follows from writing $Q_n(t+x) = sum_{k=0}^{n}a_i(t+x)^k=sum_{k=0}^{n}b_i(t)x^k$ and now $int_{0}^{1}f(t)Q_n(t-x)dt = sum_{k=0}^{n}(int_{0}^{1}b_i(t)dt)x^k$, where $b_i(t)$ are just the functions(polys) obtained by expanding each $(t+x)^k$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 12:15









          Sorin TircSorin Tirc

          1,755213




          1,755213












          • $begingroup$
            Okay, now I see why that is a polynomial: apply binomial expansion multiple times, first on $(1-(t-x)^2)^n $ and on those $(t-x)^{2i}$, then the integral will become $$sum_{i=0}^{2n}c_nk(i)left(int_0^1f(t)cdot t^{2n-i}operatorname{d}tright)x^i$$ where $k(i)$ are the merged binomial coefficients.
            $endgroup$
            – Yutong Zhang
            Dec 16 '18 at 11:46












          • $begingroup$
            But I am still wondering about the $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt$ part, would you explain this part with a little bit more details? Thanks.
            $endgroup$
            – Yutong Zhang
            Dec 16 '18 at 11:50






          • 1




            $begingroup$
            Ok, so the integral has t varying from -1 to 1, but in fact f(x+t) is $0$ for t < -x so the integral from -1 to -x of f(x+t)$Q_n(t)$ will be 0. Is this clear?
            $endgroup$
            – Sorin Tirc
            Dec 16 '18 at 16:04












          • $begingroup$
            Yes, now I get it. Thank you.
            $endgroup$
            – Yutong Zhang
            Dec 17 '18 at 2:18


















          • $begingroup$
            Okay, now I see why that is a polynomial: apply binomial expansion multiple times, first on $(1-(t-x)^2)^n $ and on those $(t-x)^{2i}$, then the integral will become $$sum_{i=0}^{2n}c_nk(i)left(int_0^1f(t)cdot t^{2n-i}operatorname{d}tright)x^i$$ where $k(i)$ are the merged binomial coefficients.
            $endgroup$
            – Yutong Zhang
            Dec 16 '18 at 11:46












          • $begingroup$
            But I am still wondering about the $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt$ part, would you explain this part with a little bit more details? Thanks.
            $endgroup$
            – Yutong Zhang
            Dec 16 '18 at 11:50






          • 1




            $begingroup$
            Ok, so the integral has t varying from -1 to 1, but in fact f(x+t) is $0$ for t < -x so the integral from -1 to -x of f(x+t)$Q_n(t)$ will be 0. Is this clear?
            $endgroup$
            – Sorin Tirc
            Dec 16 '18 at 16:04












          • $begingroup$
            Yes, now I get it. Thank you.
            $endgroup$
            – Yutong Zhang
            Dec 17 '18 at 2:18
















          $begingroup$
          Okay, now I see why that is a polynomial: apply binomial expansion multiple times, first on $(1-(t-x)^2)^n $ and on those $(t-x)^{2i}$, then the integral will become $$sum_{i=0}^{2n}c_nk(i)left(int_0^1f(t)cdot t^{2n-i}operatorname{d}tright)x^i$$ where $k(i)$ are the merged binomial coefficients.
          $endgroup$
          – Yutong Zhang
          Dec 16 '18 at 11:46






          $begingroup$
          Okay, now I see why that is a polynomial: apply binomial expansion multiple times, first on $(1-(t-x)^2)^n $ and on those $(t-x)^{2i}$, then the integral will become $$sum_{i=0}^{2n}c_nk(i)left(int_0^1f(t)cdot t^{2n-i}operatorname{d}tright)x^i$$ where $k(i)$ are the merged binomial coefficients.
          $endgroup$
          – Yutong Zhang
          Dec 16 '18 at 11:46














          $begingroup$
          But I am still wondering about the $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt$ part, would you explain this part with a little bit more details? Thanks.
          $endgroup$
          – Yutong Zhang
          Dec 16 '18 at 11:50




          $begingroup$
          But I am still wondering about the $int_{-1}^{1}f(x+t)Q_n(t)dt = int_{-x}^{1-x}f(x+t)Q_n(t)dt$ part, would you explain this part with a little bit more details? Thanks.
          $endgroup$
          – Yutong Zhang
          Dec 16 '18 at 11:50




          1




          1




          $begingroup$
          Ok, so the integral has t varying from -1 to 1, but in fact f(x+t) is $0$ for t < -x so the integral from -1 to -x of f(x+t)$Q_n(t)$ will be 0. Is this clear?
          $endgroup$
          – Sorin Tirc
          Dec 16 '18 at 16:04






          $begingroup$
          Ok, so the integral has t varying from -1 to 1, but in fact f(x+t) is $0$ for t < -x so the integral from -1 to -x of f(x+t)$Q_n(t)$ will be 0. Is this clear?
          $endgroup$
          – Sorin Tirc
          Dec 16 '18 at 16:04














          $begingroup$
          Yes, now I get it. Thank you.
          $endgroup$
          – Yutong Zhang
          Dec 17 '18 at 2:18




          $begingroup$
          Yes, now I get it. Thank you.
          $endgroup$
          – Yutong Zhang
          Dec 17 '18 at 2:18


















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