Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.












0












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Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.




I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this



I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.



Where i have;
$(x,y) = (y,x)$ for all $x,y in V$



and



$(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector



I assumed it would simply be plugging in but im not entirely sure.






linear-algebra vector-spaces inner-product-space







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.




    I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this



    I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.



    Where i have;
    $(x,y) = (y,x)$ for all $x,y in V$



    and



    $(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector



    I assumed it would simply be plugging in but im not entirely sure.






    linear-algebra vector-spaces inner-product-space







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.




      I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this



      I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.



      Where i have;
      $(x,y) = (y,x)$ for all $x,y in V$



      and



      $(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector



      I assumed it would simply be plugging in but im not entirely sure.






      linear-algebra vector-spaces inner-product-space







      share|cite|improve this question











      $endgroup$





      Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.




      I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this



      I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.



      Where i have;
      $(x,y) = (y,x)$ for all $x,y in V$



      and



      $(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector



      I assumed it would simply be plugging in but im not entirely sure.






      linear-algebra vector-spaces inner-product-space




      linear-algebra vector-spaces inner-product-space






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 14:14









      Christoph

      11.9k1642




      11.9k1642










      asked Jan 7 at 13:18









      L GL G

      248




      248






















          1 Answer
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          $begingroup$

          We have
          $$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
          And we have
          $$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
          for every f by $(f(t))^2geq 0$ and the monotony of the integral.
          We now have to show that $(f,f)=0$ implies $f=0$.
          It is standard lemma that for a continuous function $ggeq 0$ we have
          $$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
          We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
          $$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
          If you don't know the lemma, I can give you a prove for it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
            $endgroup$
            – L G
            Jan 7 at 13:26






          • 1




            $begingroup$
            +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
            $endgroup$
            – Theo Bendit
            Jan 7 at 14:02













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          1 Answer
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          active

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          $begingroup$

          We have
          $$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
          And we have
          $$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
          for every f by $(f(t))^2geq 0$ and the monotony of the integral.
          We now have to show that $(f,f)=0$ implies $f=0$.
          It is standard lemma that for a continuous function $ggeq 0$ we have
          $$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
          We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
          $$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
          If you don't know the lemma, I can give you a prove for it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
            $endgroup$
            – L G
            Jan 7 at 13:26






          • 1




            $begingroup$
            +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
            $endgroup$
            – Theo Bendit
            Jan 7 at 14:02


















          2












          $begingroup$

          We have
          $$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
          And we have
          $$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
          for every f by $(f(t))^2geq 0$ and the monotony of the integral.
          We now have to show that $(f,f)=0$ implies $f=0$.
          It is standard lemma that for a continuous function $ggeq 0$ we have
          $$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
          We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
          $$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
          If you don't know the lemma, I can give you a prove for it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
            $endgroup$
            – L G
            Jan 7 at 13:26






          • 1




            $begingroup$
            +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
            $endgroup$
            – Theo Bendit
            Jan 7 at 14:02
















          2












          2








          2





          $begingroup$

          We have
          $$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
          And we have
          $$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
          for every f by $(f(t))^2geq 0$ and the monotony of the integral.
          We now have to show that $(f,f)=0$ implies $f=0$.
          It is standard lemma that for a continuous function $ggeq 0$ we have
          $$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
          We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
          $$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
          If you don't know the lemma, I can give you a prove for it.






          share|cite|improve this answer











          $endgroup$



          We have
          $$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
          And we have
          $$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
          for every f by $(f(t))^2geq 0$ and the monotony of the integral.
          We now have to show that $(f,f)=0$ implies $f=0$.
          It is standard lemma that for a continuous function $ggeq 0$ we have
          $$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
          We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
          $$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
          If you don't know the lemma, I can give you a prove for it.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 7 at 13:48

























          answered Jan 7 at 13:21









          Student7Student7

          2089




          2089












          • $begingroup$
            ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
            $endgroup$
            – L G
            Jan 7 at 13:26






          • 1




            $begingroup$
            +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
            $endgroup$
            – Theo Bendit
            Jan 7 at 14:02




















          • $begingroup$
            ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
            $endgroup$
            – L G
            Jan 7 at 13:26






          • 1




            $begingroup$
            +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
            $endgroup$
            – Theo Bendit
            Jan 7 at 14:02


















          $begingroup$
          ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
          $endgroup$
          – L G
          Jan 7 at 13:26




          $begingroup$
          ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
          $endgroup$
          – L G
          Jan 7 at 13:26




          1




          1




          $begingroup$
          +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
          $endgroup$
          – Theo Bendit
          Jan 7 at 14:02






          $begingroup$
          +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
          $endgroup$
          – Theo Bendit
          Jan 7 at 14:02




















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