Coercive bilinear form for maximum norm











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Let $f$ be a differentiable function. Denote a bilinear form by $$b(f,f) = int_{0}^{1} bigg( frac{d f(x)}{dx} bigg)^{2} dx.$$ Given $f(0) = 0,$ we want to show that $$a cdot b(f,f) geq ||f||_{infty}^{2},$$ where $||f||_{infty} = max_{0 leq x leq 1} |f(x)|.$ Note that $a$ is a constant independent of $f$.



In attempting the proof (which doesn't get far), I express $$||f||_{infty}^{2} = max_{0 leq x leq 1} |f(x)| cdot |f(x)|.$$ However, I am not sure of a bound for this quantity that will lead naturally to the desired proof. Also, this might not be the first step in the correct proof.



Note: By Holder's inequality we have $$bigg( int_{0}^{1} frac{d f (x)}{dx} dx bigg)^{2} leq int_{0}^{1} bigg( frac{d f(x)}{dx} bigg)^{2} dx.$$ The left hand side equals $(f(1) - f(0))^{2},$ which simplifies to $(f(1))^{2}$ since $f(0) = 0.$










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    Let $f$ be a differentiable function. Denote a bilinear form by $$b(f,f) = int_{0}^{1} bigg( frac{d f(x)}{dx} bigg)^{2} dx.$$ Given $f(0) = 0,$ we want to show that $$a cdot b(f,f) geq ||f||_{infty}^{2},$$ where $||f||_{infty} = max_{0 leq x leq 1} |f(x)|.$ Note that $a$ is a constant independent of $f$.



    In attempting the proof (which doesn't get far), I express $$||f||_{infty}^{2} = max_{0 leq x leq 1} |f(x)| cdot |f(x)|.$$ However, I am not sure of a bound for this quantity that will lead naturally to the desired proof. Also, this might not be the first step in the correct proof.



    Note: By Holder's inequality we have $$bigg( int_{0}^{1} frac{d f (x)}{dx} dx bigg)^{2} leq int_{0}^{1} bigg( frac{d f(x)}{dx} bigg)^{2} dx.$$ The left hand side equals $(f(1) - f(0))^{2},$ which simplifies to $(f(1))^{2}$ since $f(0) = 0.$










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      Let $f$ be a differentiable function. Denote a bilinear form by $$b(f,f) = int_{0}^{1} bigg( frac{d f(x)}{dx} bigg)^{2} dx.$$ Given $f(0) = 0,$ we want to show that $$a cdot b(f,f) geq ||f||_{infty}^{2},$$ where $||f||_{infty} = max_{0 leq x leq 1} |f(x)|.$ Note that $a$ is a constant independent of $f$.



      In attempting the proof (which doesn't get far), I express $$||f||_{infty}^{2} = max_{0 leq x leq 1} |f(x)| cdot |f(x)|.$$ However, I am not sure of a bound for this quantity that will lead naturally to the desired proof. Also, this might not be the first step in the correct proof.



      Note: By Holder's inequality we have $$bigg( int_{0}^{1} frac{d f (x)}{dx} dx bigg)^{2} leq int_{0}^{1} bigg( frac{d f(x)}{dx} bigg)^{2} dx.$$ The left hand side equals $(f(1) - f(0))^{2},$ which simplifies to $(f(1))^{2}$ since $f(0) = 0.$










      share|cite|improve this question















      Let $f$ be a differentiable function. Denote a bilinear form by $$b(f,f) = int_{0}^{1} bigg( frac{d f(x)}{dx} bigg)^{2} dx.$$ Given $f(0) = 0,$ we want to show that $$a cdot b(f,f) geq ||f||_{infty}^{2},$$ where $||f||_{infty} = max_{0 leq x leq 1} |f(x)|.$ Note that $a$ is a constant independent of $f$.



      In attempting the proof (which doesn't get far), I express $$||f||_{infty}^{2} = max_{0 leq x leq 1} |f(x)| cdot |f(x)|.$$ However, I am not sure of a bound for this quantity that will lead naturally to the desired proof. Also, this might not be the first step in the correct proof.



      Note: By Holder's inequality we have $$bigg( int_{0}^{1} frac{d f (x)}{dx} dx bigg)^{2} leq int_{0}^{1} bigg( frac{d f(x)}{dx} bigg)^{2} dx.$$ The left hand side equals $(f(1) - f(0))^{2},$ which simplifies to $(f(1))^{2}$ since $f(0) = 0.$







      sobolev-spaces normed-spaces lp-spaces upper-lower-bounds






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          Observe
          begin{align}
          |f(x)| = left|int^x_0 f'(t) dtright| leq int^1_0|f'(t)| dt leq left(int^1_0|f'(t)|^2 dt right)^{1/2}.
          end{align}

          Since this holds for all $x in [0, 1]$ then you are done.






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            1 Answer
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            active

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            up vote
            1
            down vote



            accepted










            Observe
            begin{align}
            |f(x)| = left|int^x_0 f'(t) dtright| leq int^1_0|f'(t)| dt leq left(int^1_0|f'(t)|^2 dt right)^{1/2}.
            end{align}

            Since this holds for all $x in [0, 1]$ then you are done.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Observe
              begin{align}
              |f(x)| = left|int^x_0 f'(t) dtright| leq int^1_0|f'(t)| dt leq left(int^1_0|f'(t)|^2 dt right)^{1/2}.
              end{align}

              Since this holds for all $x in [0, 1]$ then you are done.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Observe
                begin{align}
                |f(x)| = left|int^x_0 f'(t) dtright| leq int^1_0|f'(t)| dt leq left(int^1_0|f'(t)|^2 dt right)^{1/2}.
                end{align}

                Since this holds for all $x in [0, 1]$ then you are done.






                share|cite|improve this answer












                Observe
                begin{align}
                |f(x)| = left|int^x_0 f'(t) dtright| leq int^1_0|f'(t)| dt leq left(int^1_0|f'(t)|^2 dt right)^{1/2}.
                end{align}

                Since this holds for all $x in [0, 1]$ then you are done.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Jacky Chong

                16.7k21027




                16.7k21027






























                     

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