Is $f$ Lipschitz?

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Suppose $f:mathbbR^2tomathbbR^2$ is differentiable with uniformly bounded partial derivatives $$left|dfracpartial f_ipartial x_j(x_1,x_2)right|leq 1$$ for all $(x_1,x_2)inmathbbR^2$ and for $i,j=1,2$. Show that there is a constant $L$ such that $left|f(y)-f(x)
right|leq Lleft|y-xright|$ for all $x,yinmathbbR^2.$



My idea is to use the Mean Value Theorem and I think that $L$ should be taken as the norm of the Jacobian but I don't know how to proceed.




real-analysis multivariable-calculus jacobian




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    Suppose $f:mathbbR^2tomathbbR^2$ is differentiable with uniformly bounded partial derivatives $$left|dfracpartial f_ipartial x_j(x_1,x_2)right|leq 1$$ for all $(x_1,x_2)inmathbbR^2$ and for $i,j=1,2$. Show that there is a constant $L$ such that $left|f(y)-f(x)
    right|leq Lleft|y-xright|$ for all $x,yinmathbbR^2.$



    My idea is to use the Mean Value Theorem and I think that $L$ should be taken as the norm of the Jacobian but I don't know how to proceed.




    real-analysis multivariable-calculus jacobian




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Suppose $f:mathbbR^2tomathbbR^2$ is differentiable with uniformly bounded partial derivatives $$left|dfracpartial f_ipartial x_j(x_1,x_2)right|leq 1$$ for all $(x_1,x_2)inmathbbR^2$ and for $i,j=1,2$. Show that there is a constant $L$ such that $left|f(y)-f(x)
      right|leq Lleft|y-xright|$ for all $x,yinmathbbR^2.$



      My idea is to use the Mean Value Theorem and I think that $L$ should be taken as the norm of the Jacobian but I don't know how to proceed.




      real-analysis multivariable-calculus jacobian




      share|cite|improve this question











      Suppose $f:mathbbR^2tomathbbR^2$ is differentiable with uniformly bounded partial derivatives $$left|dfracpartial f_ipartial x_j(x_1,x_2)right|leq 1$$ for all $(x_1,x_2)inmathbbR^2$ and for $i,j=1,2$. Show that there is a constant $L$ such that $left|f(y)-f(x)
      right|leq Lleft|y-xright|$ for all $x,yinmathbbR^2.$



      My idea is to use the Mean Value Theorem and I think that $L$ should be taken as the norm of the Jacobian but I don't know how to proceed.





      real-analysis multivariable-calculus jacobian





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      Habagat Maliksi

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          Write $x = (x_1,x_2), y = (y_1,y_2)$. Then $|f_i(x_1,x_2) - f_i(y_1,x_2)| le |x_1-y_1|$ and $|f_i(y_1,x_2)-f_i(y_1,y_2)| le |x_2-y_2|$ for $i=1,2$ by MVT. So



          $$||f(x)-f(y)|| le ||f(x_1,x_2)-f(y_1,x_2)||+||f(y_1,x_2)-f(y_1,y_2)||$$ $$ le sqrt2+sqrt2 le 2,$$ where this last inequality used that $a+b le sqrt2(a^2+b^2).$






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            Write $x = (x_1,x_2), y = (y_1,y_2)$. Then $|f_i(x_1,x_2) - f_i(y_1,x_2)| le |x_1-y_1|$ and $|f_i(y_1,x_2)-f_i(y_1,y_2)| le |x_2-y_2|$ for $i=1,2$ by MVT. So



            $$||f(x)-f(y)|| le ||f(x_1,x_2)-f(y_1,x_2)||+||f(y_1,x_2)-f(y_1,y_2)||$$ $$ le sqrt2+sqrt2 le 2,$$ where this last inequality used that $a+b le sqrt2(a^2+b^2).$






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              Write $x = (x_1,x_2), y = (y_1,y_2)$. Then $|f_i(x_1,x_2) - f_i(y_1,x_2)| le |x_1-y_1|$ and $|f_i(y_1,x_2)-f_i(y_1,y_2)| le |x_2-y_2|$ for $i=1,2$ by MVT. So



              $$||f(x)-f(y)|| le ||f(x_1,x_2)-f(y_1,x_2)||+||f(y_1,x_2)-f(y_1,y_2)||$$ $$ le sqrt2+sqrt2 le 2,$$ where this last inequality used that $a+b le sqrt2(a^2+b^2).$






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                Write $x = (x_1,x_2), y = (y_1,y_2)$. Then $|f_i(x_1,x_2) - f_i(y_1,x_2)| le |x_1-y_1|$ and $|f_i(y_1,x_2)-f_i(y_1,y_2)| le |x_2-y_2|$ for $i=1,2$ by MVT. So



                $$||f(x)-f(y)|| le ||f(x_1,x_2)-f(y_1,x_2)||+||f(y_1,x_2)-f(y_1,y_2)||$$ $$ le sqrt2+sqrt2 le 2,$$ where this last inequality used that $a+b le sqrt2(a^2+b^2).$






                share|cite|improve this answer















                Write $x = (x_1,x_2), y = (y_1,y_2)$. Then $|f_i(x_1,x_2) - f_i(y_1,x_2)| le |x_1-y_1|$ and $|f_i(y_1,x_2)-f_i(y_1,y_2)| le |x_2-y_2|$ for $i=1,2$ by MVT. So



                $$||f(x)-f(y)|| le ||f(x_1,x_2)-f(y_1,x_2)||+||f(y_1,x_2)-f(y_1,y_2)||$$ $$ le sqrt2+sqrt2 le 2,$$ where this last inequality used that $a+b le sqrt2(a^2+b^2).$







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                mathworker21

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