Wednesday, 20 April 2016

real analysis - Proving an inequality using the mean value theorem



I am trying to show that



|f(x)f(y)|<|xy|,



for the function f to be defined as f:[0,+)[0,+), f(x)=(1+x2)1/2, using the mean value theorem.




I have done this:



Since f is differentiable on [0,+), then there is a point x0, $x

f(x)f(y)=(xy)f(x0),



by the mean value theorem. Hence,



|f(x)f(y)|=|xy||f(x0)|=|xy||x0(1+x02)1/2||xy||x0||xy|M<|xy|




where M is a constant.



Can someone tell me if this is correct?


Answer



Hint:




  • First prove the general result: if f:RR is a differentiable function and |f(x)|<M for all xR then for all x,yR the inequality |f(x)f(y)|<M|xy|

    holds. The proof is very similar to what you have done in the question.


  • Next prove that if f(x)=1+x2 then |f(x)|<1. To do this consider f(x)2=x21+x2.



  • Combinding the two results above gives the desired result.



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