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I am trying to show that

$|f(x)-f(y)|<|x-y|$,

for the function $f$ to be defined as $f:[0,+\infty)\mapsto [0,+\infty)$, $f(x)=(1+x^2)^{1/2}$, using the mean value theorem.

I have done this:

Since $f$ is differentiable on $[0,+\infty)$, then there is a point $x_0$, $x<x_0<y$, such that

$f(x)-f(y)=(x-y)f'(x_0)$,

by the mean value theorem. Hence,

$|f(x)-f(y)|=|x-y||f'(x_0)|=|x-y||x_0 (1+{x_0} ^2)^{-1/2}|\leq|x-y||x_0|\leq|x-y|M<|x-y|$

where M is a constant.

Can someone tell me if this is correct?

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  • $\begingroup$ You need a better estimate in the last line to prove that $|f'(x_0)| < 1$. This follows from $x^2 \leq 1+x^2 \to \frac{|x|}{\sqrt{1+x^2}} < 1$ $\endgroup$ – Winther Mar 12 '15 at 0:05
  • $\begingroup$ I used the fact that ${x_0}^2 \geq 0 \Rightarrow 1+{x_0}^2 \geq 1 \Rightarrow (1+{x_0}^2)^{-1/2} < 1 \Rightarrow x_0 (1+{x_0}^2)^{-1/2} < x_0 \Rightarrow f'(x_0)< x_0 $ $\endgroup$ – P.D. Mar 12 '15 at 0:12
  • $\begingroup$ That is indeed true, but it's not a sharp enough inequality to prove the result you want: $|f(x)-f(y)| < |x-y|$. You need to prove $|f'(x_0)| < 1$. $\endgroup$ – Winther Mar 12 '15 at 0:13
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Hint:

  • First prove the general result: if $f:\mathbb{R}\to \mathbb{R}$ is a differentiable function and $|f'(x)| < M$ for all $x\in\mathbb{R}$ then for all $x,y\in\mathbb{R}$ the inequality $$|f(x)-f(y)| < M|x-y|$$ holds. The proof is very similar to what you have done in the question.

  • Next prove that if $f(x) = \sqrt{1+x^2}$ then $|f'(x)| < 1$. To do this consider $f'(x)^2 = \frac{x^2}{1+x^2}$.

  • Combinding the two results above gives the desired result.

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  • $\begingroup$ I understand the whole approach, the thing that I have stack is why $|f'(x_0)|<1$, cause I get that $f'(x_0)<x_0$, as I have said above. $\endgroup$ – P.D. Mar 12 '15 at 0:33
  • $\begingroup$ @P.D. You know that $f'(x) = \frac{x}{\sqrt{1+x^2}}$ so $f'(x)^2 = \frac{x^2}{1+x^2}$. Do you see that this is less than $1$? BTW there is no contradiction between the two estimates. We can (and do) have both $|f'(x)|<1$ and $|f'(x)| < |x|$. The only difference is that the former is much stronger than the latter for $x>1$ (try to plot the functions). $\endgroup$ – Winther Mar 12 '15 at 0:34
  • $\begingroup$ Ohh yes now I understand!! Thank you very much!! :) $\endgroup$ – P.D. Mar 12 '15 at 0:42
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hint:$f'(x) = \dfrac{1}{2}\cdot \dfrac{1}{\sqrt{1+x^2}}\cdot 2x = \dfrac{x}{\sqrt{1+x^2}} < 1 = M$.

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  • $\begingroup$ Can you give me an explicit explanation of why the last part is $<1$? $\endgroup$ – P.D. Mar 12 '15 at 0:18

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