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Show that $f(x)=\dfrac{1}{x^2+1}$ is Lipschitz on interval $[-2,2]$, $L=$?

My work:

$$f'(x)=\frac{-2x}{(x^2+1)^2}$$

$$|f'(x)|=\left| \frac{-2x}{(x^2+1)^2} \right|=\frac{2|x|}{(x^2+1)^2}$$

Is $L=\dfrac{1}{2}$? Can I say that also $L$ can be any number bigger than $\dfrac{1}{2}$?

Ex: In this case $L=2$

Because the definition says that $|f(x)-f(y)| \le L| x - y |$ or $L$ needs to be the smallest possible? (Sorry for bad English)

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    What is $\max \lvert f' \rvert$? That should help you determine $L$. And $L$ can be any finite number, though if $L < 1$ you also have a contraction. – 2017-01-29
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    $max|f'|$ is 1/2 so L can't be smaller than that but can L be bigger? – 2017-01-29

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Since $ f $ is even, the derivative on the positive and negative parts will only differ by a sign, as you stated. So you can just consider the maximum of $ f' $ on $ [0, 2] $ and take the absolute value.

$$ f' = \frac{-2x}{(x^2+1)^2} $$

$$ f'' = \frac{6x^2 - 2}{(x^2+1)^3} $$

Find where this equals 0 on [0, 2] to get the point where the absolute value of the derivative is at a maximum. $ 6x^2 - 2 = 0 $ when $ x = \frac{1}{\sqrt{3}} $ (the positive solution, though it doesn't really matter if you take the other solution because you will take the absolute value).

Using this, the maximum of $ | f' | $ is $ \frac{3 \sqrt{3}}{8} $.

Any secant line on this domain will have absolute value of slope less than or equal to this value, so this is the minimum Lipschitz constant for this function. $ L = \frac{1}{2} $ doesn't work because this maximum derivative is greater than $ \frac{1}{2} $. Any value of $ L $ which is greater than $ \frac{3 \sqrt{3}}{8} $ will still satisfy the inequality, but $ L = \frac{3 \sqrt{3}}{8} $ is the best possible bound.

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    Specifically, L cannot be 1/2 because for example x = 0.5, y = 0.55 will cause | f(x) - f(y) | <= 1/2 | x - y | to be false. – 2017-01-29
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    Thank you very much this is what i was looking for. – 2017-01-29