How do I solve the following problem concerning primitive functions?

Question

I need some help for the following problem:

Let $F:\mathbb{R}\to\mathbb{R}$ be a primitive function for $f:\mathbb{R}\to\mathbb{R}$, where $f(x)=\dfrac{1}{x^4+2}$.
Prove that:

a)$F$ is injective,

b)$|F(x_1)-F(x_2)|\leq|x_1-x_2|$, $x_1,x_2>0$.

I managed to solve "a)" but I do not have any idea how to solve "b)". Can anyone help me?

Answer 1

Thanks to mean value theorem

$F(x_2)-F(x_1)=F'(c)(x_2-x_1)=f(c)(x_2-x_1)$ for some $c\in(x_1,x_2)$

therefore

$|F(x_2)-F(x_1)|=|f(c)||x_2-x_1|\leq |x_2-x_1|$

because $f(x)\leq 1, \quad\forall x \in\mathbb{R}$

Hope this helps

Answer 2

I'm not sure how this fits in with "haven't learned integration yet," but:

$|F(x_1) - F(x_2)|$ is the area under the curve $y=f(x)$ above the $x$-axis and between $x_1$ and $x_2$. That area has to be less than the width of the interval (which is $|x_1-x_2|$) times the maximum value of $f(x)$ (which is $1/2$.)

Answer 3

Since $f(x)$ is a derivative of $F(x)$, we can say that $F$ is differentiable everywhere in the domain. Think of the problem of b) as that of finding a suitable Lipschitz coeffecient for the given function $F$. The largest possible absolute value of $f(x)$ is $\frac{1}{2}$ for $x = 0$. Thus, by the definition of differentiation, $$sup\lim_{x_2 \to 0}\frac{|F(x_1+ x_2) - F(x_1)|}{|x_1 - x_2|} = \frac{1}{2}$$ note that the supremum above is taken over all $x_1\in \mathbb{R}$. This means that $|f(x)|$ is smaller than or equal to $g(x) = \frac{1}{2}$. Now, suppose that $x_1$ and $x_2$ are arbitrarily chosen, and without loss of generality, $x_1>x_2$. Then $$\begin{align}|F(x_1)-F(x_2)|&=\left|\int_{x_2}^{x_1} f(x) dx\right|\\ &\leq \int_{x_2}^{x_1}|f(x)|dx\\ &\leq \int_{x_2}^{x_1} g(x) dx \quad (\because |f(x)|\leq g(x))\\ &=\int_{x_2}^{x_1} \frac{1}{2}dx\\ &=\frac{1}{2}|x_1-x_2|\\ &\leq\frac{1}{2}\left(|x_1+x_2|+|x_1 + x_2|\right)\\ &\leq|x_1+x_2| \end{align}$$ Now, we have the desired result $|F(x_1)-F(x_2)| \leq |x_1+x_2|$ for all $x_1,x_2 \in \mathbb{R}$ since the choice of $x_1$ and $x_2$ was arbitrary.