Inequality about functions

Question

This inequality is on the book "Semilinear Elliptic Equations for Beginners", and is on the page 42.

$f: \mathbb{R} \rightarrow \mathbb{R}$ is continuos and there exists $\sigma \in (0,1)$ and $a,b >0$ such that $$|f(t)| \leq a + b|t|^{\sigma}, \\ \forall t \in \mathbb{R}$$

Thus $f$ is no longer bounded, but is allowed to grow sublinearly $(\sigma <1)$. It follows that $F$ grows at most subquadratically, in the sense that for some $a_1, b_1 > 0$, $$|F(t)| \leq a_1 + b_1|t|^{\sigma +1} \\ \forall t\in \mathbb{R}$$

Here, $F(t) = \int_{0}^{t} f(s)ds$.

My point is, together with $a_1$ why don't appears a $|t|$? Since, in my point of view, the second inequality is obtained by the integration of the first?

Thanks for any help!

Answer 1

Let's take $t\geq0$, the same holds for $t\leq0$.

For $t\leq1$ \begin{align} at+\frac{b}{\sigma+1}t^{\sigma+1} &\leq a+\frac{b}{\sigma+1}t^{\sigma+1}\\ &\leq a+\left(a+\frac{b}{\sigma+1}\right)t^{\sigma+1} \end{align} while, for $t\geq1$ \begin{align} at+\frac{b}{\sigma+1}t^{\sigma+1} &\leq at^{\sigma+1}+\frac{b}{\sigma+1}t^{\sigma+1} \\ &=\left(a+\frac{b}{\sigma+1}\right)t^{\sigma+1}\\ &\leq a+\left(a+\frac{b}{\sigma+1}\right)t^{\sigma+1} \end{align} so if we take \begin{align} a_1 &= a\\ b_1 &= a+\frac{b}{\sigma+1} \end{align} the inequality is satisfied.

Answer 2

It is enough to verify that, for large enough $a_1$ and $b_1$, $$at+(b/(\sigma+1))t^{\sigma+1}\le a_1+b_1 t^{\sigma+1}$$ for all large $t$.