Suppose $f:\mathbb R\to \mathbb R$ is differentiable, with $f'$ bounded on $\mathbb R.$ I want to show there exist $C,D$ such that $|f(x)|\leq C|x|+D$ for all $x\in \mathbb R.$
I don't know how to get $C,D$ that works for all $x\in \mathbb R$. I can do it when $f^{\prime}$ is integrable on $[a, x]$, but I realized derivative need not be integrable
This follows from MVT.
Let $M$ be the bound of $f'$. Then on $[0,1]$, for any $x,y \in [0,1]$, there exists a $\zeta \in (0,1)$ such that \begin{align} |f(x)-f(y)| = |x-y||f'(\zeta)| \leq M|x-y| \end{align} In particular as this holds for any $x,y$, we take $y=0$, then we have \begin{align} |f(x)-f(0)| \leq M|x| \Rightarrow -Mx \leq f(x)-f(0) \leq Mx \Rightarrow |f(x)| \leq M|x| + f(0) \end{align} Let $C$ = $M$, $D=f(0)$. Then we have \begin{align} |f(x)| \leq C|x| + D \end{align} In particular as $f$ is continuous and differentiable and $f'$ is bounded on all of $ \mathbb{R}$, we see by MVT this inequality holds for any interval $ x \in [a,b] \subset \mathbb{R}$ where $0$ is contained in the interval. So given an arbitrary $x \in \mathbb{R}$, consider the interval $[-x,x]$, then we obtain the above inequality. Hence, it holds for any $x \in \mathbb{R}$