Bound for function with constant /periodic second derivative

Question

Consider a function $f : \mathbb{R}\to\mathbb{R}$ with $f''$ continuous and $f''(x)=f''(x+1)$ for all real numbers x. I need to show that there exists a real positive number $c$ such that $f(x)\leq c(1+x^2)$ for all real numbers $x$. I feel like this might come from Taylor expansion of the function but any help would be much appreciated.

Answer 1

$$f(x)=\left(\int_0^xf'(y)dy\right)+f(0)=\left(\int_0^x\left(\left(\int_0^y f''(z)dz\right)+f'(0)\right)dy\right)+f(0)$$ If $f''$ is periodic and continuous, then it is bounded hence $|f''(x)|\leq b$ for a $b\geq 0$ and we get $$f(x)\leq|f(x)|\leq \left(\int_0^x\left(\left(\int_0^y b \ dz\right)+|f'(0)|\right)dy\right)+|f(0)|\\= \left(\int_0^x by+|f'(0)|dy\right)+|f(0)|= \frac{b}{2}x^2+|f'(0)|x+|f(0)|$$ Define $c:= \max(b/2,|f(0)|)+|f'(0)|$ and we get $$f(x)\leq c(1+x^2).$$