Function with $|f(x)-\int^{\delta}_{-\delta}f(x+u)du|<\epsilon$

Question

I am looking for a function $f:\mathbb{R}\to \mathbb{R}$ and $\epsilon>0$ such that there is no $\delta>0$, for him any $x\in\mathbb{R}$: $|f(x)-\int^{\delta}_{-\delta}f(x+u)du|<\epsilon$

Clearly, $f$ must be unbounded.

Answer 1

Let $f(x)=x^2$ and $\varepsilon=1/12$. For any $\delta>0$ and any $x\in\mathbb{R}$, $$|f(x)-\int_{-\delta}^\delta f(x+u)du|=|x^2-\int_{-\delta}^\delta (x^2+2xu +u^2)du|=|x^2(1-2\delta)-(2\delta^3/3)|$$ If $\delta=1/2$, let $x=x(0)=0$. Then $|f(x)-\int_{-\delta}^\delta f(x+u)du|=1/12=\varepsilon$. If $\delta\neq 1/2$, then let $x=x(\delta)\in\mathbb{R}$ such that $x^2|1-2\delta|-2|\delta|^3/3\geq 1/12=\varepsilon$. (you can choose $x(\delta)=\dfrac{\sqrt{1/12+2|\delta|^3/3}}{\sqrt{|1-2\delta|}})$

Therefore, $f$ and $\varepsilon>0$ are such that for every $\delta>0$ there exists $x=x(\delta)\in\mathbb{R}$ satisfying $|f(x)-\int_{-\delta}^\delta f(x+u)du|\geq \varepsilon$