Let $f : \mathbb R \to \mathbb R$ a uniformly continuous function such that $\forall x > 0$ $$\lim_{n \to +\infty}f(nx) = 0$$Show that $$\lim_{x \to +\infty} f(x)= 0$$ I cannot use Baire Category theorem (I haven't studied it yet).
I have an attempt but something seems wrong with the quantifiers or with steps' logic :
- $f$ is uniformly continuous : $\forall \epsilon > 0, \exists \delta > 0 \text{ such that } \lvert x-y \rvert \leq \delta \implies \lvert f(x)-f(y) \rvert \leq \epsilon \quad{(1)}$
- $\forall x > 0, \lim_{n \to +\infty}f(nx) = 0$ : $\forall x > 0, \forall \epsilon > 0, \exists n_0 \text{ such that } \forall n > n_0 \implies \lvert f(nx) \rvert \leq \epsilon \quad{(2)}$
By (1), we have that $\exists \delta > 0 \text{ such that } \lvert x-y \rvert \leq \delta \implies \lvert f(x)-f(y) \rvert \leq \frac{\epsilon}{2}$. Let $\delta_1 = \min(1,\delta)$.
By (2), for $x=1$, $\exists n_1 \in \mathbb N, \text{ such that } \forall n > n_1, \lvert f(n) \rvert \leq \frac{\epsilon}{2}$
We want to prove $\lim_{x \to +\infty} f(x)= 0 \iff \forall \epsilon > 0, \exists x_1 > 0, \text{ such that } \forall x > x_1, \lvert f(x) \rvert \leq \epsilon$. Our goal is to find $x_1$.
By taking $x_1 > n_1$, then $\forall x > x_1, \lvert f(x) \rvert = \lvert f(x) - f(\left \lfloor{x}\right \rfloor) + f(\left \lfloor{x}\right \rfloor) \rvert \leq \lvert f(x) - f(\left \lfloor{x}\right \rfloor) \rvert + \lvert f(\left \lfloor{x}\right \rfloor) \rvert \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$
Because $\lvert x - \left \lfloor{x}\right \rfloor \rvert < 1 \leq \delta_1$ and because $\left \lfloor{x}\right \rfloor \geq \left \lfloor{x_1}\right \rfloor > n_1$.
What do you think about this proof ?