Given f such that $ f(0)=0, \lim_{x\to \infty} f(x) = 1$, is $f_n(x)=f(x+e^n)$ uniformly convergent?

Question

Let $f:\Re \rightarrow \Re$ be a bounded function with the properties $ f(0)=0, \lim_{x\to \infty} f(x) = 1$ is $f_n(x)=f(x+e^n)$ uniformly convergent? Not really sure how to go about applying the definitions of pointwise and uniform convergence in this case. Any help would be much appreciated. Thanks in advance.

Answer 1

The sequence converges point-wisely to $g(x) = 1$ forall $x$.

To find out if this convergence is also uniform, then you must consider the following sequence:

$$s_n = \sup_{x \in \mathbb{R}} \left|f_n(x)-g(x)\right| = \sup_{x \in \mathbb{R}} \left|f(x+e^{n})-1\right|.$$

Notice that this $\sup$ is always $1$ for every $n$ and it is attained at $x = -e^{n}$. Hence:

$$s_n = 1 ~\forall n.$$

Therefore, since $s_n \not \to 0$, then the convergence is not uniform.

Answer 2

No. $(f_n)$ converges pointweise to the constant function $f(x)=1$. But

$|f_n(-e^n)-1|=1$ for all $n$.