# Does $f_n(x)=e^{\frac{-x}{n}}$ converge uniformly to $1$ on $[0,\infty)$?

Given $f_n:[0,\infty)\rightarrow \mathbb R$ with $$f_n(x)=e^{\frac{-x}{n}},$$ it's easy to see that this sequence converges pointwise to $f(x)=1$.

However I am sure that if convergence is uniform also. How can I approach that?

• @Jack What was wrong with your answer? – Sahiba Arora Jun 26 '17 at 15:36
• @SahibaArora: My reasoning was wrong. Uniform convergence implies $\lim_n\int_{[0,\infty)}f_n=\int_{[0,\infty)}f$. But in this case, one can also interchange the limit and the integral. – Jack Jun 26 '17 at 15:49
• @Jack Thank you for explaining that. – Sahiba Arora Jun 26 '17 at 16:12

The sequence $(f_n)_n$ is not uniformly convergent in $[0,+\infty)$ because $$\sup_{x\geq 0}|f_n(x)-f(x)|=\sup_{x\geq 0}|e^{-x/n}-1|=\sup_{t\geq 0}(1-e^{-t})=1\not \to 0$$ where we set $t=x/n$. However, $(f_n)_n$ is uniformly convergent in $[0,a]$ in $a\geq 0$: $$\sup_{x\in [0,a]}|f_n(x)-f(x)|=\sup_{x\in [0,a]}|e^{-x/n}-1|=\sup_{t\in [0,a/n]}(1-e^{-t})=1-e^{-a/n}\to 0$$ as $n$ goes to infinity.
$$\lim_{n\to+\infty}\sup_{x\in\mathbb R}|f_n (x)-1|\ge$$
$$\lim_{n\to +\infty}|f_n (n)-1|=1-\frac {1}{e}\ne 0$$
If the convergence were uniform, there would be an index $N$ with the property that $n \ge N$ implies $|e^{-x/n} - 1| < \frac 12$ for all $x \in [0,\infty)$. This is turn would lead to $e^{-x/N} > \frac 12$ for all $x \in [0,\infty)$. This will lead to difficulty for sufficiently large $x$.