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The Problem:

Determine whether the sequence $\{ f_n \}$ defined by $f_n(x) = e^{-nx}$ converges uniformly on the following intervals:

$(a) \hspace{3mm} A = [0, \infty)$,

$(b) \hspace{3mm} B = (0, \infty)$,

$(c) \hspace{3mm} C = [1, \infty)$.

What I Have:

For $(a)$, we have that $f_n(0) = 1$, for all $n$. So $f_n(x) \to 1$, when $x=0$. But $f_n(x) \to 0$, when $x>1$. Thus, $f_n$ does not even converge pointwise (right?), so it can't converge uniformly.

For $(b)$ (this is the one I'm having trouble with), we have that $f_n(x) \to 0$, for all $x \in B$. So, we consider the behavior of the sequence $\{ M_n \}$ defined by $$ M_n = \sup_{x \in B} |e^{-nx} - 0| = \sup_{x \in B} e^{-nx}. $$ That is, if $M_n \to 0$ as $n \to \infty$, then $f_n \to 0$, uniformly. Well, since $e^{-nx} > e^{-mx}$ if $n < m$, it seems to me that it should be also true that $\sup(e^{-nx}) > \sup(e^{-mx})$; and so $M_n \to 0$. But, this assertion assumes that $x$ is fixed (I think?). Indeed, my suspicion is that the opposite is true here since, as $n$ increases, the magnitude of the slopes of each $e^{-nx}$ increases as $x$ approaches $0$; so, in a sense, each $e^{-nx}$ "goes to its supremum faster" as $n$ increases. Whatever the case, I'm not sure how to actually prove it.

For $(c)$, we can use the same reasoning as in $(b)$ since $f_n(x) \to 0$ as $n \to \infty$, for all $x \in C$. However, since $f_n(x)$ attains its supremum (at $x=1$) for all $n$, this is considerably easier to examine. Indeed, we see that $M_n \to 0$ as $n \to \infty$ in this case; and so $\{ f_n \}$ converges uniformly on $C$.

EDIT: I just noticed this post, which is a potential duplicate. But I'll leave this question up, if there are no objections.

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  • $\begingroup$ $e^{-nx}$ on $A$, $B$, and $C$ respectively is basically the same as $x^n$ on $(0,1]$, $(0,1)$, and $(0,1/2]$ just reflected over. $\endgroup$
    – user123641
    Commented Jun 4, 2018 at 21:03
  • $\begingroup$ For Part $(a)$, $f_n(x)$ does converge pointwise to a discontinuous function. The convergence is Not uniform since for $\epsilon=\frac1e$, and for any $N$, there exists a number $n>N$ and an $x\in[0,\infty)$ (take any $n>N$ and $x=1/n$) such that $|f_n(x)-f(x)|=\frac1e \ge \epsilon$. Part $(b)$ is similar. $\endgroup$
    – Mark Viola
    Commented Jun 4, 2018 at 21:04

1 Answer 1

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Regarding (b). Remember that in any $\epsilon$-$\delta$ definition, you can freely swap strong and weak inequalities. In particular, the uniform convergence of $(f_n)$ on $B$ to the function constantly equal to zero can be written as:

For each $\epsilon > 0$ there is $n_0 \in \mathbb{N}$ such that $$ \lvert e^{-nx} - 0 \rvert\le \epsilon$$ for all $n \ge n_0$ and all $x \in (0, \infty)$.

As the weak inequalities are preserved under taking suprema, from this it follows that

For each $\epsilon > 0$ there is $n_0 \in \mathbb{N}$ such that $$ \lvert \sup\limits_{x > 0} e^{-nx} - 0 \rvert \le \epsilon$$ for all $n \ge n_0$.

(Indeed, the $n_0$ is the same as above.) But this is impossible, since the supremum of $e^{-nx}$ on $(0, \infty)$ equals $1$.

Remark. If you don't like weak inequalities in the definitions of limits, you can start with

For each $\epsilon > 0$ there is $n_0 \in \mathbb{N}$ such that $$ \lvert e^{-nx} - 0 \rvert < \frac{\epsilon}{2}$$ for all $n \ge n_0$ and all $x \in (0, \infty)$.

and conclude with

For each $\epsilon > 0$ there is $n_0 \in \mathbb{N}$ such that $$ \lvert \sup\limits_{x > 0} e^{-nx} - 0 \rvert\le \frac{\epsilon}{2} < \epsilon$$ for all $n \ge n_0$.

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