Show that if the series $\sum_{n=1}^\infty f_n$ is uniformly convergent on $\Bbb R$, then the sequence $(f_n)$ converges to $0$ on $\Bbb R$.

Question

Let $f_n:\Bbb R \to \Bbb R$ be a function for which the series $\sum_{n=1}^\infty f_n$ is uniformly convergent on $\Bbb R$ for any $n \in \Bbb N$. Show that the sequence $(f_n)$ converges to $0$ on $\Bbb R$.

Definition. A sequence $(f_n)$ of functions on $A \subseteq \Bbb R$ to $\Bbb R$ converges to a function $f:A \to \Bbb R$ on $A$ if for any $\epsilon>0$ and each $x \in A$, there exists $N \in \Bbb N$ such that for all $n \ge N$, we have $|f_n(x)-f(x)|<\epsilon$.

Cauchy Criterion for Series of Functions. Let $(f_n)$ be a sequence of functions on $A \subseteq \Bbb R$ to $\Bbb R$. Then the series $\sum_{n=1}^\infty f_n(x)$ is uniformly convergent on $A$ iff for all $\epsilon>0$, there exists $K \in \Bbb N$ such that for any $m>n\ge K$, we have $$\left|\sum_{k=n+1}^m f_k(x) \right|<\epsilon,$$ for all $x \in A$.

Attempt: Since the series $\sum_{n=1}^\infty f_n$ is uniformly convergent on $\Bbb R$ for any $n \in \Bbb N$, then by the Cauchy Criterion for Series of Functions, for all $\epsilon>0$, there exists $K \in \Bbb N$ such that for any $m>n \ge K$, we have $$\left|\sum_{k=n+1}^m f_k(x) \right|<\epsilon,$$ for all $x \in \Bbb R$.

The goal is to show that $f_n \to 0$ on $\Bbb R$. Let $\epsilon>0$ and $x \in \Bbb R$ be arbitrary. Choose $N=K \in \Bbb N$ such that for all $n \ge N$, we have $$|f_n(x)-0|=|f_n(x)|<|f_{n+1}(x)+\ldots+f_m(x)|<\epsilon.$$ Hence, $f_n \to 0$ on $\Bbb R$.

Does this approach correct? Thanks in advanced.

Answer 1

Let $f:\Bbb R \to \Bbb R$ be a function such that $\sum_{n=1}^\infty f_n(x)$ converges uniformly on $\Bbb R$ to $f$. Then the series converges on $\Bbb R$ to $f$. Let $s_n:=\sum_{k=1}^n f_k(x)$ be the partial sum of this series. Notice that $s_n(x)-s_{n-1}(x) = f_n(x)$. Now, we have $s_n \to f$ and $s_{n-1} \to f$ on $\Bbb R$ as $n \to \infty$. By definition, for all $\epsilon>0$ and each $x \in \Bbb R$, there exists $N_1 \in \Bbb N$ such that for any $n \ge N_1$, we have $|s_n(x)-f(x)| < \frac{\epsilon}{4}$. Similarly, for all $\epsilon>0$ and each $x \in \Bbb R$, there exists $N_2 \in \Bbb N$ such that for any $n \ge N_2$, we have $|s_{n-1}(x)-f(x)| < \frac{\epsilon}{4}$. Let $N:=\max\{N_1,N_2\}$. Then $N \in \Bbb N$ and for all $n\ge N$, we have \begin{align*} |f_n(x)| &= |s_n(x)-s_{n-1}(x)| \\ &\le |s_n(x)-f(x)| + |s_{n-1}(x) - f(x)| \\ &< \frac{\epsilon}{4} + \frac{\epsilon}{4} \\ &= \frac{\epsilon}{2} \\ &< \epsilon. \end{align*} Thus, the sequence $(f_n)$ converges to $0$ on $\Bbb R$.

Answer 2

From Cauchy Criterion for Series of Functions (putting $m:=n+1$) one can get that for all $\epsilon>0$ there exists $K \in \Bbb N$ such that for any $n\ge K$, we have $\left|f_{n+1}(x) \right|<\epsilon$ for all $x \in A$. Therefore there exists $K \in \Bbb N$ (and equals 'previous' $K$ plus one) such that for any $n\ge K$, we have $\left|f_{n}(x) \right|<\epsilon$ for all $x \in A$. This shows that $(f_n)$ converges to zero uniformly.

Answer 3

Your use of the triangle inequality is wrong. You may think of partial sums. The proof is essentially identical to the proof of the property $$\sum_{n=1}^{\infty}a_n \text{ converges }\implies a_n\to 0$$