# Thread: Convergence of a Sum of Functions Implies Convergence of Functions to Zero

1. ## Convergence of a Sum of Functions Implies Convergence of Functions to Zero

I'm asked to prove that the uniform convergence of any sum of complex functions, $\displaystyle \{ \sum f_{n} \}$, on a set $\displaystyle S$, implies $\displaystyle f_{n} \rightarrow 0$ uniformly. However, I'm starting to doubt the truth of this, though probably naively. But what would be impermissible about a sum converging uniformly to some function, but interspersed in the sequence of functions are infinitely many functions which add a value $\displaystyle \varepsilon = f_{i}(x)$ at point $\displaystyle x$ in the domain, and also functions which subtract this same value at $\displaystyle x$?

More formally, I've been trying to tackle the argument by a $\displaystyle \varepsilon, \delta$ argument as follows and have made no progress:

Suppose $\displaystyle \displaystyle \sum^{\infty}f_{n} = f$, and suppose $\displaystyle \bigl( \exists \varepsilon' > 0 \forall M \in \mathbb{N} \exists m \geq M, x \in S \bigr) \bigl( |f_{m}(x)| \geq \varepsilon \bigr)$...

Even if we assumed (unwarrantably, I think) that the $\displaystyle x$ is the same one for every choice of $\displaystyle M$, we run into trying to show something like, for every $\displaystyle M$ there is an $\displaystyle m \geq M$ such that $\displaystyle |f_{1}(x) + ... + f_{m}(x) - f(x)| \geq \varepsilon'$, and I can't see how that'll go, since any use of the triangle inequality is going to have me splitting the absolute values by minus signs, but then I'm no longer showing that anything is arbitrarily small.

2. Originally Posted by ragnar
I'm asked to prove that the uniform convergence of any sum of complex functions, $\displaystyle \{ \sum f_{n} \}$, on a set $\displaystyle S$, implies $\displaystyle f_{n} \rightarrow 0$ uniformly. However, I'm starting to doubt the truth of this, though probably naively. But what would be impermissible about a sum converging uniformly to some function, but interspersed in the sequence of functions are infinitely many functions which add a value $\displaystyle \varepsilon = f_{i}(x)$ at point $\displaystyle x$ in the domain, and also functions which subtract this same value at $\displaystyle x$?

More formally, I've been trying to tackle the argument by a $\displaystyle \varepsilon, \delta$ argument as follows and have made no progress:

Suppose $\displaystyle \displaystyle \sum^{\infty}f_{n} = f$, and suppose $\displaystyle \bigl( \exists \varepsilon' > 0 \forall M \in \mathbb{N} \exists m \geq M, x \in S \bigr) \bigl( |f_{m}(x)| \geq \varepsilon \bigr)$...

Even if we assumed (unwarrantably, I think) that the $\displaystyle x$ is the same one for every choice of $\displaystyle M$, we run into trying to show something like, for every $\displaystyle M$ there is an $\displaystyle m \geq M$ such that $\displaystyle |f_{1}(x) + ... + f_{m}(x) - f(x)| \geq \varepsilon'$, and I can't see how that'll go, since any use of the triangle inequality is going to have me splitting the absolute values by minus signs, but then I'm no longer showing that anything is arbitrarily small.
You might as well do a direct proof like this: Suppose that $\displaystyle \sum_{k=1}^\infty f_k(x)$ converges uniformly. This amounts to the same thing as saying that the tails $\displaystyle t_n(x) := \sum_{k=n}^\infty f_k(x)$ converge uniformly to 0.

From the triangle ineqality we get

$\displaystyle |f_n(x)|=\left|\left|\sum_{k=n}^\infty f_k(x)\right|-\left|\sum_{k=n+1}^\infty f_k(x)\right|\right|\leq \left|\sum_{k=n}^\infty f_k(x)\right|+\left|\sum_{k=n+1}^\infty f_k(x)\right|=|t_n(x)|+|t_{n+1}(x)|$

Now, of course, the sum $\displaystyle |t_{n}(x)|+|t_{n+1}(x)|$ of two sequences that uniformly converge to 0 uniformly converges to 0.