Uniform/normal convergences of $\sum_{n\geq 0} f_n(x)$ and $\sum_{n\geq 0} g_n(x)$ imply the uniform/normal conv. of $\sum_{n\geq 0} f_n(x)+g(x)$

41 Views Asked by At

Consider two function series $\sum_{n\geq 0} f_n(x)$ and $\sum_{n\geq 0} g_n(x)$. I know that $$\sum_{n\geq 0} f_n(x) \mathrm{\,\, converges \,\,\, conditionally \,\,\, and } \sum_{n\geq 0} g_n(x) \mathrm{\,\, converges \,\,\, conditionally} \implies \sum_{n\geq 0} (f_n(x)+g_n(x)) \mathrm{\,\, converges \,\,\, conditionally}$$ And that $$\sum_{n\geq 0} f_n(x) \mathrm{\,\, converges \,\,\, absolutely \,\,\, and } \sum_{n\geq 0} g_n(x) \mathrm{\,\, converges \,\,\, absolutely} \implies \sum_{n\geq 0} (f_n(x)+g_n(x)) \mathrm{\,\, converges \,\,\, absolutely}$$


Are these implications extended to the cases of uniform and normal convergence.

That is, do the following implications hold?

$$\sum_{n\geq 0} f_n(x) \mathrm{\,\, converges \,\,\, uniformly \,\,\, and } \sum_{n\geq 0} g_n(x) \mathrm{\,\, converges \,\,\, uniformly} \implies \sum_{n\geq 0} (f_n(x)+g_n(x)) \mathrm{\,\, converges \,\,\, uniformly}$$

$$\sum_{n\geq 0} f_n(x) \mathrm{\,\, converges \,\,\, normally \,\,\, and } \sum_{n\geq 0} g_n(x) \mathrm{\,\, converges \,\,\, normally} \implies \sum_{n\geq 0} (f_n(x)+g_n(x)) \mathrm{\,\, converges \,\,\, normally}$$

If it is not true in general, does it hold at least for power series?

I did not find any theorem or proof on this, so it would be great to have some reference where to learn more about this.

1

There are 1 best solutions below

2
On BEST ANSWER

So you are saying that $\sum f_n$ and $\sum g_n$, defined in a set $A$ converge uniformly.

This means that $\forall \epsilon' > 0 \ \exists N'_{\epsilon'} \ / \forall n > N', \forall x \in A \ $ we have

$$ |\sum_{k=0}^n f_k - f | < \epsilon' $$

and $\forall \epsilon'' > 0 \ \exists N''_{\epsilon''} \ / \forall n > N'', \forall x \in A \ $ we have $$ |\sum_{k=0}^n g_k - g | < \epsilon''. $$

Well, suppose we are given an $ \epsilon > 0$: we apply the hypothesis to our two series taking $ \epsilon' = \epsilon'' = \frac 12 \epsilon $:

By uniform convergence we find $N', N''$ such that $$ |\sum_{k=0}^n f_k - f | < \frac 12 \epsilon \qquad |\sum_{k=0}^n g_k - g | < \frac 12 \epsilon. $$

Take $N = \max \{ N', N''\} $. Now, $\quad \forall n > N, \ \forall x \in A \ $ we have

$$ |\sum_{k=0}^n (f_k + g_k) - (f + g) | = |\sum_{k=0}^n f_k - f + \sum_{k=0}^n g_k - g | \leq |\sum_{k=0}^n f_k - f | + |\sum_{k=0}^n g_k - g | < \frac 12 \epsilon + \frac 12 \epsilon = \epsilon. $$

Now, if $\sum f_n$ and $\sum g_n$, always defined in a $A$, converge normally, we have that $ \ \exists m_n, \ \exists m'_n \ / \forall x \in A \ \forall n $ we have

$$ | f_n | \leq m_n \qquad | g_n | \leq m'_n $$

and $\exists m, m' \in \mathbb R $

$$ \sum_{k=0}^{\infty} m_k = m \qquad \sum_{k=0}^{\infty} m'_k = m'. $$

But obviously

$$ |f_n + g_n| \leq |f_n| + |g_n| \leq m_n + m'_n $$

Since $m_k, \ m'_k \ $ are positive and their sums converge, you can say that

$$ \sum_{k=0}^{\infty} ( m_k + m'_k ) = \sum_{k=0}^{\infty} m_k + \sum_{k=0}^{\infty} m'_k = m + m'. $$

You can always reorder positive convergent series.