I've tried to generalize the Weierstrass M-Test for sequences of real-valued functions to sequences of functions taking values in a Banach space, and I want to make sure my work is correct. Is the following "Theorem" true? And if so, is my proof attempt correct?
Claim: Let $X$ be a set, let $Y$ be a Banach space (over, say, $\mathbb{R}$), and let $\{f_k\}_{k=1}^{\infty}$ be a sequence of bounded functions from $X$ to $Y$. (Here, bounded means $\sup_{x \in X} \|f(x)\|_{Y} < \infty$.) Suppose there exists a sequence of nonnegative real numbers $\{M_k\}_{k=1}^{\infty}$ such that:
- $\|f_k(x)\|_{Y} \leq M_k$ for all $x \in X$, for all $k \in \mathbb{N}$, and
- $\sum_{k=1}^{\infty} M_k < \infty$.
Then there exists a bounded function $g: X \rightarrow Y$ such that $\sum_{k=1}^{\infty} f_k(x) \rightarrow g(x)$ uniformly [i.e., $\forall \epsilon > 0, \exists N_{\epsilon}$ such that $ n \geq N_{\epsilon} \implies \| \sum_{k=1}^{n} f_k(x) - g(x) \|_{Y} < \epsilon$ for all $x \in X$].
My proof attempt: Fix $\epsilon > 0$. Since $\sum_{k=1}^{\infty} M_k < \infty$, there exists some positive integer $N_{\epsilon}$ such that $\sum_{k=m}^{n} M_k < \epsilon$ whenever $n,m \geq N_{\epsilon}$. Thus, \begin{align*} n,m \geq N_{\epsilon} \implies \left\|\sum_{k=m}^{n} f_k(x) \right\|_{Y} \leq \sum_{k=m}^{n} \left\|f_k(x)\right\|_{Y} & \leq \sum_{k=m}^{n} M_k < \epsilon \quad \text{ for all } x \in S. \end{align*}
Therefore, the sequence of partial sums $g_n(x) := \sum_{k=1}^{n} f_k(x)$ is uniformly Cauchy (with respect to the norm metric $d_Y(y_1,y_2) := \|y_1-y_2\|_{Y}$) on $X$. Then since $Y$ is complete in its norm metric, this implies that the $g_n(x)$'s are uniformly convergent on $X$--that is, there exists some function $g:X \rightarrow Y$ such that for each $\epsilon > 0$, $\exists N_{\epsilon}$ such that $$n \geq N_{\epsilon} \implies \left\| g_n(x) - g(x) \right\|_{Y} < \epsilon \quad \text{ for all } x \in X.$$
Thus, $$n \geq N_{\epsilon} \implies \left\| \sum_{k=1}^{n} f_k(x) - g(x) \right\|_{Y} < \epsilon \quad \text{ for all } x \in X,$$ i.e., $\sum_{k=1}^{\infty} f_k(x) \rightarrow g$ uniformly on $X$. Finally, $g$ is bounded because \begin{align*} \sup_{x \in X} \|g(x)\|_{Y} = \sup_{x \in X} \left\| \sum_{k=1}^{\infty} f_k(x) \right\|_{Y} &\leq \sum_{k=1}^{\infty} \sup_{x \in X} \|f_k(x)\|_{Y} \leq \sum_{k=1}^{\infty} M_k < \infty. \end{align*}
Another question: Is it okay not to define a metric on $X$ in the above proof? I am reading Analysis II by Terence Tao and Tao's definition on page 51 of uniform convergence starts with: "Let $(f^{(n)})_{n=1}^{\infty}$ be a sequence of functions from one metric space $(X,d_X)$ to another $(Y,d_Y) \ldots.$" I don't see the need for imposing a metric on the space $X$. Is the metric on $X$ superfluous?
Edit 8/11/22: I realized I messed up the statement of the theorem, so I just corrected it. (I was mixing up uniform convergence of the $f_n$'s with uniform convergence of the series $\sum_{k=1}^{\infty} f_k(x)$...) I've also added a little more detail to the proof.
By your assumption we have: $sup_{X}\left\|f_{k}(x) \right\|\leq\,M_{k}$ and also
$\sum_{1}^{+\infty} sup_{X}\left\|f_{k}(x) \right\|\leq\,\sum_{1}^{+\infty}M_{k}<+\infty$
which implies by the convergence of the series that :
$sup_{X}\left\|f_{k}(x) \right\|$$\to 0$. Therefore for $f(x)=0$ we get:
$sup_{X}\left\|f_{k}(x)-f(x) \right\|$$\to 0$ which is what you wanted to prove!!