A question about $f:(0,1]\times[0,1]\rightarrow N_{l^1(\mathbb{R}^\omega)}(0,2)$ Let $L$ be the $l^1$ topology on the subset of $\mathbb{R}^\omega$ with finite $l^1$ norm.
($\left\|x\right\|_{l^1(\mathbb{R}^\omega)}:=\sum_i|x_i|$. $l^1$ topology is the metric topology induced by this norm.)
Say $f:(0,1]\times[0,1]\rightarrow N_L(0,2)$ is any continuous function s.t. $f(t,0)\equiv 0$ and $f(t,1)\equiv(1,0,0...)$.
Then, would there always exist $\{t_n\},\{x_n\}$ s.t. $t_n\rightarrow 0$ and $f(t_n,x_n)$ converges pointwisely to a point in $N_L(0,\frac{1}{2})\setminus\{0\}$?
($N_L(0,r)$ : the open ball of radius $r$ centered at 0, with respect to the metric of $L$)
 A: This is an answer in the negative to the OP's question, that means I'll construct a function $f(t,x)$ that fulfills all the required conditions, but where no $\{t_n\},\{x_n\}$ sequences exist with the required conditions.
To remove clutter from the definitions and arguments, $n$ and $k$ are always positive integers in the follwowing.
Function values of $f$ in $\mathbb R^\omega$ will only have positive and zero components, at most two of which will be positive and then those are neighboring components. That means they trivially have a finite $l^1$ norm, which is just the sum of their positive components.
To make the description easier, I'll denote for any real $a,b$ by
$$[a]_n:=(\underbrace{0}_1,\underbrace{0}_2,\ldots,\underbrace{a}_n,0,\ldots)$$
$$[a,b]_n:=(\underbrace{0}_1,\underbrace{0}_2,\ldots,\underbrace{a}_n,\underbrace{b}_{n+1},0,\ldots)$$
elements of $\mathbb R^\omega$, where the underbraces indicate the component number. I'll use the usual interval notation $[u,v]$ as well, but since they never have a subscript, I hope no misunderstanding will happen.
$f$ will be defined separately above and below the $x \le t$ line dividing its domain.
For each $n$, a helper function $d_n: [\frac1{n+1},\frac1n] \to [0,1]$ is defined as follows
$$d_n(x)= \frac{x-\frac1{n+1}}{\frac1n-\frac1{n+1}}$$
We see that $d_n$ is linear and
$$d_n\left(\frac1n\right)=1,  d_n\left(\frac1{n+1}\right) = 0. \tag1 \label{dborders}$$
We define $D(x):(0,1] \to \mathbb R^\omega$ that will become our $f$ on the diagonal $t=x$:
$$D(x)= [d_n(x),1-d_n(x)]_n \quad \text{if } \frac1{n+1} \le x \le \frac1n.  \tag2 \label{defD}$$
$D$ is well defined and continuous in its domain.
Proof:
We have double defined $D$ on arguments of the form $x=\frac1{k+1}, k=1,2,\ldots$, where it is defined both by $n=k$ and again by $n=k+1$. But the definitions are actually the same, for $n=k$ we get from \eqref{defD}:
$$D\left(\frac1{k+1}\right)=[d_k\left(\frac1{k+1}\right),1-d_k\left(\frac1{k+1}\right)]_k=[0,1]_k = [1]_{k+1},$$
using \eqref{dborders} and for $n=k+1$ we find that
$$D\left(\frac1{k+1}\right)=[d_{k+1}\left(\frac1{k+1}\right),1-d_{k+1}\left(\frac1{k+1}\right)]_{k+1}=[1,0]_{k+1} = [1]_{k+1}, $$
again using \eqref{dborders}.
Since $(0,1]= \cup_{n=1}^{\infty} [\frac1{n+1},\frac1n]$, $D$ is well defined in it's domain. But in each interval $[\frac1{n+1},\frac1n]$ it is linear and so continuous, and also left-/right continuous on the respective end of the interval, so it is continuous on the whole domain!
(End of prof)
The definition of $f(t,x):(0,1]\times[0,1] \to \mathbb R^\omega$ is now
$$f(t,x)=
\begin{cases}
\frac xt D(x) & \text{if } x \le t,\\
D(x) & \text{if } x \ge t.\\
\end{cases} \tag3 \label{defF}
$$
The following picture illustrates the behavior of $f$ above and below the line $x=t$:

Again, we double defined $f$ on the line $x=t$, but \eqref{defF} shows $f(x,x)=D(x)$ by both definitions.
At and below the line $x=t$ $f$ is a product of the continuous $D(x)$ and the continuous $\frac xt$ ($t>0$ so the division is not a problem), so $f$ is continous there.
At and abovethe line $x=t$ $f$ is just the continuos $D(x)$. So, $f$ is continous in its whole domain.
We have $f(t,0)=\frac0tD(x)=(0,0,0,\ldots)$, using the top line definition in \eqref{defF} and $f(t,1)=D(1)=[1,0]_1=(1,0,0,\ldots)$,  using the bottom line definition there.
From \eqref{defD} we get that $\forall x \in (0,1]: \Vert D(x)\Vert=1$, so from \eqref{defF} follows that $\Vert f(t,x)\Vert=1$ for all $t,x$ in its domain with $x \ge t$ and $\Vert f(t,x)\Vert=\frac xt \le 1$ for all $t,x$ in its domain with $x \le t$.
This proves that the image of $f$ is in $N_L(0,2)$, as required.
So now we've checked that the above $f$ does fulfill all the conditions required from it under the the OP's question.
Let's assume sequences $\{t_n\},\{x_n\}$ exist with $\lim_{n \to \infty} t_n = 0$ and $f(t_n,x_n) \to G \in N_L(0,\frac12)\backslash (0,0,0,\ldots)$ as $n \to \infty$, with the limit being point-/componentwise.
Since all components of $f$ are always non-negative, the same must hold for $G$. Let $m$ be an index where the $m$-th component of $G$ ($G_m$) is non-zero, hence positive. This must exist, as otherwise $G=(0,0,0,\ldots)$.
From the inifnite number of points $(t_n,x_n)$ there must be an infinite number in at least one of the areas $x\le t$ or $x \ge t$, which would then form a subsequence also converging componentwise to $G$.
We'll show a contradiction in either case.

*

*$\forall n: x_n \le t_n$
Since $\lim_{n \to \infty} t_n = 0$, there must be an $N$ with $\forall n > N: t_n < \frac1{m+1}$. From \eqref{defD} we can see that $D(x)$ for $x <\frac1{m+1}$ will always have the $m$-th component 0, as the corresponding $n$ in the definition will be biggger than $m$.
Looking at \eqref{defF} for the $x\le t$ case, that means the $m$-th component of $f(t,x)$ is also zero for $t < \frac1{m+1}$, hence the $m$-th component of $f(t_n,x_n)$ is zero for all $n>N$. So this component can not converge to a positive $G_m$, yielding to a contradiction.

*

*$\forall n: x_n \ge t_n$
In this case $ft_n,x_n)=D(x_n)$ always. If the $m$-th component of $D(x_n)$ is to converge to $G_m>0$, there must be an $N$ such that $\forall n > N:$ the $m$-th component of $D(x_n) > \frac{G_m}2 > 0$. But we know that in any $D(x)$ at most 2 components are non-zero and they are neighboring components. That means for $n > N$, the only components that can be non-zero in $D(x_n)$ are at indices $m-1,m,m+1$.
But we know that $\Vert D(x_n)\Vert=1$, so the sum of those 3 components at indices $m-1,m,m+1$ is $1$ for $n>N$. But in their pointwise limit the sum of those 3 components in $G$ must be less than $\frac12$ (it is just a part of its $l^1$ norm), which is impossible! This argument also works for $m=1$, the "$0$-th component$ is just 0 in that case.
That finally proves that the constructed $f$ is a counter example to OP's question.
