Let $[0,1] \subset \mathbb R$ be a the compact interval in the real numbers $\mathbb R$. We know that $C([0,1] \to \mathbb R)$ (the continuous function on $[0,1]$ with values in $\mathbb R$) are dense in $L^{2} ([0,1] \to \mathbb R)$ (the usual Lebesgue space).
Now consider the space of continuous functions on $[1,2]$ taking values in $L^{2} ([0,1] \to \mathbb R)$ , i.e. $C ( [1,2] \rightarrow L^{2} ([0,1] \to \mathbb R))$.
Is the space $C ([1,2] \times [0,1] \to \mathbb R)$ dense in $C ( [1,2] \rightarrow L^{2} ([0,1] \to \mathbb R))$?
In other words: Fix an arbitrary $f \in C ( [1,2] \rightarrow L^{2} ([0,1] \to \mathbb R))$. Can we find a sequence $f_{n} \in C ([1,2] \times [0,1] \to \mathbb R)$, such that $$\sup_{z \in [1,2]} \int_{[0,1]} ( f_{n} -f )^{2} (z,x) d x \rightarrow 0 ? $$
This question is related to this question, by a different order of the spaces.
Yes. Say $f_t\in L^2[0,1]$ for $t\in[1,2]$ and $f_t$ depends continuously on $t$. Then $K=\{f_t:t\in[1,2]\}$ is a compact subset of $L^2[0,1]$.
Define $T_n:L^2[0,1]\to L^2[0,1]$ by, say, letting $T_n=f*\phi_n$, where $\phi_n$ is an approximate identity. It's easy to see that for each $n$ the function $g_n(s,t)=T_nf_t(s)$ is continuous on $[0,1]\times[1,2]$. We know $T_nf\to f$ in $L^2$ for every $f\in L^2$. And since $||T_n||\le||\phi_n||_1=1$ the family $(T_n)$ is equicontinuous on compact subsets of $L^2$. Hence $T_nf\to f$ uniformly on $K$.
??? Since the downvoter doesn't see fit to show the courtesy of explaining what he or she thinks the problem is I have to guess. There are all sorts of things above that could be read incorrectly. Take the conclusion, "Hence $T_nf\to f$ uniformly on $K$", for example. A person could think that said that a certain sequence of continuous functions on $[0,1]\times[1,2]$ converged uniformly to a discontinuous function, which would of course be absurd. But that's not what it says. $K$ is a metric space, being a subset of the metric space $L^2$. $T_n$ is a sequence of mappings from this metric space into another metric space, again $L^2$ as it happens. So saying $T_nf\to f$ uniformly on $K$ says precisely that $\sup_{t\in[0,1]}||T_nf_t-f_t||_2\to0$.
Not that I have any reason to think that that's the problem, but I suspect that whatever the problem is is based on a similar sort of incorrect reading. Can't explain the actual problem without being told what it is...