The idea is that one could in principle consider the space of functions: $$\{f:\Omega\to V\}$$ with pointwise operations and uniform convergence: $$f_\lambda\to f:\iff\|f_\lambda-f\|_\infty\to 0$$
Given a Banach space of functions: $$E:=\{f:\Omega\to V:\|f\|\text{ exists}\}$$ with $\Omega$ a suitable space and $V$ a vector space.
This then gives rise to a topological vector space.
(It has special properties due to its construction via a norm.)
Now, in principle this could be extended to: $$E_\infty:=\{f:\Omega\to V\}$$ with the same vector space structure but with new topology as follows.
Define a neighborhood base by: $$B_\varepsilon(f):=\{f'\in E_\infty:f'-f\in E,\|f'-f\|<\varepsilon\}$$ Does it give rise to a topological vector space?
If $E_\infty$ contains unbounded functions (without any continuity or similar requirements on the functions, that means $\Omega$ is infinite), then the topology of uniform convergence is not a vector space topology on $E_\infty$. One sees that then the balls around $0$, $B_\varepsilon(0)$, are not absorbing, hence the scalar multiplication is not continuous. $E_\infty$ is still a topological group in the topology of uniform convergence, however.