A Banach space is a normed vector space with completeness, i.e. all Cauchy series converge to an element within the vector space.
In the case of continuous functions on an interval in the real line, $C[a,b],$ equipped with the uniform or infinity norm, it forms a Banach space.
However, the following Cauchy sequence in the space $C[-\infty,\infty]$ seems to be a counter-example, or much more likely, a clue that I misunderstood the whole thing:
$$f^{k}(x) = \frac 1 2 +\frac 1 \pi \tan^{-1}(kx)$$
The square function is not a continuous function, yet this Cauchy sequence clearly converges to it. Therefore, the vector space of continuous functions is not complete, and shouldn't be a Banach space (?).
What am I missing? Is the reason the fact that the chosen norm (the supremum of the function is always constant at $1$) renders this not a Cauchy sequence)? And if this is the case, is the step function convergence invalidating the convergence of a Cauchy sequence still valid for other norms?
The reason cannot be that there could be unbounded functions in the interval, since the argument will hold for the space of bounded functions $C[-1,1].$
Best Answer
Note that $C(-\infty,\infty)$ is not a Banach algebra with the supremum norm, as it has unbounded functions. But that's irrelevant to your example as it can be considered in any $C[-b,b]$.
In any case, your sequence is not Cauchy. For example you have $$ |f_k(1/k^2)-f_{k^2}(1/k^2)|=\frac1\pi|\arctan1/k-\arctan1|=(\frac14-\pi\arctan\frac1k). $$ This tells you that for all $k$ big enough such that $\arctan \tfrac1k<\tfrac1{8\pi}$, $$ \|f_k-f_{k^2}\|_\infty\geq\frac18. $$