How differentiable is the function $g(x) = \sum_n 2^{-n} f(x-r_n)$ where $f(x)=x^2 \sin\frac1{x}$?

Question

This is an auxiliary enquiry (something like it may well be already discussed on MSE, but I haven't found it) resulting from a feeling of unease provoked by the question of this post. Taking the example of $$f(x)=x^2 \sin\frac1{x}$$ let us suppose $r_n$ is an enumeration of the rationals in $(0,1)$, and form the function $g$ defined on $(0,1)$: $$ g(x) = \sum_n 2^{-n} f(x-r_n) $$ I guess that $g$ is well-defined and continuous, but is not uniformly continuous. what can be said about the differentiability of $g$?

Answer 1

The function $g,$ which can actually be defined for $x\in \mathbb {R},$ is uniformly continuous on every compact subset of $\mathbb {R},$ and is differentiable everywhere.

Proof: $f$ is continuous everywhere, hence is bounded on compact sets. Thus the sum converges uniformly to $g$ on compact sets by the Weierstrass M-test. As the summands are all continuous, $g$ is continuous on compact sets, hence $g$ is uniformly continuous on compact sets.

As for differentiability, note that each summand is differentiable on $\mathbb {R}.$ Furthermore, on any compact set, the derivatives of the summands are uniformly bounded. Thus the series of derivatives converges uniformly on compact subsets $\mathbb {R},$ again by Weierstrass M. The standard theorem on differentiation and uniform convergence can now be invoked to give

$$g'(x) = \sum_n 2^{-n} f'(x-r_n), \,\, x\in \mathbb {R}.$$