Continuity of weighted infinite sum of Binary entries

Question

Consider the following function on the closed interval $[0,1]$, $$B_{k}(x) = x_k,\quad k\in \mathbb{N}$$ where $x$ has the binary representation $(0.x_1x_2\ldots)_2$. Define the function $f:[0,1]\rightarrow \mathbb{R}$ as follows, $$f(x)=\sum_{k \;\text{is odd}}\frac{B_{k}(x)}{2^k}.$$ I suspect that $f(x)$ is a continuous function, but so far failed to prove so. Any help will be appreciated.

Answer 1

In fact, $f(x)$ is not a continuous function. For instance, we have $f(\frac14) = 0$, but when $x$ is a tiny bit less than $\frac14$ then $f(x)$ is a tiny bit less than $\frac12$.

Indeed $f$ is discontinuous at many dyadic rationals (perhaps, I'm guessing, the ones that terminate after an even number of bits). The graph of $f$ (below) indicates a self-similar behavior that is probably pretty easy to prove from the definition.