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$f(x): [0, 1) \rightarrow \mathbb{R}$. Prove that the function $f(x) = \sum_{n=1}^{\infty}2^{-n}\{\frac{[2^nx]}{2}\}$ is continuous. Please, give me a hint where to start (I want to prove it using definition with $\varepsilon$ and $\delta$) P.S. {x} is fractional part of x and [x] is integer part of x. I have $|x - y| \le \delta$ and I need to tranform this thing $|\sum_{n = 1}^{\infty}(2^{-n}(\{\frac{[2^nx]}{2}\} - \{\frac{[2^ny]}{2}\}))|$ somehow to show that this is less than $\varepsilon$ for some $\delta$. I dont know how to get rid of $\{\frac{[2^nx]}{2}\} - \{\frac{[2^ny]}{2}\}$

  • Hint: Look at dyadic rationals, can you guess what $f(x)$ is? – user10354138 Nov 11 '18 at 09:33
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    Do you see what is happening here: https://math.stackexchange.com/questions/2993580/prove-the-continuity-fx-sum-n-1-infty-sin-fracxn/2993587

    Please show your effort!!

     – Robert Z Nov 11 '18 at 09:33
  • A more specific hint than user10354138: Write out some arbitrary $x$ in binary. Then figure out what $f(x)$ is for that specific value of $x$. Do you see what $f(x)$ is, now? – Paul Sinclair Nov 11 '18 at 16:07

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