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W. Rudin gives the following example $f(x)$.

Let $\varphi(x) = |x|$ on $[-1,1]$ and extend periodically to the whole real line with period $2$.

Now define

$$ f(x) = \sum_{n=0}^\infty \left(\frac{3}{4}\right)^n\varphi(4^nx).$$

How does the graph of this function, or its successive approximations, look like? Just to get an idea of how continuity happens, but not differentiability.

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I suspect that $\varphi$ is defined on $[-1,1]$, and that the function is supposed to be $$f(x)=\sum_{n=0}^\infty\left(\frac34\right)^n\varphi(4^nx)\;.$$ – Brian M. Scott Jun 8 '12 at 9:26

The image to keep in mind is to start with the V, and at each step of the approximaton you add a set of $\frac{3}{4}$-of-the-previous-size smaller V's. The $\frac{3}{4}$ is just to give convergence, and the $4^n$ is to give more and more V's close to every real number.

But this would lead you to believe that the shape is very jagged, sort of like the classic Weierstrass no-where differentiable function: Weierstrass

But that's not really true. In fact, the successive additions almost 'smooth' out the 'jags' of most of the sides of the V's. On this blog page (not my blog, but instead divisbyzero), there is a picture of an essentially-the-same function:

enter image description here

This is actually the function $\displaystyle \sum \left(\frac{1}{2}\right)^n \varphi (2^n x)$, but I think that might see the similarity. On your function, the function would get a bit larger ($3/4 > 1/2$) and so the crests and troughs would be a bit deeper, etc.

One good bit about the $1/2$ function is that the blogger of DivisionByZero (the blog I linked to above) also wrote up an applet illustrating the successive approximations of the function.

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I don't understand "that's not really true" part. There is an essential difference between $3/4$ (or anything larger than $1/2$) and the critical value $1/2$. The first graph has Hausdorff dimension greater than 1 (though precise value is unknown, I think), the second has Hausdorff dimension 1, same as any smooth curve. – user31373 Jun 8 '12 at 14:39

mixedmath has already explained most of the mathematics in his answer. Here, I'll just provide Mathematica code that people can play around with:

φ[x_] := TriangleWave[{0, 1}, -x/2 - 1/4];
rudin[x_, n_Integer] := Sum[(3/4)^k φ[x 4^k], {k, 0, n}];

Animate[Plot[rudin[x, n], {x, 0, 2}, Axes -> None, Frame -> True, 
  MaxRecursion -> 4, PlotLabel -> StringForm["n=`1`", n], 
  PlotPoints -> 105, PlotRange -> {0, 7/2}], {n, 1, 6}]

Rudin's example

I decided not to include the plots for $n > 6$; at that point, the function already looks sufficiently messy to be indistinguishable from the $n=6$ partial sum.

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And here's the blancmange curve: interactive applet by Dave Richeson. – user31373 Aug 20 '12 at 20:35

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