Title in the question. I suspect the answer is yes, but love being proven wrong.
I've always wanted to know when "a function can be drawn without taking pen off of paper", but the necessary properties of such functions are not obvious because continuity on it's own doesn't do the trick. The missing pieces to this puzzle has bugged me for some time, but now I think I might be getting close to closure, and not the topological kind- excuse the pun.
Drawing by lifting pencil from paper can still beget continuous function.
So yes, $$ f(x) =
\begin{cases}
x \sin \frac 1x, & \text{if }x\neq 0 \\
0, & \text{if }x=0
\end{cases}
$$
has infinite arc length in any interval $[a,b]$ with $a\leq0$ and $b>0$, so you can't draw this with pen and paper.
But with for example $$ f(x) =
\begin{cases}
x^2 \sin \frac 1x, & \text{if }x\neq 0 \\
0, & \text{if }x=0
\end{cases}
$$
I believe the arc length is finite in any interval $[a,b]$ with $a\leq0$ and $b>0$, so this I think you would be able to draw with pen and paper.
Setting aside the fact that paper isn't "smooth" because it is made of atoms, and other irrelevant stuff, is my proposition in the title true, or are there really weird continuous, bounded functions with finite arc length that I'm not aware of?
Also, are there "better" ways to characterise functions "which can be drawn without taking pen off paper"?
Edit: Actually, now that I think about it, "boundedness" is redundant, since finite arc length $\implies$ bounded.
Best Answer
It depends on how you want to model the act of drawing. If you say the pencil is an object of positive mass subject to Newton's laws of mechanics, the force you can exert on it is bounded, and the drawing must be completed in a finite time, then there is a finite bound on the length of the path you can draw.
EDIT: The requirement of bounded force (and therefore bounded acceleration) is more stringent than finite length. For example, consider the curve $y = x^2 \sin(1/x)$ (with $y=0$ at $x=0$) which has finite length. This passes through the points $(x_k,y_k) = (2/(2k+1)\pi, 4 (-1)^k/((2k+1)^2 \pi^2))$. At bounded acceleration, to go from (if $k$ is even) $y=4/((2k+1)^2 \pi^2$ to $y=-4/((2k+3)^2 \pi^2)$ and then back up to $y=4/((2k+5))^2 \pi^2)$ will take time approximately $c/k$ for some positive constant $c$. Since $\sum_k 1/k$ diverges, this means the curve could not be drawn in finite time.