Can a continuous, nowhere differentiable function have specified "shape" at every point? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T06:01:24Z http://mathoverflow.net/feeds/question/18319 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18319/can-a-continuous-nowhere-differentiable-function-have-specified-shape-at-every Can a continuous, nowhere differentiable function have specified "shape" at every point? Mike Hall 2010-03-15T22:44:43Z 2011-07-14T15:21:33Z <p>I'm a bit embarrassed to admit that:</p> <p>a) This is a rather unmotivated question.</p> <p>b) I can't remember whether or not I've asked this before, but searching doesn't seem to turn anything up so ...</p> <p>Consider some "shape" function $\phi: \mathbf{R} \to \mathbf{R}$. Then given some function $f: \mathbf{R} \to \mathbf{R}$, one can ask whether the "difference quotient",</p> <p>$\lim_{y\to x} \frac{f(y)-f(x)}{\phi(y-x)}$,</p> <p>exists at various points $x$. Letting $\phi(x) = x$ corresponds to taking normal derivatives, and intuitively when the limit exists this means that near $x$, the function $f$ "looks like" $\phi$ does near 0. </p> <p>However, if the ratio $\phi(x)/x$ is not bounded above or away from 0 as $x\to 0$ (I'm mostly thinking of the case when it is neither, so that $\phi$ is "wildly oscillating" in some sense), then anywhere the above limit exists and is nonzero, the function $f$ is necessarily non-differentiable. </p> <p>My question: If $\phi$ is some wildly oscillating function as described above (pick your favorite), can there be an $f$ for which this limit exists everywhere?</p> <p>(Edit: I suppose I really want $\phi$ and $f$ to be continuous functions.)</p> http://mathoverflow.net/questions/18319/can-a-continuous-nowhere-differentiable-function-have-specified-shape-at-every/18345#18345 Answer by Thomas Kragh for Can a continuous, nowhere differentiable function have specified "shape" at every point? Thomas Kragh 2010-03-16T09:02:09Z 2010-03-16T09:15:58Z <p>Assume WLOG that $\phi(x)>0$ when $x>0$. Since the limit described exists for all $x$ in the source of $f$. We get for any $x$ the bound:</p> <p>$f(x+\delta)-f(x) \leq C\phi(\delta)$</p> <p>for $0 &lt; \delta &lt; \delta_0$ for some $C,\delta_0>0$ which may depend on $x$.</p> <p>diving by $\delta$ we get by the assumptions on $\phi$ that</p> <p>$\underline{\lim}_{\delta \to 0} ( \frac{f(x+\delta) - f(x)}{\delta}) \leq 0$</p> <p>This is one the four derivatives of $f$, and proposition 2 chapter 5 in Real Analysis by H.L. Royden states that if $f$ is continuous then it is (non-strictly) decreasing. Similar for increasing. So $f$ is constant.</p> http://mathoverflow.net/questions/18319/can-a-continuous-nowhere-differentiable-function-have-specified-shape-at-every/70333#70333 Answer by Dejan Govc for Can a continuous, nowhere differentiable function have specified "shape" at every point? Dejan Govc 2011-07-14T15:21:33Z 2011-07-14T15:21:33Z <p>I'm new here, so I hope my answer is of any use and not too late.</p> <p>I was wondering: Wouldn't it be perhaps more natural to consider limits of the form</p> <p>$\lim_{y\to x}\frac{f(y)-f(x)}{\phi(y)-\phi(x)}$?</p> <p>If for example we take $f(x) = |x|$ for $x\in \bf{R}$ and $\phi = f$, the "derivative" would be equal to one everywhere, which makes sense, since $f$ and $\phi$ are really the same, therefore their shape should be the same, right?</p> <p>By the previous definition which uses $\phi(y-x)$ in the denominator, we would get limits like</p> <p>$\lim_{y\to x}\frac{|y|-|x|}{|y-x|}$</p> <p>which exists only for $x = 0$, even though our intuition tells us the shape is supposed to be the same everywhere. Any comments?</p>