If the function $f$ is defined by
$$f(x) = \begin{cases}x^2 \sin(\frac{1}{x^2}) & x \ne 0 \\
0 & x = 0
\end{cases}
$$
then $f$ is differentiable and $f'$ is unbounded on $[-1,1]$.
I know this is true. How should I go about proving this?
[Math] How to show that $f(x) = x^2 \sin(1/x^2)$ with $f(0)=0$ is differentiable and $f’$ is unbounded on $[-1,1]$
real-analysis
Related Solutions
Here's an intuitive way of thinking about the problem.
(1) The $x^2$ on the outside causes the function to vanish rapidly, but the $1/x^2$ inside the sine function causes the oscillation to be similarly rapid. This balance turns out to be just enough to produce unbounded variation, as the variation behaves similarly to the harmonic series. How so?
It will suffice to consider the interval $[0,1]$. $\sin(1/{x}^2)= 0$ when
\begin{equation}x = \frac{1}{\sqrt{\pi n}}\end{equation}
and $\sin(1/x^2) = 1$ when
\begin{equation}x = \frac{\sqrt{2/\pi}}{\sqrt{4n+1}}\end{equation}
(in both cases, make sure that the denominators are not zero)
Now, "throw" these points into a partition. In other words, create a sequence of partitions containing these values for greater and greater $n$. Now, compute the variation. For the points of the form
\begin{equation}x = \frac{1}{\sqrt{\pi n}}\end{equation}
the entire expression vanishes. Thus the variation just becomes the summation of $x^2$ at points of the form
\begin{equation}x = \frac{\sqrt{2/\pi}}{\sqrt{4n+1}}\end{equation}
which is
\begin{equation}x = \sum_{n=n_{0}}^{k} \frac{2/\pi}{4n+1}\end{equation}
which, like the harmonic series, diverges as $k \to \infty$.
(2) In this case, the function vanishes at a speed faster than which it oscillates. This will give us bounded variation, in the form similar to that of the convergent sum $\sum 1/n^2$.
It will suffice to consider the interval $[0,1]$, as the mirror case is identical.
$\sin(1/x)= 0$ when
\begin{equation}x = \frac{1}{\pi n}\end{equation}
and $\sin(1/x) = 1$ when
\begin{equation}x = \frac{2}{\pi (4n+1)}\end{equation}
again making sure that the denominator is nonzero. Using the same technique as before, we construct a sequence of partitions where the $\sin(1/x)$ term either vanishes or equals one. The variation (of a particular partition in the sequence) is then the following sum
\begin{equation}x = \sum_{n=n_{0}}^{k} \frac{4}{\pi^2 (4n+1)^2}\end{equation}
which converges as $k \to \infty$ like $\sum 1/n^2$.
This technique can be extended to the "general" case of $x^{k}\sin(1/x^{n})$ very easily, and provides an interesting parallel between the vanishing/oscillation speeds of this function, and summations of the form $\sum 1/x^{m}$.
1) Your function is clearly differentiable at $t$ if $t\ne 0$, since cosine and $t^2$ are differentiable. To show differentiability at $t=0$ use the limit-of-the-difference-quotient definition of the derivative, and use the fact that $|\gamma(t)|\le t^2$. This will quickly show that the derivative at $t=0$ is defined and is zero.
2) Consider that $\gamma(t)$ oscillates between $t^2$ and $-t^2$ infinitely many times in any open interval containing zero. Find the points where $\gamma(t)$ reaches those values, and consider the sum of the line segments between those consecutive points. If $a<0<b$ then the length of the line segment between $(a,\gamma(a))$ and $(b,\gamma(b))$ is at least $|\gamma(a)|+|\gamma(b)|$. You can show the sums of the lengths of those segments is infinite. That will happen since there are "so many points" where $\gamma(t)$ reaches those high and low points.
Answering a comment to this answer: You want to find the points where the $y$-coordinate of $\gamma(t)$ oscillates the most, i.e. where $\cos\frac{\pi}{t^2}=\pm 1$, which is
$$t=\pm\sqrt{\frac 1k}\quad\text{for $k\in\Bbb Z^+$}$$
You find infinitely many of these points in any open interval containing zero, and each one adds at least the value
$$\left|t^2\cos\frac{\pi}{t^2}\right|=\frac 1k$$
to the length of any arc between zero and this $t$. And you already know that the series
$$\sum_{k=n}^{\infty}\frac 1k$$
diverges. Thus, the curve from any such $t$ to zero is not rectifiable.
Answering a comment to a comment to the main question: The derivative at $t=0$ is not $\lim_{h\to 0} \frac{0^2 \cos (\frac{\pi}{0})-0}{0^2}=\frac{0}{0}$ but rather
$$\lim_{h\to 0}\frac{(0+h)^2\cos\frac{\pi}{(0+h)^2}-0}{h}$$
And you can show that the absolute value of the expression inside the limit is less than
$$\left|\frac{h^2}{h}\right|=|h|$$
so the limit, hence the derivative, is zero.
Best Answer
The cool point of the problem is that although for any positive value $M$ and positive $\epsilon$ we can find some value of $x$ with $|x|<\epsilon$ such that $|f'(x)| > M$, yet $f$ is differentiable at $x=0$.
In fact, $f'(0) = 0$, and we can see that by applying the definition of derivative: $$ \left. \frac{df(x)}{dx} \right|_{x=0} = \lim_{h\rightarrow 0} \frac{h^2 \sin\frac{1}{h^2}-f(0)}{h} = \lim_{h\rightarrow 0} {h \sin\frac{1}{h^2}}-0 =0 $$ since $|\sin\frac{1}{h^2}|$ is bounded by 1 and that is being multiplied by $h$.
On the other hand, look at the expression for $f'(x)$ when $x \neq 0$. At points where $\frac{1}{x^2} = (n+\frac{1}{2})\pi$,
$$ |f'(x)| > \frac{2}{x} - 2x $$ and this is unbounded as $x$ approaches zero.