Proof regard to the function $ f\left(x\right)=\begin{cases} x^{2}\sin\left(\frac{1}{x^{2}}\right) & x\neq0\\ 0 & x=0 \end{cases} $

Question

For a function $ f $ which defined on an interval $[a,b]$ with partition $ P={x_0,x_1,...,x_n} $ we define the variation in the function to be

$ V\left(f,P\right)=\sum_{i=0}^{n-1}|f\left(x_{i+1}\right)-f\left(x_{i}\right)| $

We define the general variation in the interval $[a,b]$ to be

$ V_{a}^{b}\left(f\right)=Sup\{V\left(f,P\right)|P\} $ When P is a partition of [a,b].

If $ V_{a}^{b}\left(f\right) $ exists, we say that $ f $ is of bounded bounded

Prove that

$ f\left(x\right)=\begin{cases} x^{2}\sin\left(\frac{1}{x^{2}}\right) & x\neq0\\ 0 & x=0 \end{cases} $

Is of unbounded variation

So, here's what I tried.

We know that $ f'\left(x\right)=\begin{cases} 2x\sin\left(\frac{1}{x^{2}}\right)-\frac{2}{x}\cos\left(\frac{1}{x^{2}}\right) & x\neq0\\ 0 & x=0 \end{cases} $ is the deriviative, and the deriviative is not bounded, and countinious in the segment $ (0,1] $

So let $ M>0 $ be some real number, and we'll prove that exists partition P of $[a,b]$ such

$ V\left(f,P\right)>M $

That would be enough to prove what we want. So, since $ f'(x) $ is'nt bounded, we can find $ x_{0}\in(0,1]$ such that $f'(x_0)>M+1 $. let $ 0<\varepsilon<1 $ . since $ f'(x) $ countinious in $ (0,1] $ exists $ \delta>0 $ such that for any $ x\in\left(x_{0}-\delta,x_{0}+\delta\right) $ it follows that

$ M<f'\left(x_{0}\right)-\varepsilon<f'\left(x\right) $

Now let $ P=\{0,x_{0}-\frac{\delta}{2},x_{0}+\frac{\delta}{2},1\} $ be partition of $ [0,1] $. So :

$ V\left(f,P\right)=|f\left(x_{0}-\frac{\delta}{2}\right)-f\left(0\right)|+|f\left(x_{0}+\frac{\delta}{2}\right)-f\left(x_{0}-\frac{\delta}{2}\right)|+|f\left(1\right)-f\left(x_{0}+\frac{\delta}{2}\right)| $

Now what I want to do is to argue that from Lagrange theorem, exists $ \theta_{i}\in(x_{0}-\frac{\delta}{2},x_{0}+\frac{\delta}{2}) $ such that $ |f\left(x_{0}+\frac{\delta}{2}\right)-f\left(x_{0}-\frac{\delta}{2}\right)|=\delta\cdot|f'\left(\theta_{i}\right)| $

And since $ \theta_{i}\in(x_{0}-\frac{\delta}{2},x_{0}+\frac{\delta}{2}) $ then $ f'\left(\theta_{i}\right)>M $

But, I dont know how to fix the M so it would "cancel" the $ \delta $ because $ \delta $ depends on $ x_0 $.

Any ideas would be helpful. Thanks in advance

Answer 1

Consider the partition $P_n$ with points

$$0 < \frac{1}{[\frac{\pi}{2}+ 2n\pi]^{1/2}} < \frac{1}{[\frac{\pi}{2}+ (2n-1)\pi]^{1/2}} < \ldots < \frac{1}{[\frac{\pi}{2}+ \pi]^{1/2}} < \frac{1}{[\frac{\pi}{2}]^{1/2}} < 1$$

We have

$$\begin{align}V(f,P_n) &\geqslant \sum_{k=2}^{2n}\left|\frac{1}{\frac{\pi}{2}+ (k-1)\pi}\sin\left(\frac{\pi}{2}+ (k-1)\pi\right) - \frac{1}{\frac{\pi}{2}+ k\pi}\sin\left(\frac{\pi}{2}+ k\pi\right)\right|\\ &= \sum_{k=2}^{2n}\left|\frac{1}{\frac{\pi}{2}+ (k-1)\pi}(\pm1) - \frac{1}{\frac{\pi}{2}+ k\pi}(\mp 1)\right|\\ &= \sum_{k=2}^{2n}\left[\frac{1}{\frac{\pi}{2}+ (k-1)\pi} + \frac{1}{\frac{\pi}{2}+ k\pi}\right]\\&\geqslant \sum_{k=1}^{2n}\frac{1}{\frac{\pi}{2}+ k\pi}\end{align}$$

The RHS diverges to $+\infty$ as $ n \to \infty$ as a harmonic series which implies that

$$V_0^1(f) = \sup_P V(f,P) = +\infty$$