Let $f\in C^1[0,1]$. For a partition $\mathcal {P} :0=a_0<a_1<\cdots <a_n=1$ , define $$S(\cal P)=\sum_{k=1}^n|f(x_k)-f(x_{k-1})|.$$Compute the supremum of $S(\cal P)$ over all possible partitions of $[0,1]$.
Here , $\displaystyle \sup_{\cal P}S(\cal P)$ is the total variation of he function $f$ in $[0,1]$. But if the function is unknown how we find the total variation ?
I saw this but it is NOT clear .
Best Answer
Given any subinterval $[x,y]\subset [0,1]$ one has $|f(y)-f(x)|\leq \int_x^y |f'(t)|\>dt$. Therefore we at once obtain for any partition ${\cal P}$ of $[0,1]$ the estimate $$S({\cal P}):=\sum_{k=1}^n|f(x_k)-f(x_{k-1})|\leq\int_0^1|f'(t)|\>dt\ .$$ Since the right hand side does not depend on the chosen ${\cal P}$ we can can conclude that $$V_{[0,1]}(f):=\sup_{\cal P}S({\cal P})\leq\int_0^1|f'(t)|\>dt\ .\tag{1}$$ In reality we have equality here: Looking at the graph of a reasonable $f$ we immediately realize that $V_{[0,1]}(f)$ is the sum of all "infinitesimal changes" of $f$ added up with a positive sign. For the proof, however, we need the continuity of $f'$ over the whole interval $[0,1]$.
Given an $\epsilon>0$ there is a partition ${\cal P}$ with $|f'(x)-f'(y)|\leq \epsilon$ for any two points $x$, $y$ in the same subinterval of ${\cal P}$. Using this partition ${\cal P}$ we have for each subinterval $[x_{k-1},x_k]$ the estimate $$\eqalign{\int_{x_{k-1}}^{x_k}|f'(t)|\>dt&\leq\bigl( |f'(x_k)|+\epsilon\bigr)(x_k-x_{k-1})\cr &=\left|\int_{x_{k-1}}^{x_k} f'(x_k)\>dt\right|+\epsilon(x_k-x_{k-1})\leq\left|\int_{x_{k-1}}^{x_k} f'(t)\>dt\right|+2\epsilon(x_k-x_{k-1})\cr &=|f(x_k)-f(x_{k-1})|+2\epsilon(x_k-x_{k-1})\ .\cr}$$ Summing over $k$ we obtain $$\int_0^1|f'(t)|\>dt=\sum_{k=1}^n\int_{x_{k-1}}^{x_k}|f'(t)|\>dt\leq S({\cal P})+2\epsilon\leq V_{[0,1]}(f)+2\epsilon\ .$$ Since $\epsilon>0$ was arbitrary this together with $(1)$ proves that indeed $$V_{[0,1]}(f)=\int_0^1|f'(t)|\>dt\ .$$ The above argument is valid for vector-valued functions $f$ as well, and proves that the "geometrically defined" length of a $C^1$-curve in ${\mathbb R}^n$ is given by the well-known formula.