A continuous function on $[0,1]$ that is not of a bounded veriation Suppose $f$ is continuous on [0, 1]. Must there be a nondegenerate closed subinterval $[a, b]$
of [0, 1] for which the restriction of $f$ to $[a, b]$ is of bounded variation? 
$\mathbf{My\ attempt}:$
Suppose for all $[a,b]\subseteq [0,1]$ with $a\neq b$, we have $f$ restricted to $[a,b]$ is not of a bounded variation, $i.e$ for all $n \in \mathbb{Z^+}$ there exists a partition $\mathcal{P}_{{n}}$ such that $V(f,\mathcal{P_n})\geq n$.
I am not sure if there would be some examples shows that is not true, I am using a contradiction here to conclude that there must be a nondegenerate interval of $[0,1$ such that $f$ restricted to this interval is of a bounded variation. I will appreciate it any help with that. Thank you. 
 A: It is well known that any function of bounded variation is differentiable almost everywhere and that there exists a function that is continuous but differentiable nowhere. For the first statement use the fact that any function of bounded variation is the difference of two monotone functions and monotone functions are differentiable almost everywhere: http://mathonline.wikidot.com/lebesgue-s-theorem-for-the-differentiability-of-monotone-fun
For the second statement go to https://en.wikipedia.org/wiki/Weierstrass_function
A: Hint: it is well known that functions of bounded variation are of the form $f = g-h$ with $g,h$ monotone. In particular, $f$ ought to be differentiable almost everywhere on $[0,1]$. However, functions such as the Weierstraß function (restricted to $[0,1]$) are continuous but differentiable nowhere. In particular it is not differentiable in any proper subinterval and so by the former, it is not of bounded variation there either.
Edit: the original question was modified, and this does not answer the content of the post: what follows is a construction of a continuous function $f : [0,1] \to \mathbb{R}$ which is not of bounded variation in any subinterval $[0,c]$ with $c \leq 1$, but for which $V_c^1f < +\infty$.
First, let's construct a function $[0,1] \to \mathbb{R}$ which is not of bounded variation. 
Consider $f_N : \left[\frac{1}{N+1},\frac{1}{N}\right]\to \mathbb{R}$ with interpolates linearly the points: 
$$
(1/(N+1),0),\left(\frac{1/(N+1)+1/N}{2},1/N\right) \text{ and } (1/N,0)
$$ 
A drawing should illustrate the idea: we are making spikes whose height decreases with $N$, and so does the length of the considered interval. At the endpoint of the interval $f_N$ is zero, and in the middle point it attains its maximum value which is $1/N$. Moreover, $f_N$ is piecewise monotone. Now the functions $(f_N)_{N\geq 1}$ coincide at $\{1/N : N \in \mathbb{N}\}$ and so the function
$$
f(x) = f_N(x) \text{, if } x \in \left[\frac{1}{N+1},\frac{1}{N}\right]
$$
is continuous on $(0,1]$. Technicalities aside, we are just gluing the spikes together and justifying that the resulting function is indeed contiunous. Finally, we would like to extend $f$ to $[0,1]$. A necessary condition is for $f(0)$ to be zero, since we already have $f(1/N) = 0$ and $1/N \to 0$. So let's check that 
$$
f(x) = \cases{f(x) \ x \neq  0 \\ 0 \text{ otherwise}}
$$
is continuous on $[0,1]$. Take $x_n \to 0$ and $\varepsilon > 0$. Since $x_n$ goes to zero, at some point $n_0$ we have that $|x_n| < \varepsilon$ if $n \geq n_0$. Now, if $n \geq n_0$, each $x_n$ is either zero, lies on some interval $\left[\frac{1}{M_n+1},\frac{1}{M_n}\right]$ with $M_n > 1/\varepsilon$, and so by the definition of each $f_N$ then, we have that $|f(x_n)| = f(x_n) \leq 1/M_n < \varepsilon$. In any case, we have that $|f(x_n)| < \varepsilon$ if $n \geq n_0$, which proves that $f$ is continuous at zero.
So far we have only defined $f$. We still ought to show that it is not of bounded variation, despite being continuous. Intuitively this is because the spikes add each a variation of no less that $1/N$ and the harmonic series diverges. Concretely, if $f$ were of bounded variation, we would have that for each $N \geq 1$,
$$
V_0^1f = V_0^{1/(N+1)}f + \sum_{i=1}^NV_{1/(i+1)}^{1/i}f = \dots \\
\dots = V_0^{1/(N+1)}f + \sum_{i=1}^NV_{1/(i+1)}^{1/i}f_i \geq V_0^{1/(N+1)}f + \sum_{i=1}^N\frac{1}{i} \geq \sum_{i=1}^N\frac{1}{i}.
$$
Taking limit when $i \to \infty$ we have $V_0^1f = +\infty$. Moreover, $f$ is not of bounded variation in any interval of the form $[0,c]$: if it were so, by refining partitions and taking limits we would have that
$$
V_0^1f = V_0^cf + V_c^1f,
$$
and since $f|_{[c,1]}$ is of bounded variation (we are interpolating linearly a finite amount of times), then $V_c^1 < + \infty$ which would imply that $V_0^1f < + \infty$, a contradiction.
