Proving continuity of the identity among several norms for the space of bouden variation functions. A function $f:[a,b] \to  \mathbb{R}$ it is said is of bounded variation if there is $K \geq 0$ such that for every partition  $a=x_0 \leq x_1 \leq \cdots \leq x_n=b$, $\sum_{i=1}^n|f(x_i) - f(x_{i-1})|\leq K$. We define $\operatorname{VT}(f)$ as the infimum of the set of all $K \geq 0$ such that  $\sum_{i=1}^n|f(x_i) - f(x_{i-1})|\leq K$ for every partition $P$ of $[a,b]$. Now lets denote as $\operatorname{VA}[a,b]$ the set of all bounded variation functions  $f:[a,b] \to \mathbb{R}.$ Consider the following norms of $VA[0,1]$:
$$\|f\|_1:= \sup \lbrace |f(x)|: x \in [0,1] \rbrace,$$
$$\|f\|_2:= \|f\|_1+VT(f),$$
$$\|f\|_3:= |f(0)|_1+VT(f).$$
Let $VA_{1}$, $VA_{2}$ and $VA_{3}$ be three normed spaces where  $VA[0,1]$ is the vector space for this normed and space, and $\|\cdot\|_1, \| \cdot \|_2 $ and $\|\cdot\|_3$ the norms respectively. For $i,j \in \lbrace 1,2,3\rbrace$, consider the identity function $\operatorname{id}_{i,j}:VA_i \to VA_j$. Prove the following:
(1) Show that $VT(f)= \sup \lbrace \sum_{i=1}^n|f(x_i) - f(x_{i-1})|: n \in \mathbb{N}; \text{ and } 0= x_0 < x_1 < \cdots < x_n=1 \rbrace,$
(2) $id_{1,2}$ is not continuous,
(3) $id_{2,3}$ is continuous,
(4) $id_{3,2}$ is continuous.
For (1), lets rename $A=\sup \lbrace \sum_{i=1}^n|f(x_i) - f(x_{i-1})|: n \in \mathbb{N}; \text{ and } 0= x_0 < x_1 < \cdots < x_n=1 \rbrace$. On the other hand, lets take $K'=VT(f)$ which by definition is the smaller bound of all $K \geq 0$ such that $\sum_{i=1}^n|f(x_i) - f(x_{i-1})| \leq K$, but since $A$ is a supremum for all this sums, we have that $A \leq K'$. However, I cant prove $K' \leq A$. For (2),(3) and (4) I was thinking about considering the metrics that came from this norms, lets say $d_1,d_2$ and $d_3$. And consider a sequences of function such that $d_i(f_n,f) \to 0$ as $n$ approaches to infinity. Then showing that $d_j(id_{i,j}(f_n),id_{i,j}(f))$ converges or not to $0$ as $n$ approaches to infinity.
 A: For $(1)$, you are almost there. You have observed that $\sum_{i=1}^n|f(x_i)-f(x_{i-1})| \le K$ for any such sum, and so taking suprema, you have $A\le K$. Taking infimum of the $K$, you get $A\le K'$. As 0XLR points out in the comments, the inequality $K'\le A$ is even easier, as $K=A$ is an upper bound of all the sums.
For $(2)$, $(3)$, and $(4)$, you want to use the following fact:

Let $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$ be two normed spaces, and $T:X\to Y$ a linear map. Then $T$ is continuous with respect to $\|\cdot\|_X$ and $\|\cdot\|_Y$ if and only if there exists $C>0$ such that $\|Tx\|_Y\le C\|x\|_X$ for all $x\in X$.

If you are not familiar with this fact, prove it! In this case, $X=Y=VA[0,1]$, $T=\text{id}$, and the norms are given in each individual problem.
The other helpful thing to note is that, if $f$ is continuously differentiable, $VT(f)$ represents the arc length of the graph of $f$. So for $(2)$, you can show that $\frac{\|f\|_2}{\|f\|_1}$ is unbounded, and so in particular $\text{id}_{1,2}$ is discontinuous, by finding a sequence bounded functions whose arc length is unbounded. Explicitly, consider
$$f_n(x):=\sin(nx).$$
Obviously, $\|f_n\|_1=1$ for all $n\ge2$. Consider the partition $0=x_0<\ldots<x_N=1$ given by $x_i=\frac{\pi i}{2n}$ for $i<N:=\lfloor\frac{2n}\pi\rfloor + 1$. Then $|f_n(x_i)-f_n(x_{i-1})|=1$ for all $i<N$, and so
$$\|f_n\|_2\ge VT(f_n)\ge\sum_{i=1}^N|f_n(x_i)-f_n(x_{i-1})| \ge \frac{2n}\pi-1,$$
which is unbounded. Hence there cannot exist $C>0$ such that $\|f\|_2\le C\|f\|_1$ for all $f\in VA[0,1]$, and so $\text{id}_{1,2}$ is discontinuous.
For $(3)$, we obviously have $\|f\|_3\le\|f\|_2$, and so this is trivial. For $(4)$, observe that for any $x\in[a,b]$, $|f(x)-f(0)|\le VT(f)$, and so in particular $|f(x)|\le|f(0)|+VT(f)$. Taking supremum, we get $\|f\|_1 \le |f(0)|+VT(f)$, and so
$$\|f\|_2=\|f\|_1+VT(f)\le|f(0)|+2VT(f) \le 2\|f\|_3,$$
showing $(4)$ and finishing the proof.
