Let $H$ be a separable Hilbert space and $f:[0,1] \to H$ be a norm-decreasing map in the sense that \begin{equation} \lVert f(t_2) \rVert_H \leq \lVert f(t_1) \rVert_H \end{equation} if $0 \leq t_1 \leq t_2 \leq 1$.
Suppose further that $f$ is right-continuous in the sense that \begin{equation} \lVert f(t+\epsilon)-f(t) \lVert_H \to 0 \text{ as } \epsilon \to 0^+ \end{equation} for each $t \in [0,1)$.
Then, I wonder if $f$ is of bounded variation with respect to the norm of $H$. That is, do we have \begin{equation} \sup\Bigl\{ \sum_{i=1}^n \bigl\lVert f(t_i) -f(t_{i-1})\bigr\rVert_H \mid n \in \mathbb{N}, 0=t_0 < \cdots < t_n=1 \Bigr \} < \infty? \end{equation}
I cannot find a way myself to bound the sum inside the supremum by using the above condiitons since the norm of $H$ is not necessarily an absolute value...
Could anyone please help me?
It suffices to come up with a continuous function $f:[0,1]\to \mathbb{T}$ with unbounded variation. Then we can take $\mathcal{H}=\mathbb{C}.$ For $h(t)=t\sin (t^{-1})$ we have $|h(t)|\le 1.$ Let $$ f(t)=h(t)+i\sqrt{1-h^2(t)}$$ Then $f$ is continuous and has unbounded variation as its real part has unbounded variation.