# Uniformly bounded variation of the Picard iteration

Let $$I:=[0,1]$$ and $$(X\|\cdot\|)$$ a Banach space. Recall that $$f:I\longrightarrow X$$ is said to be of bounded variation if $$V_{I}(f):=\sup \sum_{i=1}^{n} \| f(t_{i})-f(t_{i-1})\| <+\infty ,$$ where the supremum is taken over all finite $$0=t_{0}<\ldots < t_{n} =1$$ partitions of $$I$$. Likwise, we say that a set $$\mathcal{F}$$ of mappings from $$I$$ into $$X$$ is of uniformly bounded variation if there is $$M\geq 0$$ such that $$V_{I}(f)\leq M$$ for all $$f\in \mathcal{F}$$.

Now, consider $$T:C(I,X)\longrightarrow C(I,X)$$ continuous, where $$C(I,X)$$ is the Banach space of the continuous mappings from $$I$$ into $$X$$, and the Picard iteration

$$f_{n}:=T\big( f_{n-1} \big) ,$$

where $$f_{0}\in C(I,X)$$ of bounded variation has been prefixed. Then, Under that conditions the above sequence is of uniformly bounded variation?

It seems clear that if $$T$$ is Lipschitzian with Lipschitz constant less or equal than one, then the Picard iteration is a sequence of uniformly bounded variation.

• That Lipschitz condition may be sharp. I mean, if $T$ has a Lipschitz constant bigger or equal to one, then I conjecture that $f_n$ needs not be of uniformly bounded variation. This is easily proved if the Lipschitz constant is strictly bigger than one; just consider $Tf=2f$ and $f_0(x)=x$, so that $f_n$ has variation $2^n$, not uniformly bounded. The interesting question is what happens for $T$ with Lipschitz constant exactly one. – Giuseppe Negro Nov 8 at 16:41
• Also $f_0$ has to be of bounded variation. – daw Nov 8 at 21:04
• The iteration requires that $T$ is a map from $X$ to $X$. otherwise $(Tf)(t)$ has no relation to $f(t)$. – daw Nov 8 at 21:16
• Thanks for your observation. The notation was confusing (even incorrect), I have edited: $f_{n}:=T(f_{n-1})$ and so $T$ is defined in $C(I,X)$ – user123043 Nov 9 at 8:29
Let $$f_0$$ be of bounded variation, $$T:X\to X$$ be Lipschitz with Lipschitz constant $$\le 1$$, i.e., $$T$$ is a contraction. Then $$\sum_i \|(Tf)(t_i)-(Tf)(t_{i-1})\| \le \sum_i \|f(t_i)-f(t_{i-1})\|.$$ Hence $$f_0$$ is of bounded variation, then the sequence $$(f_k)$$ is of uniformly bounded variation.