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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.

Many thanks in advance for your comments.

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  • $\begingroup$ 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. $\endgroup$ – Giuseppe Negro Nov 8 at 16:41
  • $\begingroup$ Also $f_0$ has to be of bounded variation. $\endgroup$ – daw Nov 8 at 21:04
  • $\begingroup$ The iteration requires that $T$ is a map from $X$ to $X$. otherwise $(Tf)(t)$ has no relation to $f(t)$. $\endgroup$ – daw Nov 8 at 21:16
  • $\begingroup$ 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 $C(I,X)$ $\endgroup$ – user123043 Nov 9 at 8:29
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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.

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  • $\begingroup$ Thanks for your reply. Yes, under these conditions, the Picard iteration is of uniformly bounded variation but, There are more general conditions for T? $\endgroup$ – user123043 Nov 9 at 8:24

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