Let $(C[0, 1], d_∞)$ be the metric space of continuous functions on $[0, 1]$ , where the distance function is defined by $$d_∞(f,g)= \sup_{x\in [0,1]}|f(x)-g(x)|$$
Consider the function $T:(C[0, 1], d_∞)\rightarrow (C[0, 1], d_∞)$ defined by
$$(Tf)(x):=\int_{0}^{x}f(t)dt$$
Prove that $T$ is not a contraction, i.e. there does not exist $0 < K < 1$ such that
$$d_∞(Tf,Tg)\leq K\cdot d_∞(f,g)$$ holds for any $f, g ∈ C[0, 1]$ .
Well, this problem asks to prove it, but I could disprove it (in a wrong way, but I don't know what is wrong).
My solution
- $d_∞(Tf,Tg)= \sup_{x\in [0,1]}|Tf(x)-Tg(x)|= \sup_{x\in [0,1]}|\int_{0}^{x}f(t)dt-\int_{0}^{x}g(t)dt|= \sup _{x\in [0,1]}\left | \int_{0}^{x}\left \{ f(t)-g(t) \right \}dt \right |$
Then, for some $a\in [0,1]$, $\sup _{x\in [0,1]}\left | \int_{0}^{x}\left \{ f(t)-g(t) \right \}dt \right |=\left | \int_{0}^{a}\left \{ f(t)-g(t) \right \}dt \right |$
- $d_∞(f,g)= \sup_{x\in [0,1]}|f(x)-g(x)|$
Then, for some $b\in [0,1]$, $\sup_{x\in [0,1]}|f(x)-g(x)|=|f(b)-g(b)|$
Note that for some partition $P$, $$d_∞(Tf,Tg)= \left | \int_{0}^{a}\left \{ f(t)-g(t) \right \}dt \right |\leq \int_{0}^{a}\left | f(t)-g(t) \right |dt=U(|f-g|)=U(|f-g|,P)$$ $$=\sum_{k=1}^{n}(t_{k-1}-t_{k}) \sup_{x\in [t_{k-1},t_{k}]}\left | f(x)-g(x) \right |\leq a\left | f(c)-g(c) \right | \; \text {for some c}\in [0,a]$$ $$\leq a\left | f(b)-g(b) \right |= a d_∞(f,g)$$
Thus, there exist $K$ since $a$ is arbitrary number in $[0,1]$
Why this is a wrong disproof?
P.S. I haven't learned about the Banach space (which I've seen in other proof.)