Let $T:C[0,1] \to C[0,1]:Tx(t) = \int_0^t x(s)ds$. Is it a contraction? Is any power of T a contraction?

Question

Let $T:C[0,1] \to C[0,1]:Tx(t) = \int_0^t x(s)ds$. Is it a contraction? Is any power of T a contraction?

$T: X \to X$ is a contraction if $\exists \alpha < 1: d(T_x,T_y) \leq \alpha d(x,y),\ \forall x,y \in X$

I couldn't figure out how to take integral so that I can compare it with d(x,y).

Answer 1

No, $T$ is not a contraction; an example has already been given. But it is a "weak contraction", in that $d(Tx,Ty)\le d(x,y)$.

And $T^2$ is a contraction; hence so is $T^k$ for any $k\ge2$, by induction.

Define $||x||=\sup_{t\in[0,1]}|x(t)|$ as usual. Then $$|Tx(t)|\le\int_0^t||x||=t||x||,$$so $$||Tx||\le||x||\sup_{t\in[0,1]}=||x||.$$ And$$|T^2x(t)|\le||x||\int_0^ts\,ds\le||x||/2.$$Since $d(x,y)=||x-y||$ and $T$ is linear this shows that $d(T^2x,T^2y)\le d(x,y)/2$.

Answer 2

Hint:

1.) Consider $x\equiv 1$ and $y\equiv 0$ and compute and compare $d(Tx,Ty)$ and $d(x,y)$ for the first question.

2.) For the second question, I refer to the other answers here...

Answer 3

Since $\|x\|=d(0,x)$ and $T$ is linear (so that $T0=0$), we can talk about norms instead of distances.

Now, $\|Tx\| = \sup\limits_{t\in[0,1]}|(Tx)(t)| = \sup\limits_{t\in[0,1]}\left|\int\limits_0^t x(s)ds\right|$.

Let $x(s)=1$. Then, $(Tx)(t)=\int\limits_0^t 1ds=t$. In this case, $\|x\|=1$ and $\|Tx\|=1$, so $T$ is not a contraction.

For $T^2$, we have $(T^2x)(h)=\int\limits_0^h (Tx)(t)dt=\int\limits_0^h \left(\int\limits_0^t x(s)ds\right)dt$. Then,

$$|(T^2x)(h)| = \left|\int\limits_0^h \left(\int\limits_0^t x(s)ds\right)dt\right| \leq \int\limits_0^h \left(\int\limits_0^t |x(s)|ds\right)dt \leq \\ \leq \int\limits_0^h \left(\int\limits_0^t \|x\|ds\right)dt = \|x\| \cdot \int\limits_0^h \left(\int\limits_0^t 1ds\right)dt = \frac{\|x\|}{2}$$.

Thus, $\|T^2x\| \leq \frac{\|x\|}{2}$, so $T^2$ is a contraction.

I think you can get what happens for $T^k$.

Answer 4

$T^2$ can be simplified using integration by parts: $$ T^2f(x) = \int_{0}^{x}\int_{0}^{y}f(z)dzdy \\ = -\int_{0}^{x}\frac{d}{dy}(x-y)\int_{0}^{y}f(z)dzdy \\ = \int_{0}^{x}(x-y)f(y)dy. $$ In general $$ T^n f(x) = \frac{1}{(n-1)!}\int_{0}^{x}(x-y)^{n-1}f(y)dy $$ So, $$ \|T^nf\|_{\infty} \le \frac{1}{(n-1)!}\int_{0}^{1}(1-y)^{n-1}dy\|f\|_{\infty} = \frac{1}{n!}\|f\|_{\infty}. $$ Therefore $T^n$ is a contraction for $n =2,3,\cdots$. However, $T$ is not a contraction because $T1 =x$, which gives $\|T1-T0\|=1=\|1-0\|$.