Finding the fixed point of $T(\phi)=\int_{0}^{x} \phi (t) dt$

Question

Define $T: C[0, 1]\rightarrow C[0, 1]$ as follows: for $\phi \in C[0, 1]$

$$T(\phi)=\int_{0}^{x} \phi (t) dt$$

How to show that $T$ is not a contraction but have a fixed point.

Thought: $$\begin{align} \bigl\lvert(Tf)(x)-(Tg)(x)\bigr\rvert &= \biggl\lvert \int_{0}^{x}f(t)dt-\int_{0}^{x}g(t)dt\biggr\rvert \\ &\le d_\infty(f,g) \end{align}$$ But why it is not a contraction. and is that zero function is fixed point. Please help.

Answer 1

Assume that $T$ is contraction then there exists $0<k<1$ such that, for all $f\in C(0,1)$

$$\|Tf\|_\infty \le k\|f\|_\infty $$

For $f\equiv 1$ we have $T1 =1$ $$ 1= \|T1\|_\infty \le k\|1\|_\infty =k<1 $$ which is absurd .

Hence $T$ is not a Contraction.

Now Let show the existence of fix point Obvioulsy we have $$|Tf(x)|=\int_{0}^{x} \|f\|_\infty dt \le x\|f\|_\infty$$

Therefore, Since $0\le x\le 1$ we have $$|T^2f(x)| \le\int_{0}^{x} \left| Tf(t) \right| dt \le\int_{0}^{x} t\|f\|_\infty dt \le \frac{x^2}{2}\|f\|_\infty \le \frac{1}{2}\|f\|_\infty $$

that is, for all $f\in C(0,1)$

$$\|T^2f\|_\infty \le \frac{1}{2}\|f\|_\infty $$

This meas that $T^2$ has a unique fix point $\phi$

But $$ \phi = T^2\phi\implies T\phi = T^2(T\phi)$$

**this means that $T\phi$ is also a fix point of $T^2$ and by unicity of $\phi** $ we have, $$\color{blue}{T\phi =\phi}$$

Answer 2

If $T\phi=\phi$ , you get $$ \phi(x)=\int_0^x\phi(t)\,dt. $$ Since $\phi$ is continuous, the right-hand-side tells you that $\phi$ is differentiable. After differentiating, you get $$ \phi'=\phi. $$ So $\phi(x)=ce^x$, but since $T\phi(0)=0$, we get $c=0$ and $\phi=0$. So $T$ does not have a fixed point (counting zero as a fixed point is not very interesting, as in that case every linear map has a fixed point).

As for $T^2=T$, that clearly false since $T$ maps constants to non-constants; as mentioned by José Carlos Santos, $T1=x$, while $T^21=x^2/2$.

The map $T$ is a contraction, since $$ |T\phi(x)|=\left|\int_0^x\phi\right|\leq\int_0^x|\phi|\leq x\,\|\phi\|\leq\|\phi\|. $$ It is not a strict contraction, since for instance $\|T1\|=1$.