Is the operator $Tf = f(x) - f(x) \int_0^1 f(y) dy$ a contraction?

Question

Is the operator $$Tf = f(x) - f(x) \int_0^1 f(y) dy$$ in $C[0,1]$ (with the uniform norm) a contraction and what is the possible fixed point?

Answer 1

Hint 1. Note that $Tf=f$ if and only if $f(x) \int_0^1 f(y) dy=0$ for all $x\in [0,1]$.

Hint 2. We have that $$|(Tf)(x)|= |f(x) - f(x) \int_0^1 f(y) dy|=|f(x)|\left|1-\int_0^1 f(y) dy\right|$$ so, for example, if $f=3$ and $g=0$ then $$\|(Tf)-(Tg)\|=\left|1-3\right|\|f-g\|=2\|f-g\|$$ where $\|\cdot\|$ is the uniform norm in $C[0,1]$.

Hint 3. Is $T$ a contraction in the closed ball $\{f\in C[0,1]: \|f\|\leq r \}$ for some $r>0$?