proof that the operator of integral equation is a contraction

Question

The part I am reading is similar to the following questions, so if you are not familiar with integral operator and contraction, you could refer to them.

Show integral operator is contraction mapping

Reiterate Volterra integral operator is a contraction

I am reading a part about this.

$h(x),K(x,t)$ are continuous functions on $[0,1]$ and $[0,1]^2$. $|K(x,t)| \le B$. $T[f](x) = \int_0^x K(x,t) f(t)dt+h(x)$. Show that $T$ is a contraction mapping.

The proof:

define a metric: $M$ is a positive real number, $$d_M(f,g)=\sup_{x\in[0,1]}|f(x)-g(x)| e^{-Mx} $$ we can show that $d(T[f],T[g])<B/Md(f,g)$. Then we can show that, if $M>B$, the map $T$ is a contraction on $(C([0,1]),d_M)$ where $C([0,1])$ is the space of all continuous functions whose domain is $[0,1]$.

The book is written in this way. I have typed everything. If it is confusing to you, I could not do anything about it because I am confused. I understand that this is a way to construct a $c\in [0,1)$ to make it satisfies the definition of a contraction.

Assume that everything written in the book is clear to you, my question is I am not sure how to use the $e^{-Mx}$, meaning how to link it with $B/M$. I am also confused whether the metric $d_M$ preserve metric $d$, meaning if $T$ is a contraction on $d_M$ metric, then it is definitely a contraction on $d$ metric, or the other way around, or they are equivalent.

Answer 1

First of all, $T$ is not a contraction mapping on $(C[0,1], d)$. For example, if $h(x)=0$ and $K(x, y) = 1$. Then

$$ T[f](x) = \int_0^x f(s)ds.$$ If $f(x) = c_0$, $g(x) = c_1$ (that is, $f, g$ are constant functions). Then

$$ d (T[f], T[g])= d(c_0 x, c_1 x) = |c_1 - c_0| = d(f, g).$$ So there is no $c\in [0,1)$ so that $d(T[f], T[g]) \le c d(f, g)$ for all $f, g\in C([0, 1])$.

What you can show is that $T$ is a contraction on $C([0,1], d_M)$ when $M>B$. Indeed, for any $f, g\in C([0,1])$,

\begin{align} d_M (T[f], T[g]) &= \sup_{x\in [0,1]} e^{-Mx} \left| \int_0^xK(x, t) (f(t)-g(t)) dt\right| \\ &\le B\sup_{x\in [0,1]} e^{-Mx}\int_0^x |f(t)-g(t)| dt \\ &=B\sup_{x\in [0,1]} e^{-Mx}\int_0^x e^{Mt} e^{-Mt}|f(t)-g(t)| dt \\ &\le B d_M (f, g) \sup_{x\in [0,1]} e^{-Mx}\int_0^x e^{Mt} dt \\ &=\frac BM d_M (f, g) \sup_{x\in [0,1]} e^{-Mx}(e^{Mx}-1) dt\\ &\leq\frac BM d_M (f, g) . \end{align}

Of course, the above calculation also answers your questions: that $T$ is a contraction with respect to $d_M$ does not mean that it is a contraction with respect to $d$. But once you show that $T$ is a contraction with respect to some complete metric, you can use the contraction mapping theorem to conclude that $T$ has a fixed point.