Use contraction mapping theorem to show that the integral equation has a unique continuous solution on $t \in [0,3]$

Question

I have to use the contraction mapping theorem to prove that the integral equation with continuous functions $K(t,s)$ and $f(t)$,

$\begin{equation*} x(t) = \lambda \int_{0}^{3} K(t,s) x(s)ds + f(t), \, 0 \leq t \leq 3 \end{equation*}$

has a unique continuous soltuion on $t \in [0,3]$ for sufficiently small values of real parameter $\lambda$. Also, I need to find the interval of the admissible values of $\lambda $ if $\displaystyle \sup_{0 \leq t, \, x \leq 3}|K(t,s)|=A$.

The following is what I have done so far. What I am worried about is that I did not use the Cauchy-Schwarz inequality correctly. Also, I was told that the bounds on $\lambda $ should be $|\lambda|\leq \frac{1}{3A}$, but I got only $\frac{1}{A}$, and I'm not sure where the $3$ should have popped out from (the length of the interval, perhaps?). Please let me know what is wrong with this, and how to fix it:

Suppose we have $x(t) = \lambda \int_{0}^{3}K(t,s)x(s)ds + f(t)$ with all the conditions given above.

Then, let $T(x(t))=\lambda \int_{0}^{3} K(t,s)x(s) ds + f(t)$.

So, $\Vert T_{x_{1}} - T_{x_{2}}\Vert_{\infty} = |\lambda| \Vert \int_{0}^{3}(K(t,s)x_{1}(s)-K(t,s)x_{2}(s))ds\Vert_{\infty}\\ = |\lambda| \Vert \int_{0}^{3}(K(t,s)x_{1}(s)ds-\int_{0}^{3}K(t,s)x_{2}(s)ds\Vert_{\infty} \\ = |\lambda|\Vert \langle K(t,s),x_{1}(s)\rangle - \langle K(t,s),x_{2}(s)\rangle \Vert_{\infty} \\ = |\lambda| \Vert\langle K(t,s), x_{1}(s)-x_{2}(s) \rangle \Vert_{\infty} \\ \leq |\lambda| \Vert K(t,s) \Vert_{\infty} \Vert x_{1}(s)-x_{2}(s)\Vert_{\infty} \,\,\text{(by Cauchy-Schwarz)} \\ = |\lambda| \cdot A \cdot \Vert x_{1}(s)-x_{2}(s)\Vert_{\infty},\, \text{which is a contraction mapping if}\,|\lambda|A \leq 1 \implies |\lambda| \leq \frac{1}{A}$.

I don't know where the $3$ in the denominator is supposed to come from.

Answer 1

Indeed, you didn't apply Cauchy-Schwarz, but a strange (and false) "version". Cauchy-Schwarz would tell you that

$$\left| \int_0^3 K(t,s) x(s) ds \right| \leq \| K(t,.) \|_2\|x\|_2$$

That is not very interesting here.

What you want to apply here is Hölder inequality :

$$\left| \int_0^3 K(t,s) x(s) ds \right| \leq \|x\|_\infty\int_0^3 | K(t,s) |ds$$

Then, you use the majoration of $|K(t,s)|$ to have

$$\leq \|x\|_\infty\int_0^3 \| K \|_\infty ds$$

And the $3$ appear

$$\leq 3 \|x\|_\infty \| K \|_\infty ds$$

PS : Ok, Hölder is not really required , as the majoration $\int|f(s)||g(s)|ds \leq \int \|f\|_\infty \|g\|_\infty ds$ comes directly from the basic properties of the integral