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Let

$T:f\mapsto (x\mapsto \frac{2}{5}\int_0^1 (x^2+t^5)f(t) dt + \sin(x))$ for any $x\in[0,1]$, $f\in C([0,1])$.

I want to show that that there is a uniqu $\tilde{f}$ that solves that equation $f(x)=\frac{2}{5}\int_0^1 (x^2+t^5)f(t) dt + \sin(x)$, i.e. by using the BNF to show there is a $C\in[0,1]$ such that $||Tf-Tg|| for all $f,g\in C([0,1])$ and $x\in[0,1]$

By definition, I get

$||Tf-Tg||\leq\frac{4}{5}||f-g||+\sin(1) \leq \sin(1)(||f-g||+1)$

What am I doing wrong?

  • 0
    From whence $\sin(1)$?2012-10-30

2 Answers 2

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A related problem.

You are doing fine, the $\sin(x)$ should be cancelled and you are left with

$ ||Tf-Tg|| \leq \frac{2}{5}\int_0^1 |x^2+t^5||f(t)-g(t)| dt \leq \frac{4}{5}||f-g||_{\infty}\,$

which proves your operator is an attractive operator on the Banach space you are given.

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very easy ! $|\frac 2 5\int_{0}^{1}(x^2+t^2)f(t)dt-\frac 2 5\int_{0}^{1}(x^2+t^2)g(t)dt|\leq|\frac 2 5\int_{0}^{1}(x^2+t^2)|f(t)-g(t)|dt| \\\leq \max|f(t)-g(t)|\frac 2 5\int_{0}^{1}(x^2+t^2)|dt|\leq\max|f(t)-g(t)|\frac 8 {15}$