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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 $||Tf-Tg||<C||f-g||$ 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?

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  • $\begingroup$ Where does the $\sin(1)$ appear? You have $Tf - Tg = \int \ldots +\sin - \int \ldots - \sin = \int (\ldots - \ldots)$ $\endgroup$ – martini Oct 30 '12 at 14:58
  • $\begingroup$ From whence $\sin(1)$? $\endgroup$ – M. Strochyk Oct 30 '12 at 15:04
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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 ! $$|\frac 2 5\int_{0}^{1}(x^2+t^2)f(t)dt-\frac 2 5\int_{0}^{1}(x^2+t^2)g(t)dt|\leq|\frac 2 5\int_{0}^{1}(x^2+t^2)|f(t)-g(t)|dt| \\\leq \max|f(t)-g(t)|\frac 2 5\int_{0}^{1}(x^2+t^2)|dt|\leq\max|f(t)-g(t)|\frac 8 {15}$$

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