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Be $\lambda \in \mathbb{R}$ such that $\left | \lambda \right |> \left \| \kappa \right \|_{\infty }(b-a)$.

Prove that the solution $f^*$ of the integral equation of Fredholm $$\lambda f -\int_{a}^{b}\kappa (x,y)f(y)dy=g(x)$$ for all $x\in [a,b]$ satisfies $$\left \| f^*-\sum_{m=1}^{k}\frac{1}{\lambda ^m}\Im ^{m-1}g \right \|_{\infty }\leq \frac{\alpha ^{k}}{(1-\alpha) \left | \lambda \right |} \left \| g \right \|_{\infty }$$ for all $k\in \mathbb{N}$ where $$\alpha :=\frac{\left \| \kappa \right \|(b-a)}{\left |\lambda \right | }$$

i´m really stuck in this problem, I know that I can use that $$\Im :C_{\infty }^{0}[a,b]\rightarrow C_{\infty }^{0}[a,b]$$ is Lipschitz continuous and that linear Fredholm´s operator is linear, can anybody just give a hint please? thanks!

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  • $\begingroup$ I've edited your post to hopefully fix the MathJax - it just needed dollar signs (\$) inserted around the code. Please verify that it's correct. $\endgroup$ – user61527 Oct 9 '13 at 1:42
  • $\begingroup$ after $"for all x\in[a,b] satisfies"$ in the right side of the inequality, the denominator should be $"((1-\alpha )\left | \lambda \right |)"$ $\endgroup$ – k73586 Oct 9 '13 at 1:51
  • $\begingroup$ It should be fixed now. $\endgroup$ – user61527 Oct 9 '13 at 1:52
  • $\begingroup$ This is an interesting problem. Where does it come from? $\endgroup$ – Antonio Vargas Oct 9 '13 at 2:13
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Since $\lambda \ne 0$, we may write \begin{align*} f(x) - \frac{1}{\lambda}\int_{a}^{b} \kappa(x,y) f(y)\, \mathrm{d}y = \frac{g(x)}{\lambda} \end{align*} Let $\Im\colon C^{\infty}_{0} \to C^{\infty}_{0}$ be defined by \begin{align*} \Im[f](x):=\int_{a}^{b} \kappa(x,y) f(y) \, \mathrm{d}y \end{align*} Then by standard integral inequalities \begin{align*} \frac{\left|\Im[f](x) \right|}{\lambda} \le \frac{\left\|\kappa\right\|_{\infty}}{\lambda}(b-a)\left\|f\right\|_{\infty} = \alpha\left\|f\right\|_{\infty} <\left\|f\right\|_{\infty}, \end{align*} which shows that $\left\|\Im \right\|/|\lambda|\le \alpha < 1$, i.e, that $\Im/\lambda$ is a contractive linear operator on the Banach space $C_{0}^{\infty}$.

Symbolically, we can now write the equation as \begin{align*} (I-\Im/\lambda)f = g/\lambda. \end{align*} In elementary algebra, we would write $(1-x)^{-1} = \sum_{k=0}^{\infty} x^{k}$ if $|x|<1$. A similar identity holds here: \begin{align*} (I-\Im/\lambda)\sum_{m=0}^{k} \Im^{m}/\lambda^{m} &= I-(\Im/\lambda)^{k+1} \to I \text{ as } k\to \infty \end{align*} as $\left\|\Im\right\|_{\infty}/|\lambda| <1$. Hence we may write \begin{align*} (I-\Im/\lambda)^{-1} = \sum_{m=0}^{\infty} \frac{\Im^{m}}{\lambda^{m}} \end{align*} Applying this operator gives \begin{align*} f &= (1-\Im/\lambda)^{-1}g/\lambda = \sum_{m=0}^{\infty} \frac{\Im^{m}}{\lambda^{m+1}}g \\ &= \sum_{m=1}^{\infty} \frac{\Im^{m-1}}{\lambda^{m}}g \end{align*} Now the problem is easy: \begin{align*} \left\|f^{*} -\sum_{m=1}^{k} \frac{\Im^{m-1}}{\lambda^{m}}g \right\|_{\infty} &= \left\| \sum_{m=1}^{\infty} \frac{\Im^{m-1}}{\lambda^{m}}g -\sum_{m=1}^{k} \frac{\Im^{m-1}}{\lambda^{m}}g \right\|_{\infty} \\ &\le \left\|\sum_{m=k+1}^{\infty} \frac{\Im^{m-1}}{\lambda^{m-1}}\right\| \frac{\left\|g\right\|_{\infty}}{|\lambda|} \\ &\le \sum_{m=0}^{\infty} \left(\frac{\left\|\Im \right\| }{|\lambda|}\right)^{m+k} \frac{ \left\|g\right\|_{\infty} }{ |\lambda| } \\ &\le \sum_{m=0}^{\infty} \alpha^{m}\alpha^{k}\frac{\left\|g\right\|_{\infty}}{|\lambda|} \\ &= \frac{\alpha^{k}\left\|g\right\|_{\infty}}{(1-\alpha)|\lambda|} \end{align*}

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