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 November 19th, 2011, 11:13 AM #1 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 Applying the Contraction Mapping Principle Problem: Prove that there exists a continuous function $f: [0,1] \rightarrow \mathbb{R}$ such that: $f(x)= sin(x) + \int _{0} ^{1} \frac{f(y)}{e^{x+y+1}} dy$. This is what I have done: Define the function $T:C[0,1] \rightarrow C[0,1]$ by: $T(f)= sin(x) + \int _{0} ^{1} \frac{f(y)}{e^{x+y+1}} dy$, and observe that $T(f)= f$ is and only if $f$ is a solution to the above. We next note that: $d_{\infty}(f_{1},f_{2})= \parallel T(f_{1} - T(f_{2}) \parallel _{\infty}$ $\hspace{85mm}=\sup_{0 \leq x \leq 1} \| (sin(x) + \int _{0} ^{1} \frac{f_{1}(y)}{e^{x+y+1}} dy) - (sin(x) + \int _{0} ^{1} \frac{f_{2}(y)}{e^{x+y+1}} dy) \|$ $\hspace{85mm}=\sup_{0 \leq x \leq 1} \| \int _{0} ^{1} \frac{f_{1}(y)}{e^{x+y+1}} dy - \int _{0} ^{1} \frac{f_{2}(y)}{e^{x+y+1}} dy \|$ $\hspace{85mm}=\sup_{0 \leq x \leq 1} \int _{0} ^{1} \| \frac{f_{1}(y)-f_{2}(y)}{e^{x+y+1}}\| dy$ $\hspace{85mm} \leq \sup_{0 \leq x \leq 1} \int _{0} ^{1} \frac{\parallel f_{1}(y)-f_{2}(y) \parallel _{\infty}}{e^{x+y+1}} dy$ $\hspace{85mm}= \parallel f_{1}(y)-f_{2}(y) \parallel _{\infty} \sup_{0 \leq x \leq 1} \int _{0} ^{1} \frac{1}{e^{x+y+1}} dy$ $\hspace{85mm}= \parallel f_{1}(y)-f_{2}(y) \parallel _{\infty} \int _{0} ^{1} \frac{1}{e^{y+1}} dy$ $\hspace{85mm}= \alpha d_{\infty} (f_{1},f_{2})$, where $\alpha= \int _{0} ^{1} \frac{1}{e^{y+1}} dy \approx 0.232544 < 1=$ Since $C[0,1]$ is complete with the metric $d_{\infty}$, we invoke the Contraction Mapping Principle to conclude that $T$ has a unique fixed point. In particular, there exists a function $f:[0,1] \rightarrow \mathbb{R}$ such that $f$ satisfies: $f(x)= sin(x) + \int _{0} ^{1} \frac{f(y)}{e^{x+y+1}} dy$.
 November 19th, 2011, 12:23 PM #2 Member   Joined: Nov 2009 From: France Posts: 98 Thanks: 0 Re: Applying the Contraction Mapping Principle It seems correct, except that the first line of your "align", it should be $d(Tf_1,Tf_2)$.

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