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I wanted to show that in general, T has a unique fixed point and to determine that fixed point of T such that $Tf = f$

$T : (C[0,1], \left\lVert.\right\rVert_{inf} \rightarrow C[0,1], \left\lVert.\right\rVert_{inf} )$, with $\left\lVert f\right\rVert_{inf} = \sup_ {x\in [0,1]}{|f(x)|}$ be defined by

$(Tf)(x) = \int_0^1\frac{1}{2}(xy-1)(f(y) + 1)\mathrm{d}y$

Thanks for the help.

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  • $\begingroup$ Welcome to MSE. You will have more chances to have your question answered if you show what you have done so far towards solving the problem, where you have encountered difficulties, etc. $\endgroup$ – user539887 May 20 '18 at 21:02
  • $\begingroup$ Contraction Mapping Theorem? $\endgroup$ – glowstonetrees May 20 '18 at 21:03
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Existence and Uniqueness:

\begin{align} & \lVert (Tf)(x) - (Tg)(x) \rVert \\ = & \bigg\lVert \int_0^1\frac{1}{2}(xy-1)(f(y) + 1)dy - \int_0^1\frac{1}{2}(xy-1)(g(y) + 1)dy \bigg\rVert \\ = & \sup_{x \in [0,1]} \bigg| \int_0^1\frac 12 (xy-1)(f(y)-g(y))dy \bigg| \\ ≤& \sup_{x \in [0,1]} \int_0^1 \bigg| \frac 12 (xy-1)(f(y)-g(y))\bigg| dy \\ =& \sup_{x \in [0,1]} \frac 12 \int_0^1 |xy-1| \bigl|f(y)-g(y)\bigl|dy \\ ≤ & \frac 12 \int_0^1 \bigl|f(y)-g(y)\bigl|dy && x,y\in[0,1] \implies |xy-1|≤1\\ ≤ & \frac 12 \int_0^1 \sup_{y \in [0,1]}|f(y)-g(y)|dy \\ = & \frac 12 \sup_{y \in [0,1]}|f(y)-g(y)| \\ = & \frac 12 \lVert f-g \rVert \end{align}

Hence, this map is a contraction with constant $\frac 12$, and so by the Contraction Mapping Theorem, a unique fixed point exists.


Construction:

\begin{align} (Tf)(x) = f(x) \implies f(x) & = \int_0^1 \frac 12 (xy-1)(f(y)+1)dy \\ & = \frac x2 \int_0^1y(f(y)+1)dy - \frac 12 \int_0^1 (f(y)+1)dy \\ & = Ax+B \end{align}

for some constants $A$ and $B$.

i.e. $f$ must be a linear function of the form $f(x)=Ax+B$. Plug this back into the above:

\begin{align} & (Tf)(x) \equiv f(x) \\ \implies & Ax+B \equiv \int_0^1 \frac 12 (xy-1)(Ay+B+1)dy \\ \implies & Ax+B \equiv \biggl( \frac A6 + \frac B4 + \frac 14 \biggl)x + \biggl(-\frac A4 - \frac B2 - \frac 12 \biggl) \end{align}

Comparing coefficients, we get

$$\begin{pmatrix} \frac 56 & -\frac 14 \\ \frac 14 & \frac 32 \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = \begin{pmatrix} \frac 14 \\ -\frac 12 \end{pmatrix}$$

This gives $A = \dfrac{4}{21}$ and $B = -\dfrac{23}{63}$, and so the unique fixed point is

$$f(x) = \dfrac{4}{21} x -\dfrac{23}{63}$$

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  • $\begingroup$ Thanks a lot !! In addition how can I get that fixed point of T $\endgroup$ – Dereje May 20 '18 at 21:32
  • $\begingroup$ The fixed point doesn't look so nice, not sure if I made some sort of error, but that is the idea $\endgroup$ – glowstonetrees May 20 '18 at 22:10
  • $\begingroup$ Thanks again. I don't see any error ! $\endgroup$ – Dereje May 20 '18 at 23:25

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