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I don't know how to verify this function if I can't use the Banach-Caccioppoli Contraction Principle. Please help

Amy12
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2 Answers2

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$g$ has a fixed point $\hat x$ if and only if $g(\hat x) = \hat x$, hence $$ g(\hat x) = \hat x \iff 1 + \hat x - \left(\frac{\hat x}{2} \right)^3 = \hat x \iff 1-\left(\frac{\hat x}{2} \right)^3 = 0 $$

Can you continue from here? Does exist $\hat x$? Is it unique?

Sewer Keeper
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A fixed point for $g(x)$ should be a point $\hat{x}$ with the property that $g(\hat{x})=x$. I.e., we want to solve $$1+x-\frac{1}{8}x^3=x$$ for $x$. We rewrite this to $$1-\frac{1}{8}x^3=0,$$ and finally to $$x^3=-8.$$ It is clear that this last equation has $-2$ as its only solution.

EBP
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