A question on Contraction and contractive map

Question

how to prove this map is not contraction and have no fixed point and I am proved contractive by using mean value theorem

Answer 1

Use MVT to prove that $|T(x)-T(y)|<|x-y|$ if $x \neq y$. Suppose $|T(x)-T(y)|\leq c|x-y|$ for all $x,y$ with $c <1$. take $y=0$ to get $|\frac {1-e^{x}} x|<c$ for $x \in (-\infty ,0)$. You get a contradiction by letting $x \to 0$ and using L'Hopital's Rule. The last part is false. The function $ye^{y}$ is continuous on $[0,\infty)$, it is $0$ at $0$ and tends to $\infty$ as $ y \to \infty$. Hence there exists $y>0$ such that $ye^{y}=1$. If $x =-y$ then we get $-e^{x}=x$ or $T(x)=x$.

Answer 2

$T$ is contractive: For $x\ne y$, we have (by the Mean Value Theorem) $$T(x)-T(y)=(x-y)T'(\xi) $$ for some $\xi$ betawwen $x$ and $y$. As that makes $\xi$ negative, it follows that $|T'(\xi)|=|e^\xi|<1$.

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$T$ is not a contraction: Let $k$ be such that $|T(x)-T(y)|\le k|x-y|$ for all $x,y$. Letting $y\to x$, we get $$|T'(x)|= \left|\lim_{y\to x}\frac{T(y)-T(x)}{y-x}\right|\le k.$$ As $|T'(0)|=1$, we cannot have $k<1$. (Even if we exclude $0$ from the domain, $|T'(x)|\to 1$ as $x\to 0$ and we still could not have $k<1$).

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$T$ does have a fixed point: The function $f(x)=T(x)-x$ is continuous, $f(0)=-1$, $f(-1)=1-\frac1e>0$, hence by the Intermediate Value Theorem, $f(x)=0$ for some $x\in(-1,0)$.