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Is there any way to prove this with only the MVT and IVT? We showed this using the Banach Fixed Point theorem and a cauchy sequence however we were told there is an easier way. Any help would be much appreciated.

Suppose $f$ is differentiable on $(-\infty,\infty)$ and there is a constant $k<1$ such that $|f'(x)|\leq k$ for all real $x$. Show that $f$ has a fixed point.

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Only using MVT and IVT:

If $f(0) = 0$ then we're done. Suppose that $f(0) = a$, where WLOG we can assume $a > 0$ (else take $f \rightarrow -f$).

What can we say about $f(x)$ for $x > 0$? By the MVT, we have:

$$ \displaystyle\frac{f(x) - f(0)}{x} = f'(c) $$

for some $c \in (0,x)$

Now, we have $f'(c) \leq k$, and so we can conclude that $f(x) \leq kx + a$ for all positive $x$.

Taking $y>0$ such that $ky+a < y$, we find that $$f(y) < y$$

But now consider the function $g(x) = f(x)-x$. We have $g(0) = a > 0$ and $g(y) = f(y) - y < 0$.

So by IVT, there is some $z \in (0,y)$ with $g(z) = 0$, i.e. $f(z) = z$ as required.

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Suppose $f(0) > 0$. Then by the mean value theorem $\frac{f(x) - f(0)}{x} = f'(\xi) \le k$ for all $x \ge 0$, so especially $f(x) \le kx + f(0)$. This implies $f(x_0) < x_0$ for any $x_0 > \frac{f(0)}{1-k}$, so the continuous map $x \mapsto f(x) - x$ has different signs at $0$ and $x_0$. The intermediate value theorem now yields the assertion.

The case $f(0) = 0$ is trivial and the case $f(0) < 0$ works in a similar way, by considering a sufficiently small negative number $x_0$.

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