Given a strictly convex function $f(x)$ with strictly increasing first derivative and positive second derivative on the interval $[0,1]$. $f$ ist also stricly increasing and has a fixpoint in $x_0=1$. Why is it, that if $f'(1)\leq1$ that there cannot be a root $r$ of the fixpoint equation $f(r)=r$ in the interval $0\leq r<1$?
I know that the mean value theorem implies that if $f(r)=r$ and $1=f(1)$ that there has to be a point between those two fix points with $f'(b)=1$ for $r<b<1$. In my proof this is the end, so it says that this is impossible. Now my question would be why is that not possible?
Best Answer
Think about it geometrically: If there exists $0 \leq r < 1$ such that $f(r) = r$, then strict convexity implies that the graph of $f$ must lie under the line $y=x$ on the entire interval $(r,1)$. If the graph of $f$ lies under $y=x$ just to the left of $x=1$ (say at point $x = 1-\delta$ for some small $\delta>0$), it must increase at a faster rate than the line $y=x$ to reach 1 at $x=1$. This is the same as saying that the slope of $f$ at $x=1$ has to be larger than 1, i.e. $f'(1)>1$. So the existence of a fixed point $0 \leq r < 1$ implies $f'(1)>1$, which is equivalent to saying that $f'(1) \leq 1$ implies no such fixed point can exist.
See the following link for an illustration: https://ars.els-cdn.com/content/image/3-s2.0-B9780123814166000034-f03-07-9780123814166.jpg
In mathematical terms, note that for sufficiently small $\delta>0$, you can use the approximation $$ f(1-\delta) \approx f(1)-\delta \cdot f'(1) = 1-\delta \cdot f'(1) $$ (think about the definition of the deriative). If $f$ lies below the line $y=x$ at $x=1-\delta$, we must have $f(1-\delta) < 1-\delta$ i.e. $1-\delta \cdot f'(1) < 1-\delta$. This implies $\delta \cdot f'(1) > \delta$ or $f'(1) > 1$ since $\delta>0$. This argument can be made more precise by using the Taylor series approximation of $f$ at $x=1$.