Show that $f$ is a strong contraction when $f$ is continuously differentiable.

Mathematics Asked on January 7, 2022

Let $f: [a,b] to R$ be a differentiable function of one variable such that $|f'(x)| le 1$ for all $xin [a,b]$. Prove that $f$ is a contraction. (Hint: use MVT.) If in addition $|f'(x)| < 1$ for all $x in [a,b]$ and $f’$ is continuous, show that $f$ is a strict contraction.

Using MVT, $|f(x) – f(y)| = |f'(c)(x-y)| le |x-y|$ for $c$ between $x$ and $y$.

I don’t know the proof for a strict contraction. I guess that I need to use the continuity of $f’$, but I am not sure how to use it. Any help would be appreciated.

2 Answers

You're almost there! A (weak) contraction is defined as a function $f: A to mathbb{R}$ such that $$|f(x) - f(y)| leq k | x - y| ~forall x,y in A,::: 0 leq k leq 1$$ (in your case $A = [a,b]$.)

This might look a little familiar, as you have already concluded that $|f(x) - f(y)| < |x - y|$ (by removing the middle inequality). We just need to make sure that $k = |f(x)|$ meets the required bounds. Combine $0 leq |f(x)|$ (a property of absolute value) with the given $|f(x)| leq 1$ for all $x in [a,b]$, you can conclude $0 leq |f(x)| leq 1$, so $1geq k geq |f(c)| geq 0$ (by continuity and $c in[a,b]$).

So you have effectively shown that $f$ is a weak contraction.

Showing that if $|f(x)| < 1$, then $f$ is a strict contraction follows a similar path. @bitesizebo is correct in noting that $f'$ continuous is essential here. A strict contraction replaces the weak inequality with a strong inequality. A function is a strict contradiction if $$|f(x) - f(y)| < k | x - y| ~forall x,y in A,::: 0 leq k leq 1$$ (in your case $A = [a,b]$.)

You know that $|f(x) - f(y) leq |f(c)(x-y)|$ by MVT. You can separate the RHS of the equation and get $|f(c)(x-y)| = |f(c)||x-y|$. Using the fact that $f'$ is continuous and $|f'(x)| < 1$, you need to prove that $|f'(c)| < 1$ (level of detail here depends on your class), as @bitesizebo says.

There are a few ways to show that $f$ satisfies $sup_{[a,b]} |f'(c)| < 1$. My favorite method is a bit more broad than @Michael Hardy's approach and proves the Extreme Value Theorem along the way. I use prefer the 'fact' that the image of a compact set under a continuous function is also compact. This works in any topological space, not just $mathbb{R}$. So $f'([a,b])$ is a closed set, so it must attain all of it's limit points (the limit of any sequence of numbers in $f'([a,b])$ must also be a point in $f'([a,b])$). So the infimum and supremum of $f'([a,b])$ must be obtained and be in the set. So as $|f'(c)| < 1$ for all $c in[a,b]$, you must have also hit the supremum and infimum of $f'([a,b])$. So $sup_{[a,b]}|f'(c)| < 1$. You can fill in the algebraic details and inequalities.

Once you have $sup_{[a,b]}|f'(c)| < 1$, you can plug this in to the statement we got from the MVT, and conclude $|f(x) - f(y) < k |y-x|$ where $|f'(c)| < sup_{[a,b]}|f'(c)| leq k < 1$ (I've never seen the assumption $k>0$ used to define a strict contraction since it could give a 'strongest' possible contraction step, although I don't have my copy of Rudin handy), and conclude that $f$ is a contraction map.

Answered by M Kupperman on January 7, 2022

Besides the mean-value theorem, another "MVT" is the "maximum-value theorem": A continuous function on a closed bounded interval has an absolute maximum. If $|f'|$ is continuous, then there is some point $cin[a,b]$ for which $|f'|$ is at least as big as it is at any point in that interval. And that value must be less than $1,$ by hypothesis. Then apply the mean value theorem again.

(You probably won't see that other theoem called the "MVT", for "maximum-value theorem", since it is instead called the extreme-value theorem, applying to both maxima and minima.)

Answered by Michael Hardy on January 7, 2022

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