# 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.

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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