Can we find f such that |(x+x−1)−(a+a−1)|&lt;f(|x−a|)Proposition. Assume x and a are elements of R≠0. Then if x is near to a, we can deduce that x−1 is near to (a+a−1)−x.

We cannot. In other words, g(x)=x+x−1 fails to be uniformly continuous over BR≠0.For a fixed δ&gt;0 and any x, we can take a=x+δ. We find that limx→0+|g(x)−g(a)|=∞, so that there is no value for f(|x−a|)=f(δ) for which |g(x)−g(a)|&lt;f(|x−a|) holds for all x∈BR≠0.For a,x in a closed subinterval [p, q] of BR≠0, it suffices by the mean value theorem to take f(|a−x|)=maxt∈[p,q]|g′(t)|⋅|a−x|.

Check the solution to "Can we find f such that |(x+x−1)−(a+a−1)|<f(|x−a|)Proposition. Assume..."

High School
Algebra
Algebra I
Quadratic function and equation

Annabella Lopez

Answered question

2022-03-16

Can we find f such that

| (x + x^{- 1}) - (a + a^{- 1}) | < f (| x - a |)

Proposition. Assume x and a are elements of

R_{\neq 0}

. Then if x is near to a, we can deduce that

x^{- 1}

is near to

(a + a^{- 1}) - x .

Answer & Explanation

Quentin Olsen

Beginner2022-03-17Added 5 answers

We cannot. In other words,

g (x) = x + x^{- 1}

fails to be uniformly continuous over

B R_{\neq 0}

.
For a fixed

δ > 0

and any x, we can take

a = x + δ

. We find that

lim_{x \to 0^{+}} | g (x) - g (a) | = \infty

, so that there is no value for

f (| x - a |) = f (δ)

for which

| g (x) - g (a) | < f (| x - a |)

holds for all

x \in B R_{\neq 0}

.
For a,x in a closed subinterval [p, q] of

B R_{\neq 0}

, it suffices by the mean value theorem to take

f (| a - x |) = max_{t \in [p, q]} | g^{'} (t) | \cdot | a - x |

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