Table of Contents
Is 1 sqrt x uniformly continuous on 1 Infinity?
No. A function that is continuous on a compact set is uniformly continuous (cf Theorem 4.19 in Rudin’s Principles).
Is square root function uniformly continuous?
Use this and part (a) to prove that the square root function is uniformly continuous on [0, ∞). < δ 2 = ϵ. Therefore, f is uniformly continuous no [1, ∞) as required. Since [0, 2] is a closed, bounded interval, f is uniformly continuous on [0, 2].
Is root x uniformly continuous on 0 infinity?
| √ x − √ y|2 ≤ | √ x − √ y|| √ x + √ y| = |x − y| < ϵ2, hence | √ x − √ y| < ϵ. This shows that f(x) = √ x is uniformly continuous on [0, ∞). Since f is uniformly continuous, there exists some δ > 0 such that d2(x, y) < δ implies d3(f(x),f(y)) < ϵ for all x, y ∈ M2.
Is root x log x uniformly continuous?
Is uniformly continuous for? Yep. The functions and are both continuous everywhere in their respective domains, and , except at the left end points, where they are only right-continuous.
Is X uniformly continuous?
Identity function f (x) = x is uniformly continuous on R. Since f (x) − f (y) = x − y, we have |f (x) − f (y)| < ε whenever |x − y| < ε. The sine function f (x) = sinx is uniformly continuous on R. It was shown in the previous lecture that | sinx − sin y|≤|x − y| for all x, y ∈ R.
Is square root function Lipschitz?
is absolutely continuous on [0,1]. So Lipschitz is a stronger condition than absolutely continuous.
Is 1 x uniformly continuous?
A uniformly continous function is obviously continuous. But the converse is not true. For example, if A = (0,1) and f(x)=1/x, then f is continuous on A, but it is not uniformly continuous on A. The point is that if x is close to zero, then δ needs to be chosen smaller than if x is not close to zero.
Is X log X uniformly continuous?
The function f(x)=xlog(x) is clearly continuous on [0,1] and thus uniformly continuous.
How do you show not uniformly continuous?
If f is not uniformly continuous, then there exists ϵ0 > 0 such that for every δ > 0 there are points x, y ∈ A with |x − y| < δ and |f(x) − f(y)| ≥ ϵ0. Choosing xn,yn ∈ A to be any such points for δ = 1/n, we get the required sequences.
How do you find uniformly continuous?
A function f:(a,b)→R is uniformly continuous if and only if f can be extended to a continuous function ˜f:[a,b]→R (that is, there is a continuous function ˜f:[a,b]→R such that f=˜f∣(a,b))….Answer
- f(x)=xsin(1x) on (0,1).
- f(x)=xx+1 on [0,∞).
- f(x)=1|x−1| on (0,1).
- f(x)=1|x−2| on (0,1).
Are all uniformly continuous functions Lipschitz?
Any Lipschitz function is uniformly continuous. for all x, y ∈ E. The function f (x) = √x is uniformly continuous on [0,∞) but not Lipschitz.