Continuous limited function, thus uniformly continuous

[icantadd]

Homework Statement

suppose f : [0,infinity) -> R is continuous, and there is an L in R, s.t. f(x) -> L, as x -> infinity. Prove that f is uniformly continuous on [0,infinity).

Homework Equations

limit at xo: |x-xo| < delta then |f(x) -L| < epsilon

continuous |x-x0| < delta then |f(x) - f(xo) | < epsilon

uniformly continuous: |x-y| < delta then |f(x) - f(y)| < epsilon.

The Attempt at a Solution

Let A be a number s.t.
if |x - y| < delta1 , |f(x) - L | < epsilon for all x in [a,infinity).
b/c f is continuous then |x-xo| < delta1 |f(x) - f(y) | < epsilon.
Thus f is uniformly continuous on [a,infinity).

Then,
b/c f is continuous it is uniformly continuous on any compact domain, namely [0,a].
|x-y| < delta2 ...

Then pick delta = min{delta1,delta2}, then the rest should follow.

Does that seem sound?

 

[mighty2000]

Yes, your proof is sound. Here are some additional explanations for clarity:

First, we need to show that for any ε > 0, we can find a δ > 0 such that for all x, y in [0,∞), if |x-y| < δ, then |f(x)-f(y)| < ε. Since f is continuous on [0,∞), we know that for any x in [0,∞), there exists a δx > 0 such that if |x-y| < δx, then |f(x)-f(y)| < ε. This is the definition of continuity.

Next, we need to show that f is uniformly continuous on [0,∞), which means that for any ε > 0, we can find a δ > 0 such that for all x, y in [0,∞), if |x-y| < δ, then |f(x)-f(y)| < ε. Since f is continuous on [0,∞), we know that for any x in [0,∞), there exists a δx > 0 such that if |x-y| < δx, then |f(x)-f(y)| < ε. However, this δx may vary depending on the choice of x. We need to find a δ that works for all x in [0,∞). This is where the condition f(x) → L as x → ∞ comes in.

Since f(x) → L as x → ∞, we know that for any ε > 0, there exists a M > 0 such that if x > M, then |f(x)-L| < ε/2. Now, let δ = min{δx, M}. Then for any x, y in [0,∞), if |x-y| < δ, we have two cases:

1) If both x and y are greater than M, then |f(x)-f(y)| < ε since |f(x)-f(y)| ≤ |f(x)-L| + |L-f(y)| < ε/2 + ε/2 = ε.

2) If one of x or y is less than or equal to M, say x, then |f(x)-f(y)| < ε since |f(x)-f(y)| ≤ |f(x)-L| + |L-f(y)| < ε/2 +