Question

Show that the function \[f(x) = x^{2}\]is not uniformly continuous on [0,∞)

Answer 1

So I am trying to show that as x->∞, |f(x)-f(y)| gets larger, i.e. there exists an epsilon > 0 such that for all delta >0, there exists an x and a y in the domain D such that |x-y|<delta AND |f(x)-f(y)|>= epsilon. I can do the proof D = (0,1) for uniform continuity, but am unsure what to do with the ∞ in this question.

Answer 2

\[|f(x) - f(y)| = |x^{2}-y ^{2}|=|x+y|.|x-y|\]If I had an upper limit, say 1, I know that the sum of x and y cannot exceed 2, so have the next step as < 2|x-y| ...,but can't do this with ∞

Answer 3

can a sequence
0,1,4,9,16...show that the change is not uniform?

Answer 4

uncommon difference

Answer 5

isnt it an equation of parabola....?

Answer 6

which is non uniformly continuous ...

Answer 7

Does this work?
let y= x-h such that |x -y| = |x - x +h| = h ≤ delta
now find the limit as x-> inf of |f(x)- f(y)|
where f(x)= x^2

| x^2 - (x-h)^2|= |2x*h + h^2|
lim as x->inf of |2x*h + h^2| = inf