Uniform convergence of a sequence of functions

[AllRelative]

Homework Statement

This is a translation so sorry in advance if there are funky words in here[/B]
f: ℝ→ℝ a function 2 time differentiable on ℝ. The second derivative f'' is bounded on ℝ.
Show that the sequence on functions $$ n[f(x + 1/n) - f(x)] $$ converges uniformly on f'(x) on ℝ.

Homework Equations

In my attempt I used the definition of differentiation:
$$\lim_{x\to\infty} \frac{f(x+h) - f(x)}{h} $$

and I used a criteria for uniform convergence of sequences of functions:

f_n fonverges to f uniformely on A
if and on if
for all ε > 0, ∃ N ∈ ℕ, for which
$$\lim_{x\to\infty} sup|fn - f(x)| \leqslant \varepsilon$$
for all n≥ N, for all x ∈ A

The Attempt at a Solution

[/B]
I arrived to an answer but I fear I got sidetracked somewhere because I did not use the bounded second derivative.

I rewrote
$$ n[f(x + 1/n) - f(x)] = \frac{[f(x + 1/n) - f(x)]}{1/n} $$

Now this looks awefully like the derivative of f_n for all x which is:
$$\lim_{n\to\infty} \frac{[f(x + 1/n) - f(x)]}{1/n}$$

And now I applied the definition of the uniform convergence which is:
$$\lim_{n\to\infty} sup| \frac{[f(x + 1/n) - f(x)]}{1/n} - f'(x)| \leqslant \varepsilon $$

And therefore, I proved the uniform convergence to f'(x) on ℝ.

(I am missing a few for all ε belonging to...and stuff, I just wanted to write it quickly)
Thank you

 

[FactChecker]

1) Your definition of differentiation is wrong. It should have h -> 0.
2) You have not used the bound on f''. I think you need that to prove uniform convergence.

 

[AllRelative]

1) Your definition of differentiation is wrong. It should have h -> 0.
2) You have not used the bound on f''. I think you need that to prove uniform convergence.

Thanks for the reply.
You're right my definition is not good when I wrote down the definitions at the start. But in the problem I think it is.
I did not write it but here I implied that h = (1/n)
So n -> ∞ ⇒ h -> 0 because

 

[NickP]

We need to be somewhat careful about whether or not f' is well defined. We have to be certain that the limit exists before we claim that it is equal to anything. The existence of some limits is trivial, while others are a little more questionable. Suppose, for example,
$$f''=\frac{1}{\sqrt{x}}$$