# real analysis – Please check my solution(Uniformly convergence of the functions)

Here is the question in my real analysis textbook.

$$f$$ is a continuous on $$mathbb{R}$$

Say $$f_n(x) = {n over 2}int_{x-{1 over n }}^{x+{1 over n }} f(t) dt$$

Show the $$f_n$$ is a uniformly converge to $$f(x)$$ on $$(0,1)$$

In fact, the textbook itself solve this by M.V.T for integral.

But I tried a different way like the below.

$$(sol)$$ For $$forall x in (0,1)$$, consider the $$I_n = (x- {1 over n }, x+{1 over n })$$

Since $$f$$ is a continuous on $$I_n$$, f is uniformly continuous on $$I_n$$

Therefore, by the definition of the uniform continuity

$$(1)$$ $$exists {2 over n} leq delta$$ s.t. $$Vert x-y Vert < delta$$ $$Rightarrow$$ $$Vert f(x) – f(y) Vert <{2epsilon over n }$$

$$(2)$$plus, By Archimedes $$exists k in mathbb N s.t. n geq k Rightarrow {2epsilon over n } < epsilon$$

By $$(1)$$ and $$(2)$$
$$Vert f_n(x) – f(x) Vert = Vert {n over 2}int_{x-{1 over n }}^{x+{1 over n }} (f(t) – f(x)) dt Vert leq {n over 2}int_{x-{1 over n }}^{x+{1 over n }} Vert f(t) – f(x) Vert dt leq {n over 2 } {2 over n } epsilon =epsilon$$

Hence, $$exists k s.t. n geq k Rightarrow Vert f_n(x) – f(x) Vert < epsilon$$ (Uniformly convergence.)

I don’t have a confidence my solution is right or not. Please check my idea and solution.

Thank you.