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.