Check Uniform Convergence of f_n(x): Solutions & Explanations

[estro]

Homework Statement

f(x) is defined within [a,b].
$f_n(x)=\frac{\big\lfloor nf(x) \big\rfloor}{n}$

Check if $f_n(x)$ is uniform convergent.

The Attempt at a Solution

This one seems to be easy however since I didn't touch calculus for quite a time I'm not confident with my solution.

$|\frac {nf(x)-1} {n}| \leq |\frac {\big\lfloor nf(x) \big\rfloor }{n}| \leq |\frac {nf(x)+1} {n}|$ so by the squeeze theorem: $\lim_{n \rightarrow \infty}f_n(x)=f(x)$.

Now, let $x_0 \in [a,b]$ if $f(x_0) \geq 0$ then: $|f_n(x)-f(x)|=|\frac {\big\lfloor nf(x) \big\rfloor } {n}-f(x)| \leq |\frac {nf(x)+1} {n}|=\frac {1}{n} \rightarrow 0$, similarly we can show that the same equation hold when f(x_0)<0 what in turns means that we can choose N that is not dependent on x and satisfies $|f_n(x)-f(x)|< \epsilon$

Looks good?

 

[mighty2000]

I would like to provide some feedback on your solution. It is correct to use the squeeze theorem to show that \lim_{n \rightarrow \infty}f_n(x)=f(x). However, your argument for choosing N that is not dependent on x is not entirely clear.

To show that f_n(x) is uniformly convergent, we need to show that for any given \epsilon > 0, there exists an N \in \mathbb{N} such that for all x \in [a,b] and all n \geq N, we have |f_n(x)-f(x)| < \epsilon.

In your solution, you only showed that for a fixed x_0 \in [a,b], there exists an N \in \mathbb{N} such that for all n \geq N, we have |f_n(x_0)-f(x_0)| < \epsilon. This is not enough to show uniform convergence, as N depends on the choice of x_0.

To fix this, you can use the fact that [a,b] is a bounded interval and f is continuous on [a,b]. This means that f is uniformly continuous on [a,b], which implies that for any given \epsilon > 0, there exists a \delta > 0 such that for all x,y \in [a,b] with |x-y| < \delta, we have |f(x)-f(y)| < \epsilon.

Now, for any given x \in [a,b], we can choose N \in \mathbb{N} such that |nf(x)-nf(x_0)| < \delta for all n \geq N. This implies that |f_n(x)-f_n(x_0)| < \epsilon for all n \geq N. Since x_0 was arbitrary, this shows that |f_n(x)-f(x)| < \epsilon for all x \in [a,b] and all n \geq N, which is what we needed to show for uniform convergence.

Overall, your solution is on the right track, but it would benefit from being more explicit about the choice of N and how it is not dependent on x.