(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

f(x) is defined within [a,b].

[itex]f_n(x)=\frac{\big\lfloor nf(x) \big\rfloor}{n}[/itex]

Check if [itex]f_n(x)[/itex] is uniform convergent.

3. 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.

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

Now, let [itex]x_0 \in [a,b] [/itex] if [itex] f(x_0) \geq 0 [/itex] then: [itex] |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[/itex], 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 [itex] |f_n(x)-f(x)|< \epsilon[/itex]

Looks good?

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# Uniform convergence

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