# Does $\frac{\lfloor nf(x) \rfloor}{n}$ uniformly converge?

I tried solving the following exercise, and want to ask if my attempt is correct:

Say we have some function $f(x)$ defined in $[a,b]$, and we are asked whether $\frac{\lfloor nf(x)\rfloor}{n}$ uniformly converges in that interval, where $\lfloor nf(x) \rfloor$ is the whole part of $nf(x)$.

Attempt

I think that it does uniformly converge. Here's what I tried:

For every $x$ in $[a,b]$, we have that $|f(x)-f_n(x)|=f(x)-f_n(x)=f(x)-\frac{\lfloor nf(x) \rfloor}{n}<f(x)-\frac{nf(x)-2}{n}=\frac{2}{n}$

And since $\frac{2}{n}$ goes to 0, we have that the function series uniformly converges.

What do you think? Is this correct? I'm kind of not sure since the exercises in this section are usually more elaborate, and this seems kind of 'obvious'. Much thanks!

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It is important to remark that the bound $2/n$, which I think could (not that it matters) be replaced by $1/n$, is independent of $x$. –  André Nicolas Feb 4 '12 at 15:01
I think this is correct. –  Joel Cohen Feb 4 '12 at 15:17
Okay, thanks! I guess the question is closed then. @AndréNicolas: I actually used 2 there because I didn't feel like looking up the 'less or equal to' sign code on Latex :). –  ro44 Feb 4 '12 at 15:56
@ro44: It is \leq. And you didn't need to look it up, since $nf(x)-1< \lfloor nf(x)\rfloor$. Indeed the problem was not hard, but many students who are not fazed by a mildly complicated "formula" have difficulties with the floor function. –  André Nicolas Feb 4 '12 at 16:08