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Let $f_n$ be a sequence of functions in the set of all bounded functions from $X$ to $F$ where $F$ is the real or complex numbers. The sequence is said to converge uniformly to a bounded function $f:X \to F$ if, given $\epsilon>0$, there exists an $N\in \mathbb N$ s.t. $\sup\{|f_n(x)-f(x)| : x \in X \}<\epsilon$ for $n\ge N$

I want to study the uniform convergence of $$ f_n(x) = \begin{cases} n-2n^2|x-\frac{1}{2n}| & \quad \text{if } 0\leq x \leq \frac{1}{n}\\ 0 & \quad \text{if } x\geq \frac{1}{n} \end{cases} $$

Using the definition, it should be easy to see that $\sup\{|0-0| : x\geq\frac{1}{n} \}<\epsilon$ and therefore $f_n$ converges uniformly in $[\frac{1}{n},\infty)$.

Now, to see if $f(n)$ converges uniformly, I need to calculate his pointwise limit $\text{if } 0\leq n \leq \frac{1}{n}$:

$$\lim_{n\to \infty}n-2n^2|x-\frac{1}{2n}|$$ I see that if $x\neq \frac{1}{n}$, $\lim_{n\to \infty}n-2n^2|x-\frac{1}{2n}|=-\infty$.Is this enough to prove $f_n$ does not converge uniformly?

I would really appreciate if someone could tell me if this is correct and, in case it is wrong, correct it.

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    $\begingroup$ Note that $n$ runs. So, if you fix some small $x$, there will be an $N$ such that for $n > N$ you will have $1/n < x$. $\endgroup$ Commented Nov 5, 2017 at 17:28
  • $\begingroup$ thé séquence does not converge poinwise $\endgroup$ Commented Nov 5, 2017 at 17:33

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Assume that it is uniform, then some $N\in{\bf{N}}$ is such that for all $n\geq N$, $|f_{n}(x)|<1$ for all $x\in[0,1]$. Plugging $x=1/2n$ yields $n<1$ for all such $n\geq N$, a contradiction.

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    $\begingroup$ Why don't you let him just think a moment by himself? This is not very helpful. $\endgroup$ Commented Nov 5, 2017 at 17:29
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Hint. We have that for $x=1/(2n)<1/n$, $$f_n(x)=n-2n^2\left|x-\frac{1}{2n}\right|=n$$ So your limit is wrong (note also that $f_n(x)\geq 0$ for all $x\geq 0$). But the above observation is useful for the purpose.

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