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Suppose we have a sequence of function $(f_n(x))_{n \in \mathbb{N}}$ the converges pointwise. Now I want disprove that it converges uniformely. I know I can use a sequence $(x_n)_{n\in \mathbb{N}}$.
In all of the cases I used and saw used the sequence $(x_n)_{n \in \mathbb{N}}$ depended solely on $n$, for example $x_n=\frac{1}{n}$, my question would be if I could use a sequence $(x_n)_{n \in \mathbb{N}}$ that would also depend on $x$? For example, would $(x_n)= \frac{1}{x^n}$ be acceptable?

EDIT: Concerning what exactly is $x$.
If $(f_n)_{n \in \mathbb{N}} : \mathbb{R} \rightarrow \mathbb{R}, x \mapsto nx^n(1-x) $. Then I would like to use $(x_n)=\frac{1}{x^n}$, thus giving me $f(x_n)=n(x_n)^n(1-x_n)$

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    So what is $x$? – 2017-02-09
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    @NiklasHebestreit I made an edit explaining what is $x$. – 2017-02-09
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    Still it is **not** clear what $x$ is since $f$ is a function in $x$. – 2017-02-09
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    When you are specifying the sequence $(x_n)$, you are specifying a sequence of $x$-values to be inserted into your function $f$. In the usual notation the independent variable (or argument) of a function is called $x$, so it makes sense to call the sequence of numbers you want to use as arguments $(x_n)$. But in all of this there is no specific $x$ to refer to, and that's why Niklas repeatedly asked you what $x$ is. In your EDIT, you didn't use any $x$, you simply put the term $x_n$ into your function definition. – 2017-02-09

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You can disprove the uniform convergence of $f_n \to f$ in different ways. These two methods might help you in the future:

  1. If $f_n \to f$ pointwise, all $f_n$ are continuous and $f$ is not continuous then $f_n \to f$ is not uniform convergent.
  2. Show $\|f_n -f\|_\infty \geq \text{const}>0$ that is $\|f_n -f\|_\infty \nrightarrow 0$. Therefore $f_n \to f$ is not uniform convergent.