# Why uniform convergence of a series doesn't preserve boundedness?

There is a series of functions $f_n:(0,1) \rightarrow R$ defined as $f_n(x)=\frac{n}{nx+1}$.

The limit of the series is $f(x) = \lim_{n\rightarrow \infty}f_n(x) = \lim_{n\rightarrow \infty}\frac{1}{x+\frac{1}{n}} = \frac{1}{x}$. So $f_n\rightarrow f$ pointwise on $(0,1)$.

1) $f$ isn't bounded on $(0,1)$, because $\lim_{x\rightarrow0} \frac{1}{x} = \infty$. Why are $f_n$ bounded on $(0,1)$ (should we consider $n$ strictly less than $\infty$?

2) Why in general pointwise convergence doesn't preserve boundedness, but uniform convergence does? Is the preservation of boundedness by uniform convergence the reason for the possibility of interchanging limits of uniformly convergent series?

• Did you try to sketch the graphs of $f_n$ and $f$? That will help a lot – ThePortakal Feb 9 '17 at 20:30

For a fixed $n$, $$|f_x(x)|\leq n,$$ so $f_n$ is bounded.
Uniform convergence preserves boundedness because if $f_n\to f$ uniformly and you advance enough along the sequence, you can find some $f_n$ with $|f_n(x)-f(x)|<1$ for all $x$. Then $$|f(x)|<|f_n(x)|+1\leq M_n+1$$ (where $M_n$ is a constant such that $|f_n(x)|<M_n$ for all $x$), and so $f$ is bounded.
• Thank you for the great explanation! If we select $n=\infty$ in your first formula we still have a bounded function? Or we just can't/shouldn't select $n=\infty$ in it? – Konstantin Feb 9 '17 at 20:40