For which values of $x \in I$ is $(f_n)$ differentiable term by term?

Question

Let $f_n(x) = \frac{1}{n}e^{-nx}$ on $x\in[0,1] =I$.

1. Discuss the pointwise and uniform convergence of $(f_n)$ on $I$.
2. For which values of $x \in I$ is $(f_n)$ differentiable term by term?

My attempt:

1.

Pointwise Convergence:

$\displaystyle \lim_{n \to \infty} f_n(x) = 0$ for all $x \in I$.

Hence $(f_n)$ converges pointwise to $f(x)=0$ on $I$.

Uniform Convergence:

$\|f_n-0\| = \sup \bigg\{\bigg|\frac{1}{n}e^{-nx} \bigg| : x \in [0,1] \bigg\} = \bigg| \frac{1}{n} \bigg| = \frac{1}{n} \to 0$ as $n \to \infty$.

Hence $(f_n)$ converges uniformly to $f(x) = 0$ on $I$.

Is this correct?

2.

This is the question which I am stuck at.

I know that, in order for $(f_n)$ to be term by term differentiable

• $(f_n)$ must converge pointwise

• $f'_n$ must be continuous and

• $(f'_n)$ must converge uniformly.

However, this does not occur on the entire interval $I = [0,1]$. Can anyone please show me how I can go about finding the the values of $x$ in $I$ for which this is true?

Answer 1

$f'_n(x)=e^{-nx}$ is a geometric series that converges uniformly on $[a,1]$ for $0<a<1$. It follows that $\{f_n\}$ is differentiable term by term on $(0,1]$.