# Prove that $\sum_{n=1}^\infty f_n$ converges uniformly on $\mathbb{R}$ where $f_n$ is defined piecewise.

Prove that the series of functions $\sum_{n=1}^\infty f_n$ converges uniformly on $\mathbb{R}$, where $$f_n: \mathbb{R} \to \mathbb{R}: x \mapsto \begin{cases}0 \quad x \neq n \\\frac{1}{x} \quad x =n\end{cases}$$

My attempt:

Let $x \in \mathbb{R}$ be fixed. Let $n \in \mathbb{N}$ with $n > x$. Then, it is clear that:

$$\sum_{k=1}^n f_k(x) = \begin{cases}0 \quad x \notin \mathbb{N_0}\\\frac{1}{x} \quad x \in \mathbb{N_0}\end{cases}$$

Hence, letting $n \to \infty$, we have that the given series converges pointwise to the function $$f: \mathbb{R} \to \mathbb{R}: x \mapsto \begin{cases}0 \quad x \notin \mathbb{N_0}\\\frac{1}{x} \quad x \in \mathbb{N_0}\end{cases}$$

We now show uniform convergence. Let $x \in \mathbb{R}$ and $n \in \mathbb{N}$.

If $x \in \mathbb{N_0}$, we consider two cases:

(i) $n\geq x$: then $\left|\sum_{k=1}^nf_k(x) - f(x)\right| = 0$

(ii) $n < x$: then $\left|\sum_{k=1}^nf_k(x) - f(x)\right| = |f(x)| = \frac{1}{x} < \frac{1}{n}$

If $x \notin \mathbb{N_0}$, then $\left|\sum_{k=1}^nf_k(x) - f(x)\right| = 0$

So, we have proven that $\forall n \in \mathbb{N}, \forall x \in \mathbb{R}: \left|\sum_{k=1}^nf_k(x) - f(x)\right| < \frac{1}{n} \to 0$

Let then $\epsilon > 0$. Choose $n_0: \forall n \geq n_0: \frac{1}{n} < \epsilon$. Then, for $n \geq n_0: \left|\sum_{k=1}^nf_k(x) - f(x)\right| < \frac{1}{n} < \epsilon$

So, the convergence is uniform. Is this correct?

• Congratulations! That's true and precise... – Mostafa Ayaz Jan 3 '18 at 12:44
• Are you claiming that $\left|\sum_{k=1}^nf_k(x) - f(x)\right| = 0$ for all $x$ and all $n$? – Martin R Jan 3 '18 at 12:46
• Yes, I think that I'm claiming that... Unless I made a mistake? – user370967 Jan 3 '18 at 12:48
• That cannot be correct, it would imply that all partial sums are equal to the limit functions. – Martin R Jan 3 '18 at 12:49
• I see what you mean. Let me think how to fix it. – user370967 Jan 3 '18 at 12:50

You are mostly there. With the $f$ you have found as the limiting function, for any $x\in \mathbb{R}$, $$\left|\sum_{k=1}^nf_k(x)-f(x)\right|\leq \frac{1}{n+1}$$ since if $x\not\in \mathbb{N}$ or if $x\in \{ 1,\dots,n\}$ the difference is zero. Otherwise, the worst this difference could be is the difference at $x=n+1$. This gives you $$\sup_{x\in \mathbb{R}}\left|\sum_{k=1}^nf_k(x)-f(x)\right|\leq \frac{1}{n+1}\to 0$$ as required.