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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?

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

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  • $\begingroup$ I just fixed my answer, while your answer was posted! Is it correct now? $\endgroup$ – user370967 Jan 3 '18 at 13:03
  • $\begingroup$ looks good to me! $\endgroup$ – qbert Jan 3 '18 at 13:04
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    $\begingroup$ Great! Your help is much appreciated! $\endgroup$ – user370967 Jan 3 '18 at 13:05

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