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For the series of functions $f_{k}(x)$ defined by
$$f_{k}(x)=\begin{cases}k \quad 0<x\leq\frac{1}{k},\\0 \quad \text{otherwise}.\end{cases}$$ Is this pointwise converging to the zero function? I would say so (graphically), but I'm not sure how I would show formally. This is what I did:
We want
$$\forall x \in \mathbb{R}, \forall \epsilon >0, \exists K>0: k\geq K \implies |f_k(x) - 0| < \epsilon.$$

So for $0<x\leq \frac{1}{k}$, $|f_k(x) - 0| = k.$ I'm not sure if it actually does pointwise converge now, seeing as if I pick $K= k = 10$ and $\epsilon = 1$, then the inequality obviously does not hold for all $\epsilon$... but my intuition keeps telling me it does, when I look at it graphically.

Twenty-six colours
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1 Answers1

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Yes, this converges pointwise to $f=0$, and the line in the middle of your question is a proof for this statement.

You must not pick some $k$ and then some $\varepsilon$. For pointwise convergence you first pick some (arbitrary) $x$ and some $\varepsilon >0$ and then you have to show there is some $K$ such that for $k\ge K$ the the necessary inequality holds.

Thomas
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  • Ah right, how silly. What $K$ would we choose in this case (a $K$ that would depend on $x$ and $\epsilon$? I can't think of one – Twenty-six colours Jun 09 '17 at 09:05
  • @Twenty-sixcolours $K$ would usually depend on $x$ and $\varepsilon$. Due to the special nature of your sequence it only depends on $x$. (For $x>0$) you have to choose $K$ such that for $k\ge K$ you know that $x>\frac {1}{k}$. Then $f_k(x) = 0$ is automatically smaller than any positive $\varepsilon$. If $x\le 0$ $K$ can be chosen arbitrarily. If $x>0$ you can just choose the smallest integer $K$ such that $1/K < x$ Then $k\ge K$ implies $1/k\le 1/K < x$ – Thomas Jun 09 '17 at 09:54
  • Thank you!
    Then this means $f$ does not converge uniformly right? Since we require that it HAS to depend on $x$?     – Twenty-six colours Jun 09 '17 at 10:56
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    @Twenty-sixcolours $f$ does not converge. The sequence $f_k$ does. And it, in fact, does not converge uniformly. – Thomas Jun 09 '17 at 12:06
  • Thanks. A slightly general question -- how did we know that we couldn't just find an expression for $K$ that depends on $\epsilon$ but not on $x$? As in, what if there does exist an expression, but it's just that I'm not mathematically insightful enough to find one. Is there a way to show it's not uniformly in another way? Iirc, uniform limit of continuous functions is continuous, but I don't think that helps since we already start with a discontinuous function. – Twenty-six colours Jun 09 '17 at 12:16
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    @Twenty-sixcolours the only possible limit of that sequence is the zero function. In order to have uniform convergence, the sequence has to be everywhere, independent of $x$, close to that function. But the maximum of the $f_k$ is $k$, which immediately implies this is not possible. In a general case you may need a more sophisticated argument. There are no general recipees for this kind of question. – Thomas Jun 09 '17 at 12:45