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Let $(a_n)$ be a decreasing sequence of continuous functions on the interval $[a,b]$ which converges pointwise to $0$. Show that it converges uniformly to $0$.

This is my attempt at the proof, could someone tell me if it is close to being right and if it's not (which I don't think that it is) could someone tell me how to approach proving this.

(I think this is a special case of Dini's theorem. However this is for a course where we have not done compactness or openness.)

Thanks for any help!

Proof.

From the definition of pointwise convergence, we have that:

$$\forall \epsilon > 0 \quad \forall x \in [a,b] \quad \exists N: \quad \forall n \geq N: \quad |f_n(x)| < \epsilon $$

Now let the max of $f_n(x)$ be at $x_0$. Then we have that:

$$ \forall \epsilon > 0 \quad \exists N_0: \quad \forall n \geq N_0: \quad |f_n(x_0)| < \epsilon $$

Now as $x_0$ is the max and $f_{N_0}>f_{N_0+1}> \cdots$, we have that:

$$\forall \epsilon > 0 \quad \exists N_0 : \quad \forall n \geq N_0 \quad \forall x \in [a,b]: \quad |f_n(x)| < \epsilon $$

So we have uniform convergence.

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    There is a problem, since $x_0$ may depend on $n$. Fix $\varepsilon >0$, and consider the open set $O_n=\left\{x\in\left[a,b\right],f_n(x)<\varepsilon\right\}$. THe family $\{O_n\}$ is an open cover of the compact $\left[a,b\right]$, and you can extract a finite subcover.2011-10-31
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    When you say "let the max of $f_n$ be at $x_0$", you don't have any definition of $n$ in range, so essentially you're assuming that there's a single $x_0$ that is the maximum of _every_ $f_n$. This obviously isn't always true.2011-10-31
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    Also, it is bad form to mix words and symbols as you do. Write **either** "$\forall\epsilon>0.\forall x\in[a,b].\exists N.\forall n\ge N.|f_n(x)|<\epsilon$", **or** "For all $\epsilon>0$ and all $x\in[a,b]$ there is an $N$ such that for all $n\ge N$ it holds that $|f_n(x)|<\epsilon$". Once you've started writing a symbolic formula, do not switch back to English before that formula is finished.2011-10-31
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    Thanks for the help, however we have not done open cover,subcovers,.... so I think i should be looking for an alternative solution. Is there anyway to modify this proof to solve that problem? Thanks for any help2011-10-31
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    Thanks for the point about notation, fixed it now.2011-10-31
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    You missed some instances of "s.t." after some of the $\exists$s.2011-10-31
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    I slightly improved the Texing. I added a space after the quantifiers since that seemed to improve the readability. Hope that's ok.2011-11-01
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    @hmmmm Following the idea of my first comment, but without using open covers, we will show that there is a $n_0$ such that $O_{n_0}=\left[a,b\right]$. If it's not the case, we can pick $x_n\in\left[a,b\right]\setminus O_n$ for all $n$, and by Bolzano-Weierstrass extract a converging subsequence $\{x_{n_k}\}$ which converges to a $x\in\left[a,b\right]$. What about $f_{n_k}(x)$?2011-11-01

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