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Before asking this question, I have scoured the stackexchange for a satisfactory answer, but could not find one. Some answers mentioned using the $\epsilon/3$ trick. I have attempted a proof but could not find a use for this trick. Below is my proof:

Let $f_n(x)$ be a sequence such that $\forall\epsilon>0\exists N>0,\forall m,n>N,|f_m(x)-f_n(x)|<\epsilon$. Prove uniform convergence for this sequence.

Define $f(x)=lim_{n\rightarrow\infty}f_n(x)$. This is well defined as $f_n(x)$ is a cauchy sequence for all x.

For fixed m > N and a given $\epsilon>0$, $\forall n>N, |f_m(x)-f_n(x)|<\epsilon$.

Or $f_m(x)-\epsilon<f_n(x)<f_m(x)+\epsilon$.

$f_m(x)-\epsilon<\lim_{n\rightarrow\infty}f_n(x)<f_m(x)+\epsilon$.

Or $f_m(x)-\epsilon<f(x)<f_m(x)+\epsilon$.

i.e, $|f_m(x)-f(x)|<\epsilon$.

Since m was chosen arbitrarily, we have

$\forall m>N, |f_m(x)-f(x)|<\epsilon$

Which proves uniform convergence.

Is there any mistake in above reasoning. I would be grateful for anyone to point it out.

symplectomorphic
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somitra
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1 Answers1

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It's almost perfect. Two caveats:

1) The line starting with "For fixed $m > N$ and a given $\epsilon>0$, ..." should be changed to "Let $\epsilon>0$. Choose $N$ so that $|f_n(x)-f_m(x)|<\epsilon$ for all $n,m>N$ and for all $x$.".

The $\epsilon$ is chosen first, and then you find your corresponding $N$. Also, you need to state your inequality holds for all $x$.

2) The line

$\ \ \ \ f_m(x)-\epsilon<\lim_{n\rightarrow\infty}f_n(x)<f_m(x)+\epsilon$

should be changed to

$\ \ \ \ f_m(x)-\epsilon\le\lim_{n\rightarrow\infty}f_n(x)\le f_m(x)+\epsilon$.

When passing to the limit, you might lose the strict inequality. The foregoing lines should be changed accordingly.

In the end, you'll get $|f_m(x)-f(x)|\le\epsilon$ for all $m>N$ and all $x$. This is ok; but if you're concerned about the non-strict inequality, just start the proof with $N$ chosen for $\epsilon/2$.

David Mitra
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