# Thread: Is there something wrong with this proof?

1. ## Is there something wrong with this proof?

Someone please tell me if this proof is wrong:

QUESTION: Let $\displaystyle f_n:X\to\mathbb{R}$ be a sequence of functions such that $\displaystyle f_n(x)\neq 0$ for all $\displaystyle n\in\mathbb{N}$ and for all $\displaystyle x\in X$. If there exists $\displaystyle c\in\mathbb{R}$ and $\displaystyle N\in\mathbb{N}$ such that $\displaystyle |\frac{f_{n+1}(x)}{f_n(x)}|\leq c<1$ for all $\displaystyle n\geq N$ and all $\displaystyle x\in X$ then $\displaystyle \sum |f_n(x)|$ and $\displaystyle \sum f_n(x)$ converge uniformly on X.

PROOF: Let $\displaystyle x\in X$ and $\displaystyle n\geq N$. Then $\displaystyle |\frac{f_{N+1}(x)}{f_N(x)}|\leq c$, $\displaystyle |\frac{f_{N+2}(x)}{f_{N+1}(x)}|\leq c$, ... , $\displaystyle |\frac{f_{n}(x)}{f_{n-1}(x)}|\leq c$. Multiplying all these inequalities we get $\displaystyle |\frac{f_{n}(x)}{f_N(x)}|\leq c^{n-N}\Rightarrow |f_n(x)|\leq c^{n-N}|f_N(x)|$. If we put $\displaystyle a_n=c^{n-N}|f_N(x)|$ then the conclusion follows from the Weierstrass M-Test, since $\displaystyle \sum a_n$ is convergent.

2. Originally Posted by math9
Someone please tell me if this proof is wrong:

QUESTION: Let $\displaystyle f_n:X\to\mathbb{R}$ be a sequence of functions such that $\displaystyle f_n(x)\neq 0$ for all $\displaystyle n\in\mathbb{N}$ and for all $\displaystyle x\in X$. If there exists $\displaystyle c\in\mathbb{R}$ and $\displaystyle N\in\mathbb{N}$ such that $\displaystyle |\frac{f_{n+1}(x)}{f_n(x)}|\leq c<1$ for all $\displaystyle n\geq N$ and all $\displaystyle x\in X$ then $\displaystyle \sum |f_n(x)|$ and $\displaystyle \sum f_n(x)$ converge uniformly on X.

PROOF: Let $\displaystyle x\in X$ and $\displaystyle n\geq N$. Then $\displaystyle |\frac{f_{N+1}(x)}{f_N(x)}|\leq c$, $\displaystyle |\frac{f_{N+2}(x)}{f_{N+1}(x)}|\leq c$, ... , $\displaystyle |\frac{f_{n}(x)}{f_{n-1}(x)}|\leq c$. Multiplying all these inequalities we get $\displaystyle |\frac{f_{n}(x)}{f_N(x)}|\leq c^{n-N}\Rightarrow |f_n(x)|\leq c^{n-N}|f_N(x)|$. If we put $\displaystyle a_n=c^{n-N}|f_N(x)|$ then the conclusion follows from the Weierstrass M-Test, since $\displaystyle \sum a_n$ is convergent.
In the last line of the proof, you can't apply the M-test in that way, because $\displaystyle a_n$ depends on $\displaystyle x$. The M-test only applies if $\displaystyle (a_n)$ is a sequence of constants. If the function $\displaystyle f_N(x)$ is bounded, say $\displaystyle |f_N(x)|\leqslant K$ for all $\displaystyle x\in X$, then you can take $\displaystyle a_n = c^{n-N}K$. In that case, the M-test does apply and the conclusion is correct. In particular, if the space $\displaystyle X$ is compact, then $\displaystyle f_N(x)$ will automatically be bounded and the result is correct.

But suppose for example that $\displaystyle X=\mathbb{R}$ and $\displaystyle f_n(x) = 2^{-n}x$. Then $\displaystyle \sum_{n=1}^\infty f_n(x) = x$ for all $\displaystyle x\in\mathbb{R}$, but the convergence is not uniform.

3. Originally Posted by Opalg
In the last line of the proof, you can't apply the M-test in that way, because $\displaystyle a_n$ depends on $\displaystyle x$. The M-test only applies if $\displaystyle (a_n)$ is a sequence of constants. If the function $\displaystyle f_N(x)$ is bounded, say $\displaystyle |f_N(x)|\leqslant K$ for all $\displaystyle x\in X$, then you can take $\displaystyle a_n = c^{n-N}K$. In that case, the M-test does apply and the conclusion is correct. In particular, if the space $\displaystyle X$ is compact, then $\displaystyle f_N(x)$ will automatically be bounded and the result is correct.

But suppose for example that $\displaystyle X=\mathbb{R}$ and $\displaystyle f_n(x) = 2^{-n}x$. Then $\displaystyle \sum_{n=1}^\infty f_n(x) = x$ for all $\displaystyle x\in\mathbb{R}$, but the convergence is not uniform.
I see.... Do you have any suggestions as to how I can fix my proof? Or maybe another proof?

4. Originally Posted by math9
I see.... Do you have any suggestions as to how I can fix my proof? Or maybe another proof?
No. Unless you have some extra condition to ensure that $\displaystyle f_N(x)$ is bounded, the result is false (as the example at the end of my previous comment shows).

5. Originally Posted by Opalg
No. Unless you have some extra condition to ensure that $\displaystyle f_N(x)$ is bounded, the result is false (as the example at the end of my previous comment shows).
Well.. the example you gave doesn't really show the result is false since your sequence of functions take on the value 0 at x=0 and one of the hypothesis of the problem is that $\displaystyle f_n(x)\neq0$ for all $\displaystyle x \in X$.

6. Originally Posted by math9
Well.. the example you gave doesn't really show the result is false since your sequence of functions take on the value 0 at x=0 and one of the hypothesis of the problem is that $\displaystyle f_n(x)\neq0$ for all $\displaystyle x \in X$.
I can easily fix that by restricting the domain to the interval $\displaystyle X = [1,\infty)$. The hypotheses of the question are then all satisfied, but $\displaystyle \sum f_n(x)$ does not converge uniformly on $\displaystyle X$.