# 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 $\displaystyle X$.
The content structure involves a mathematical question about the convergence of a series of functions, a proof provided for the question, and a critique of the proof by another user, followed by a discussion on the validity of the proof and the conditions under which it holds.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 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 $\displaystyle X$.
user
The content structure involves a mathematical question about the convergence of a series of functions, a proof provided for the question, and a critique of the proof by another user, followed by a discussion on the validity of the proof and the conditions under which it holds.

TYPE: SINGLE_QUESTION_WITH_ANSWER