Thread: vexingly simple proof (sum periodic functns)

1. vexingly simple proof (sum periodic functns)

I have a feeling this proof is supposed to be more complex, is this legitimate?
Let f(x) and g(x) be continuous, periodic functions defined for -infinity<x<+infinity. If limx->+infinity (f(x)-g(x))=0 prove that f(x):=g(x).

Proof:
Let h(x)=f(x)-g(x).

-g(x) is continuous as it is the product of continuous functions t(x)=-1 and g(x). h(x) is continuous as it is the sum of continuous functions f(x) and -g(x).

Therefore, (f(x)-g(x))=limx->+infinity (f(x)-g(x))=0 for all x>N where N is the N defined in "if limx->+infinity (f(x)-g(x))=0 <=> for any epsilon>0, if x>N then abs((f(x)-g(x))-0)<epsilon."

Now, f(x)-g(x)=0 for all x>N.

If f(x) has period p then f(x+p)-g(x+p)=0 for all x+p>x>N.

Now, f(x)-g(x)=0=f(x+p)-g(x+p) for all x+p>x>N.

Hence f(x)-f(x+p)-g(x)=-g(x+p) for all x+p>x>N.

But f(x)=f(x+p) as p is a period of f(x); therefore, -g(x)=-g(x+p) i.e. g(x)=g(x+p) for all x+p>x>N.

All x+p>x>N includes all (x+(m/q)b)+p>x>N where (m/q) is any rational number and b is the base period of g(x).Therefore, p must be a period of the periodic function g(x).

Now, for any integer k such that (x-kp)<N i.e. k>((x-N)/p), f(x)=f(x-kp) and g(x)=g(x-kp) [as p is a period of both f(x) and g(x)].

Therefore, 0=f(x)-g(x)=f(x-kp)-g(x-kp) for all x>N, i.e. for all (x-kp)<N.

Therefore, f(x-kp)-g(x-kp)=0 for all x>N, i.e. for all (x-kp)<N; and f(x)-g(x)=0 for all x>N.

Note that the same logic follows if k is replaced by some integer z such that (x-z)<<N i.e. z>>((x-N)/p) [just to reaffirm that this proof holds for x approaching -infinity aswell].

Therefore, f(x)=g(x) for all x.

Finally, as f(x) and g(x) have the same domain, -infinity<x<+infinity, and as f(x) and g(x) also have the same values at each x in their domain, i.e. f(x)=g(x) for all x in -infinity<x<+infinity, it must be true that f(x):=g(x). /QED

Is my proof faulty because I didn't prove f(N)=g(N) so I didn't really prove f(x)=g(x) for ALL x? Could I fix this by defining k as any integer such that k>or=((x-N)/p)?

2. Originally Posted by 234578
I have a feeling this proof is supposed to be more complex, is this legitimate?
Let f(x) and g(x) be continuous, periodic functions defined for -infinity<x<+infinity. If limx->+infinity (f(x)-g(x))=0 prove that f(x):=g(x).

Proof:
Let h(x)=f(x)-g(x).

-g(x) is continuous as it is the product of continuous functions t(x)=-1 and g(x). h(x) is continuous as it is the sum of continuous functions f(x) and -g(x).

Therefore, (f(x)-g(x))=limx->+infinity (f(x)-g(x))=0 for all x>N where N is the N defined in "if limx->+infinity (f(x)-g(x))=0 <=> for any epsilon>0, if x>N then abs((f(x)-g(x))-0)<epsilon."

Now, $\displaystyle f(x)-g(x){\color{red}=}0$ for all x>N.
How so?

If f(x) has period p then f(x+p)-g(x+p)=0 for all x+p>x>N.

Now, f(x)-g(x)=0=f(x+p)-g(x+p) for all x+p>x>N.

Hence f(x)-f(x+p)-g(x)=-g(x+p) for all x+p>x>N.

But f(x)=f(x+p) as p is a period of f(x); therefore, -g(x)=-g(x+p) i.e. g(x)=g(x+p) for all x+p>x>N.

All x+p>x>N includes all (x+(m/q)b)+p>x>N where (m/q) is any rational number and b is the base period of g(x).Therefore, p must be a period of the periodic function g(x).

Now, for any integer k such that (x-kp)<N i.e. k>((x-N)/p), f(x)=f(x-kp) and g(x)=g(x-kp) [as p is a period of both f(x) and g(x)].

Therefore, 0=f(x)-g(x)=f(x-kp)-g(x-kp) for all x>N, i.e. for all (x-kp)<N.

Therefore, f(x-kp)-g(x-kp)=0 for all x>N, i.e. for all (x-kp)<N; and f(x)-g(x)=0 for all x>N.

Note that the same logic follows if k is replaced by some integer z such that (x-z)<<N i.e. z>>((x-N)/p) [just to reaffirm that this proof holds for x approaching -infinity aswell].

Therefore, f(x)=g(x) for all x.

Finally, as f(x) and g(x) have the same domain, -infinity<x<+infinity, and as f(x) and g(x) also have the same values at each x in their domain, i.e. f(x)=g(x) for all x in -infinity<x<+infinity, it must be true that f(x):=g(x). /QED

Is my proof faulty because I didn't prove f(N)=g(N) so I didn't really prove f(x)=g(x) for ALL x? Could I fix this by defining k as any integer such that k>or=((x-N)/p)?

3. Originally Posted by 234578
I have a feeling this proof is supposed to be more complex, is this legitimate?
Let f(x) and g(x) be continuous, periodic functions defined for -infinity<x<+infinity. If limx->+infinity (f(x)-g(x))=0 prove that f(x):=g(x).
Here is a short proof (and I'm like Failure, I don't understand yours).

Let $\displaystyle T_f,T_g$ be the periods of $\displaystyle f,g$ respectively. For any $\displaystyle x$, we have $\displaystyle f(x)-g(x+k T_f)=f(x+k T_f)-g(x+k T_f)\to_k 0$ (where $\displaystyle k\in\mathbb{N}$), hence $\displaystyle g(x+k T_f)\to_k f(x)$. Therefore, we have $\displaystyle g((x+T_g)+kT_f)\to_k f(x+T_g)$ (taking $\displaystyle x+T_g$ as $\displaystyle x$ in the previous formula). But we also have $\displaystyle g((x+T_g)+kT_f)=g(x+kT_f)\to_k f(x)$ using the definition of $\displaystyle T_g$. As a consequence, comparing the limits, we have $\displaystyle f(x+T_g)=f(x)$.

Thus $\displaystyle T_g$ is a period for $\displaystyle f$ as well. Then the conclusion is straightforward: for all $\displaystyle x$, $\displaystyle f(x)-g(x)= f(x+kT_g)-g(x+kT_g)\to_k 0$ hence $\displaystyle f(x)=g(x)$.

NB: I did not even use the continuity of $\displaystyle f,g$...

4. Thanks Just to clarify, does the symbol$\displaystyle f(x)-g(x)\to_k 0$ mean the limit of f(x)-g(x) as x approaches infinity is zero? Or does it mean as x approaches k, in which case I was wrong to think I understood this proof--why is k in the limit sign? (sorry if I sound ignorant)

Originally Posted by Laurent
Here is a short proof (and I'm like Failure, I don't understand yours).

Let $\displaystyle T_f,T_g$ be the periods of $\displaystyle f,g$ respectively. For any $\displaystyle x$, we have $\displaystyle f(x)-g(x+k T_f)=f(x+k T_f)-g(x+k T_f)\to_k 0$ (where $\displaystyle k\in\mathbb{N}$), hence $\displaystyle g(x+k T_f)\to_k f(x)$. Therefore, we have $\displaystyle g((x+T_g)+kT_f)\to_k f(x+T_g)$ (taking $\displaystyle x+T_g$ as $\displaystyle x$ in the previous formula). But we also have $\displaystyle g((x+T_g)+kT_f)=g(x+kT_f)\to_k f(x)$ using the definition of $\displaystyle T_g$. As a consequence, comparing the limits, we have $\displaystyle f(x+T_g)=f(x)$.

Thus $\displaystyle T_g$ is a period for $\displaystyle f$ as well. Then the conclusion is straightforward: for all $\displaystyle x$, $\displaystyle f(x)-g(x)= f(x+kT_g)-g(x+kT_g)\to_k 0$ hence $\displaystyle f(x)=g(x)$.

NB: I did not even use the continuity of $\displaystyle f,g$...

5. Originally Posted by 234578
Thanks Just to clarify, does the symbol$\displaystyle f(x)-g(x)\to_k 0$ mean the limit of f(x)-g(x) as x approaches infinity is zero? Or does it mean as x approaches k, in which case I was wrong to think I understood this proof--why is k in the limit sign? (sorry if I sound ignorant)
This is the limit of the sequence indexed by $\displaystyle k$. Another notation is $\displaystyle a_k \to_{k\to\infty,\,k\in\mathbb{N}}\ell$.

In the case of my last limit (i.e. $\displaystyle f(x)-g(x)\to_k 0$), no $\displaystyle k$ appears in the "sequence"; therefore it is a constant sequence, and its limit is its common value, $\displaystyle f(x)-g(x)$. The idea is that by periodicity $\displaystyle f(x)-g(x)=f(x+kT_g)-g(x+kT_g)$, which now depends on $\displaystyle k$. And since $\displaystyle x+kT_g\to_k+\infty$, we may apply the hypothesis: we have $\displaystyle f(x+kT_g)-g(x+kT_g)\to_k 0$ (the limit when $\displaystyle x\to\infty$ is equal to the limit along any sequence $\displaystyle (x_k)_k$ tending to $\displaystyle +\infty$: $\displaystyle \lim_{x\to\infty}\phi(x)=\lim_k \phi(x_k)$ if the limit $\displaystyle \lim_{x\to\infty}\phi(x)$ exists).