Supremum of f(x) on [a,b] vs. supremum of f(x+c) on [a+c, b+c]

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Discussion Overview

The discussion revolves around the relationship between the supremum of a function \( f(x) \) over the interval \([a,b]\) and the supremum of the function \( f(x+c) \) over the interval \([a+c,b+c]\). Participants explore whether these two supremums are equal and the implications of a potential typo in the original question.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant initially proposes that \( \alpha = \sup_{x\in [a,b]} f(x) \) should equal \( \sup_{x\in [a+c,b+c]} f(x+c) \).
  • Another participant challenges this assertion, providing a counterexample with specific values for \( a, b, c, \) and \( f(x) \).
  • Subsequent posts clarify that the original question contained a typo, suggesting it should be \( \sup_{x\in [a,b]} f(x) = \sup_{x\in [a+c,b+c]} f(x-c) \).
  • One participant suggests proving the equality of two sets derived from the function values over the specified intervals to establish the relationship between the supremums.
  • Another participant expresses gratitude for the responses but indicates a misunderstanding due to internet issues, reiterating their request for suggestions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the original claim regarding the supremums. There is disagreement about the validity of the initial assertion, and the discussion remains unresolved regarding the correct relationship between the supremums.

Contextual Notes

The discussion includes a potential typo that affects the interpretation of the problem, leading to confusion among participants. The assumption that the supremum exists is also noted but not resolved.

OhMyMarkov
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Hello everyone!

I'm really stuck on this one, it looks so obvious, but I can't prove it:

Let $\alpha = \sup _{x\in [a,b]} f(x)$, how can I show that $\alpha = \sup _{x\in [a+c,b+c]} f(x+c)$?

Thanks!
 
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Re: Supremum of $f(x)$ on $[a,b]$ vs. supremum of $f(x+c)$ on $[a+c,b+c]$

OhMyMarkov said:
Hello everyone!

I'm really stuck on this one, it looks so obvious, but I can't prove it:

Let $\alpha = \sup _{x\in [a,b]} f(x)$, how can I show that $\alpha = \sup _{x\in [a+c,b+c]} f(x+c)$?

Thanks!
Don't think that's true. Put $a=0,b=1,c=1,f(x)=x$.

I think you mistyped the question. It should be $\sup_{x\in[a,b]}f(x)=\sup_{x\in[a+c,b+c]}f(x-c)$
 
Re: Supremum of $f(x)$ on $[a,b]$ vs. supremum of $f(x+c)$ on $[a+c,b+c]$

Yes that was a typo... Any answers?
 
Re: Supremum of $f(x)$ on $[a,b]$ vs. supremum of $f(x+c)$ on $[a+c,b+c]$

OhMyMarkov said:
Yes that was a typo... Any answers?
Try proving $\{f(x):x\in [a,b]\}=\{f(x-c):x\in [a+c,b+c]\}$
 
Re: Supremum of $f(x)$ on $[a,b]$ vs. supremum of $f(x+c)$ on $[a+c,b+c]$

Yes that was a typo, any suggestions?
 
Re: Supremum of $f(x)$ on $[a,b]$ vs. supremum of $f(x+c)$ on $[a+c,b+c]$

OhMyMarkov said:
Yes that was a typo, any suggestions?
In post#4 I suggest you try proving equality of two sets. Why those sets are the same is quite straight-forward.

Write $A=\{f(x):x\in[a,b]\}$ and $B=\{f(x-c):x\in[a+c,b+c]\}$. Let $y\in A$.
Then $\exists r\in[a,b]$ such that $f(r)=y\Rightarrow f((r+c)-c)=y$.
But $r+c$ is in $[a+c,b+c]$. This proves $y\in B$.

We conclude $A\subseteq B$. Show the reverse inclusion too. Then it easily follows that $\sup A=\sup B$ simple because $A$ and $B$ are the same. Of course this assumes that the supremum exists.
 
Re: Supremum of $f(x)$ on $[a,b]$ vs. supremum of $f(x+c)$ on $[a+c,b+c]$

Hello caffeinemachine, thanks for replying (twice)

I may have sounded rude repeating "any suggestions?" twice, but that was due to an internet problem. I did exactly what you suggested the 1st time :)
 
Re: Supremum of $f(x)$ on $[a,b]$ vs. supremum of $f(x+c)$ on $[a+c,b+c]$

OhMyMarkov said:
Hello caffeinemachine, thanks for replying (twice)

I may have sounded rude repeating "any suggestions?" twice, but that was due to an internet problem. I did exactly what you suggested the 1st time :)
(Yes)
 

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