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

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The discussion centers on proving that the supremum of a function f(x) over the interval [a, b] is equal to the supremum of f(x+c) over the interval [a+c, b+c]. A user initially posed the question but later acknowledged a typo, clarifying that the correct relationship involves f(x-c) instead. Another participant suggested proving the equality of two sets derived from these intervals to establish the relationship between the suprema. The conversation emphasizes the importance of demonstrating that the sets of function values are equivalent, leading to the conclusion that their suprema are also equal. The discussion remains focused on mathematical proofs and clarifications regarding the supremum concept.
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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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