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

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SUMMARY

The discussion centers on proving the equality of the supremum of a function f(x) over the interval [a,b] and the supremum of f(x+c) over the interval [a+c,b+c]. The key conclusion is that if α = sup_{x∈[a,b]} f(x), then α = sup_{x∈[a+c,b+c]} f(x+c) holds true under the condition that the supremum exists. The proof involves demonstrating the equality of the sets A = {f(x): x∈[a,b]} and B = {f(x-c): x∈[a+c,b+c]}, establishing that both sets contain the same elements.

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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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