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- #1

#### OhMyMarkov

##### Member

- Mar 5, 2012

- 83

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!

- Thread starter OhMyMarkov
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- Thread starter
- #1

- Mar 5, 2012

- 83

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!

- Mar 10, 2012

- 835

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

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!

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

- Thread starter
- #3

- Mar 5, 2012

- 83

Yes that was a typo... Any answers?

- Mar 10, 2012

- 835

Try proving $\{f(x):x\in [a,b]\}=\{f(x-c):x\in [a+c,b+c]\}$Yes that was a typo... Any answers?

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- #5

- Mar 5, 2012

- 83

Yes that was a typo, any suggestions?

- Mar 10, 2012

- 835

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

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.

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- #7

- Mar 5, 2012

- 83

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

- Mar 10, 2012

- 835

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