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Rudin 5.2

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[SOLVED] Rudin 5.2

Homework Statement


Suppose f'(x)>0 in (a,b). Prove that f is strictly increasing in (a,b), and let g be its inverse function. Prove that g is differentiable, and that

g'(f(x))=1/f'(x)

when a<x<b.


Homework Equations





The Attempt at a Solution


I can prove that f is strictly increasing. I cannot prove that the inverse exists. Do I have to go back to the epsilon-delta definition of the derivative for this or is there some clevel way I can manipulate the limits...
 

Answers and Replies

  • #2
Dick
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You've proved f is strictly increasing. f is continuous since f' exists, yes? And you can't prove the inverse exists???
 
  • #3
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You've proved f is strictly increasing. f is continuous since f' exists, yes? And you can't prove the inverse exists???
Sorry. I meant I cannot prove that the derivative of the inverse exists.
 
  • #4
Dick
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Define 'cannot'. Did you try? If f(x)=y and f is invertible then f^(-1)(y)=x. The difference quotient is (f(x)-f(y))/(x-y). Try using the notion of an inverse. Note f' is non-zero.
 
  • #5
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OK. So, let y be in the range of f (domain of g). Let f(x) = y. Then I want to show that

[tex]\lim_{t\to y}\frac{f(t) - x}{t-f(x)} = \lim_{s\to x} \frac{s-x}{f(s)-f(x)}[/tex]

As you pointed out the right limit exists and equals 1/f'(x). I am just stuck trying to prove that equality.

And cannot is defined as a lack of success after about 20 minutes of trying.
 
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  • #6
Dick
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If you've been trying to prove that equality for 20 minutes and failing, maybe it's not true. The limit on the left side is (f(y)-x)/(y-f(x)). That's not even a difference quotient. Try and set that up again.
 
  • #7
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[tex]
\lim_{t\to y}\frac{g(t) - x}{t-f(x)} = \lim_{s\to x} \frac{s-x}{f(s)-f(x)}
[/tex]
 
  • #8
Dick
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[tex]
\lim_{t\to y}\frac{g(t) - x}{t-f(x)} = \lim_{s\to x} \frac{s-x}{f(s)-f(x)}
[/tex]
Brilliant. This is far from the most difficult exercise in Rudin. If that's the closest you can come to actually expressing the problem in symbols, I don't think I can help you.
 
  • #9
Dick
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[tex]
\lim_{t\to y}\frac{g(t) - x}{t-f(x)} = \lim_{s\to x} \frac{s-x}{f(s)-f(x)}
[/tex]
Apologies for the tone of my previous post. I had forgotten you defined g=f^(-1) and COULD NOT figure out why you changed it like that. That's actually exactly what you want. Since f(x)=y, g(y)=x. So rewrite it as (g(t)-g(y))/(t-y). Sorry again.
 
  • #10
HallsofIvy
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I would think using the chain rule on f(g(x))= x would be simplest.
 
  • #11
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We need to prove that the inverse of f is differentiable first.
 
  • #12
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Apologies for the tone of my previous post. I had forgotten you defined g=f^(-1) and COULD NOT figure out why you changed it like that. That's actually exactly what you want. Since f(x)=y, g(y)=x. So rewrite it as (g(t)-g(y))/(t-y). Sorry again.
Well, I wrote it as (g(t)-g(y))/(t-y) at first but then I rewrote it as (g(t)-x)/(t-f(x)). I still have to prove the two limits above are equal.
 
  • #13
Dick
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Well, I wrote it as (g(t)-g(y))/(t-y) at first but then I rewrote it as (g(t)-x)/(t-f(x)). I still have to prove the two limits above are equal.
You still don't see it? Why not pick t=f(s), s=g(t)? Then the two sides ARE equal, even before the limits are taken.
 
  • #14
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You still don't see it? Why not pick t=f(s), s=g(t)? Then the two sides ARE equal, even before the limits are taken.
Sorry, I still do not see it. What do you mean pick t=f(s)? t and s are totally independent symbols. How does that even prove that

[tex] \lim_{t\to y}\frac{g(t) - g(y)}{t-y}[/tex]

even exists? For each \epsilon > 0, we need to find [itex]\delta>0[/itex] so that [itex]|t-y|<\delta[/itex] implies

[tex]\left|\frac{g(t) - g(y)}{t-y} - \lim_{s\to x} \frac{s-x}{f(s)-f(x)}\right|<\epsilon[/tex]


What is really hard is that limit inside the limit thing.

I don't see why the "two sides being equal before taking the limit" is a meaningful statement. Are they supposed to "stay" equal as you take the limit? What in the world does that mean?
 
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  • #15
Dick
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You are complicating this to death. DON'T consider t and s totally independent symbols. DO take t=f(s). NOW the two sides are equal. NOW take the limit s->x. The right side is a difference quotient and converges to 1/f'(x). The left side is a difference quotient for g'(y). What else is there to say?
 
  • #16
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OK. I agree that

[tex]\lim_{s\to x} \frac{g(f(s))-g(y)}{f(s)-y} = \frac{1}{f'(x)}[/tex]

But the heart of the problem is proving the left hand side is equal to

[tex] \lim_{t \to y} \frac{g(t)-g(y)}{t-y}[/tex]

Hmm. Maybe that is not so hard actually. By Theorem 4.2 in Rudin we just need to prove for every sequence converging to y, the difference quotient above converges to 1/f'(x). But if you give me a sequence {t_n} converging to y, I can find a sequence {s_n} converging to x such that f(s_n) = t_n so...yes that works. The reason {s_n=g(t_n)} converges to x is that by g is continuous (Rudin Thm 4.17).
 
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  • #17
Dick
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Yes! All of the functions here are continuous.
 

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