Prove that a sequence of functions has a convergent subsequence

In summary, to prove that \{ f_{n} \}_{n=1}^{\infty} has a convergent subsequence, we can use the fact that C[0,1] is a compact metric space and show that the sequence is bounded and equicontinuous. The Taylor polynomial and its remainder term can be used to show that the sequence is bounded, and the definition of equicontinuity can be used to show that it is equicontinuous. This allows us to conclude that there exists a convergent subsequence of \{ f_{n} \}_{n=1}^{\infty}.
  • #1
jeckt
19
0

Homework Statement



Let [tex] \{ f_{n} \}_{n=1}^{\infty} \subset C[0,1] [/tex] be twice differentiable, and satisfying [tex] 0 = f_{n}(0) = f'_{n}(0) [/tex] and [tex] \| f''_{n}\|_{\infty } [/tex]. Prove that [tex] \{ f_{n} \}_{n=1}^{\infty} [/tex] has a convergent subsequence.

Homework Equations



So since [tex] C[0,1] [/tex] is a compact metric space. If [tex] \{ f_{n} \}_{n=1}^{\infty} \subset C[0,1] [/tex] is bounded and equicontinuous, then it has a convergent subsequence.

The Attempt at a Solution



I can show that it is bounded. The hint the lecturer gave us was to consider the remainder term of the taylor polynomial. That is what I used to show it is bounded. As for equicontinuous - I'm having a little difficult, I'm trying to prove it using the definition and the remainder term of the taylor polynomial.

Thanks for the help guys!
 
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  • #2


Hello,

Thank you for your post. It seems like you are on the right track with using the Taylor polynomial and the remainder term to show that the sequence is bounded. To prove equicontinuity, you can start by using the definition of equicontinuity: for any \epsilon > 0, there exists a \delta > 0 such that for all x, y \in [0,1], if |x-y| < \delta, then |f(x) - f(y)| < \epsilon for all f \in \{ f_{n} \}_{n=1}^{\infty}.

From there, you can use the fact that the sequence is twice differentiable to show that the remainder term of the Taylor polynomial can be made arbitrarily small for any given \epsilon. This will allow you to choose a corresponding \delta and prove equicontinuity.

Once you have shown that the sequence is bounded and equicontinuous, you can use the fact that C[0,1] is compact to conclude that there exists a convergent subsequence of \{ f_{n} \}_{n=1}^{\infty}. I hope this helps and good luck with your proof!
 

1. What is a sequence of functions?

A sequence of functions is a list of functions, where each function is associated with a specific value or term in the sequence. The functions can have different mathematical expressions, but they all share a common domain and range.

2. What does it mean for a sequence of functions to have a convergent subsequence?

A convergent subsequence of a sequence of functions is a subset of the original sequence where the functions approach a single limit as the terms in the subset approach infinity. In other words, the functions in the subsequence get closer and closer to each other as the terms increase, eventually converging to a single value.

3. Why is proving the existence of a convergent subsequence important?

Proving the existence of a convergent subsequence is important because it allows us to make conclusions about the behavior of the original sequence of functions. It also helps us determine if the sequence is bounded or unbounded, and if it has a limit or not.

4. How do you prove that a sequence of functions has a convergent subsequence?

One method is to use the Bolzano-Weierstrass theorem, which states that any bounded sequence of real numbers has a convergent subsequence. This can be applied to a sequence of functions by showing that the functions are bounded and then finding a convergent subsequence using the theorem.

5. Can a sequence of functions have multiple convergent subsequences?

Yes, a sequence of functions can have multiple convergent subsequences. In fact, a sequence can have infinitely many convergent subsequences, as long as there are enough terms in the sequence. Each convergent subsequence will converge to a different limit.

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