Applying the Mean Value Theorem to sequences of function

[jdinatale]

As usual, I typed up the problem and my attempt in LaTeX:

Maybe I'm not applying the MVT correctly, but my result does not seem to help me solve the problem in anyway. What are your thoughts?

 

[micromass]

Maybe you shouldn't apply the mean value theorem on [a,b] but rather on [x,x₀]??

 

[jdinatale]

Maybe you shouldn't apply the mean value theorem on [a,b] but rather on [x,x₀]??

Thanks, that would help. Ok I tried your suggestion and here is the result. The problem is now making use of the $f'_n(c) - f'_m(c)$ since they do not appear in the triangle inequality. And also, I am not using the assumption that $f'_n$ converges uniformly.

 

[micromass]

In your triangle inequality, you have a term

$$|(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))|$$

Isn't this the same as the numerator of what you got after applying the mean value theorem??

 

[jdinatale]

In your triangle inequality, you have a term

$$|(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))|$$

Isn't this the same as the numerator of what you got after applying the mean value theorem??

Well, no, it's slightly different:

$$|f_n(x) + f_m(x) - f_n(x_0) - f_m(x_0)|$$

We need it to be $$f_n(x) - f_m(x)$$ instead of +

Oh btw, I noticed in your profile that it says you are 15. Is that true? Because that's incredible that you know so much math at that age, really astonishing.

 

[micromass]

Oh, how did you get that + in the mean value theorem? That isn't correct. The mean value theorem says that

$$(f_n-f_m)^\prime(c)(x-x_0)=(f_n-f_m)(x)-(f_n-f_m)(x_0)$$

You seemed to have applied the mean-value theorem to $f_n+f_m$...

 

[jdinatale]

Micromass, I believe I have found the solution! Here is my completed proof, what do you think?

 

[micromass]

Hmm, you know that that c is depend on n and m right?? You act like there is only one c, but there are multiple c's.

And to be sure you really get it: where exactly do you need uniform convergence of $(f^\prime_n)_n$??

 

[jdinatale]

Hmm, you know that that c is depend on n and m right?? You act like there is only one c, but there are multiple c's.

And to be sure you really get it: where exactly do you need uniform convergence of $(f^\prime_n)_n$??

Oh, well it was my understanding that the c was only dependent on the $$[x, x_0]$$. I guess I didn't understand that c could be different for different m's and n's.

We need uniform convergence of $(f^\prime_n)_n$ because we have to use the Cauchy Criterion for Uniform Convergence.

Well, since my proof is incorrect then, could you point me in the right direction so that I can try to complete this?

 

[micromass]

The mean value theorem on a function actually implies:

$$|g(a)-g(b)|\leq |a-b| \sup_{c\in [a,b]}|g^\prime(c)|$$

I suggest you use this. It will become apparent where the uniform convergence is used.

 

[jdinatale]

The mean value theorem on a function actually implies:

$$|g(a)-g(b)|\leq |a-b| \sup_{c\in [a,b]}|g^\prime(c)|$$

I suggest you use this. It will become apparent where the uniform convergence is used.

I guess I'm not quite sure I understand. My book has no mention of this variation to the MVT. Are you certain that it is necessary to solve this problem?

 

[micromass]

I guess I'm not quite sure I understand. My book has no mention of this variation to the MVT.

It's quite easy to prove from the MVT.

Are you certain that it is necessary to solve this problem?

I think it's the easiest and cleanest way...