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The Question: Let ${(f_{n}): \mathbb{R} \rightarrow \mathbb{R}}$ be a sequence of continuously differentiable functions such that the sequence of derivatives ${f'_{n}: \mathbb{R} \rightarrow \mathbb{R} }$ is uniformily convergent and the sequence ${f_n(0)}$ is convergent. Prove that $f_{n}(x)$ is pointwise convergent.

The attempt:

Here is what I have. I need to show $\lim_{n \rightarrow \infty} f_{n}(x) = f(x)$ , or equivently, $\lim_{n \rightarrow \infty} f_{n}(x) -f(x) = 0$

Let $\epsilon > 0$ and let $x_{0} \in \mathbb{R}.$

Then,

$\lim_{n \rightarrow \infty} (f_{n}(x) -f(x)) = \lim_{n \rightarrow \infty} (x-x_{0})\frac{(f_{n}(x) -f(x))}{(x-x_{0})} = \lim_{n \rightarrow \infty} (x-x_{0})\frac{(f_{n}(x) -f_{n}(x_{0}) +f_{n}(x_{0}) -f(x))}{(x-x_{0})}$.

This is where I am stuck. I am not sure where to go at this point. Am I on the right track?

Thank you very much!!

3 Answers 3

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HINTS:

  1. Use $\epsilon{-}N$ Cauchy characterization of convergent sequences.

  2. After write an upper bound of $|f_n(x)-f_m(x)|$ using the convergency of $(f_n(0))$.

  3. Last step: use the mean value theorem over the function $f_n-f_m$.

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Notice that

$$f_n(x) = f_n(0) + \int_0^x f'_n(t) dt $$

Now let's prove that $(f_n(x))_n$ is a Cauchy sequence :

$$ \left| f_n(x) - f_m (x) \right| \leq \left| f_n(0) - f_m (0) \right| + \left| \int_0^x f'_n(t) - f'_m(t) dt \right| $$

$$\leq \left| f_n(0) - f_m (0) \right| + \int_0^x \| f'_n- f'_m \| dt $$

$$\leq \left| f_n(0) - f_m (0) \right| + |x| \| f'_n- f'_m \| $$

$$\leq \left| f_n(0) - f_m (0) \right| + |x| \left( \| f'_n- f'\| + \| f'_m - f' \| \right) $$

Now, as $f_n(0)$ is convergent, it a Cauchy sequence, so for $m$ and $n$ big enough, $\left| f_n(0) - f_m (0) \right| \leq \epsilon$

And for $m$ and $n$ big enough, you also have $ |x| \left( \| f'_n- f'\| + \| f'_m - f' \| \right) \leq \epsilon$

Hence for $m$ and $n$ big enough, $ \left| f_n(x) - f_m (x) \right| \leq 2\epsilon$ : it is a Cauchy sequence, so it converge

0

I think you should try to reformulate your problem with what you know.

($f_n$) is continously differentiable, one can write:

$$f_n (x)=\int_{0}^{x} f'_{n}(t) dt +f_n(0)$$

Likewise, ($f'_n$) is uniformly convergent toward a function that i call f' (continuous thanks to uniform convergence) ; let:

$$f(x)=\int_{0}^{x}f'(t)dt+l$$ with $l=lim (f_n(0) )$

Now you want to consider :

$$f_n(x)-f(x) = (f_n(0)-l) + \int_{0}^{x}[f'_n(t)-f'(t)]dt$$

And you need to prove that both right side terms converge, to zero. The first term is obvious by definition, the integrale is all that's left :) ... and you might want to consider the uniform convergence here