My question is how to have uniform convergence in first theorem.

The statement of the two theorems.

Theorem 1: Let $\{f_n\}$ be a sequence of real valued functions on an open interval $I\in \mathbb{R}$, each having continuous derivative. Suppose that $\{f'_n\}$ is uniformly convergent on $I$ and that for some $a\in I$ $f_n(a)$ converges. Then $\lim_{n \to \infty}f_n$ exists, is differentiable, and $$(\lim_{n \to \infty}f_n)'=\lim_{n \to \infty} f_n'$$

Theorem 2: Let $\{f_n\}$ be a sequence of continuously differentiable on $[a,b]$ and $f_n(x_0)$ converges for some $x_0 \in [a,b]$. If $\{f_n'\}$ converges uniformly on $[a,b]$ , then $\{f_n\}$ converges uniformly on $[a,b]$ to some function $f$ and $$f'(x)=\lim_{n \to \infty }f'_n(x)$$

Now I noticed that the first theorem does not conclude that $\{f_n\}$ converges uniformly. In the proof of Theorem I, it breaks down.

Proof (Theorem I) By FTC $$\int_a^xf'_n(t)dt=f_n(x)-f_n(a)$$ for any $x\in I$ and any $n=1,2,3,...$. Let $\lim_{n \to \infty} f'_n=g$. Now we know that $\lim_{n \to \infty}\int_{x_0}^x f'_n=\int_{x_0}^x g$. Since $$\lim_{n\to \infty}f_n(a)=f(a)$$ $$\implies \lim_{n\to \infty} (f_n(x)-f_n(a))=\int_a^x g$$ for any $x\in I$. The limit exists it is $$\lim_{n\to \infty}f_n(x)=f(a)+\int_a^x g$$. Let $f(x)= f(a)+\int_a^x g$. Cannot get uniform convergence because for $N$ large enough $|f_n(a)-f(a)|\lt \epsilon$ and for $N^*$ large enough $|f_n(x)-g(x)|\lt \epsilon$. We choose $\max(N,N^*)=M$. So we have $$|f_n(x)-f(x)|=|f_n(a)-f(a)+\int_a^xf'_n-\int_a^xg|$$ $$\le|f_n(a)-f(a)|+\left|\int_a^xf'_n-g\right|\le \epsilon+|x-a|\epsilon$$ So we have pointwise convergence for $n\ge M$.

Is it enough to make $I$ bounded i.e $|I|\le C$ where $C\in \mathbb{R}$ so that we can do the following $|f_n(x)-g(x)|\lt \frac{\epsilon}{C}$ which would allow $$|f_n(a)-f(a)|+\left|\int_a^xf'_n-g\right|\le \epsilon+|x-a|\frac{\epsilon}{C}\le \epsilon+\epsilon$$ because $\frac{|x-a|}{C}\le 1$


Difference Between Theorems. Both theorems have the same hypotheses except that in Theorem 1 the interval is open while in Theorem 2 the interval is compact. Both theorems have the same conclusion except Theorem 1 does not have uniform convergence of $\{f_n\}$ while the Theorem 2 does.

Answer Part 1. The OP asks if assuming in Theorem 1 that the open interval $I$ is bounded is enough to get uniform convergence of $\{f_n\}$. The argument the OP suggests for this is correct. So the answer to that question is is yes.

Remark. The proof of the first theorem goes through if we only assume the $f'_n$ are integrable on any compact subinterval of $I$ rather than assuming the $f'_n$ are continuous on $I$.

Remark. In Theorem 1 we can conclude uniform convergence on any compact subinterval of $I$ by applying the second theorem to any such interval.

Remark. The hypothesis that the derivatives $f'_n$ are continuous is not necessary in the Theorem 2. See Theorem 7.17 in "Principles of Mathematical Analysis" by Rudin. See also the remark following the proof of that theorem.

Answer Part 2. No additional hypotheses are necessary in Theorem 1 to get uniform convergence of $\{f_n\}$ on $I$. In fact, the hypothesis that the derivatives $f'_n$ are continuous can be dropped. See Theorem 9.13 in "Mathematical Analysis" by Apostol. The proof is based on the mean value theorem rather than the fundamental theorem of calculus.


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