Lebesgue Fundamental calculus theorem

Question

How can I formally show that if $f:[0,1]\to \mathbb{R}$ continuous, $f'(x)$ exists for everything $x\in[0,1]$ y $\sup_x |f'(x)|=B<\infty$ then $\sup_{n,x} |g_n(x)|\leq B$ with $g_n(x)=\frac{f(x+1/n)-f(x)}{1/n}$

I know that $f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ and here I do not know if it can be done immediately that $f'(x)=\lim_n  g_n(x).$ In this way, the idea is to "turn that lim into a sup" but I do not know formally how to do it.

How would it be?

Also like to prove that if $f$ is continuous, for any $x_0\in [0,1]$ then

$\lim_n (\frac{1}{n})\int_{0}^{1/n}[f(x_0+x)]dx=f(x_0)$?

If $f(0)=0$. How proves that $f(y)=\int_{0}^{y}f'(x)dx$? I know that it must be obtained from the above but I can not see the trick

Answer 1

First part is immediate from Mean Value Theorem. For second part use the fact that $$| n \int_0^{1/n} f(x_0+x)dx-f(x_0)|=| n \int_0^{1/n} [f(x_0+x)-f(x_0)]dx|$$ $$\leq n \int_0^{1/n}|f(x_0+x)dx-f(x_0)|dx.$$ If $n$ is large enough then $|f(x_0+x)dx-f(x_0)| <\epsilon$ for all $x \in [0,1/n]$. Can you complete the proof now? [I have used the fact that $ n \int_0^{1/n} [f(x_0+x)-f(x_0)]dx=n \int_0^{1/n} f(x_0+x)dx-f(x_0)$ which follows from the fact that $f(x_0)$ is a constant].

Answer 2

First part is wrong. Take $f(x)=x^2$, then $B=2$ whereas $g_1(1)=3>B$.

Scond part is fundamental theorem of calculus: Define $F(y) := \int_0^y [f(x_0+x)] \, dx$ for $y\in [0,1-x_0]$. Then, $F$ is differentiable with $F'(y)=f(x_0+y)$, in particular $F'(0)=f(x_0)=\lim_n (\frac{1}{n})^{-1} \int_0^{1/n}[f(x_0+x)]\,dx$.

Answer 3

To address the first question in a bit more detail. In a sense the general approach to this type of questions would be to separate the set $A=\{n: n=1,2,\dots\}$ to $B=\{n:\ n=1,2,\dots,N\}$ and $C:=\{n:\ n=N+1,N+2,\dots\}$. Working with the first set is usually easy and for the second set you can use the limit result.

In more detail, we have $$ A = \{|g_{n}(x)|:\ n\in\mathbb{N},x\in[0,1]\} $$ and then split this one up in the two parts where $n$ is finite and the rest. To be precise, let $\epsilon>0$, then because $g_n(x)\to f'(x)$ there will exist an $N_{\epsilon}\in\mathbb{N}$ such that $|g_n(x) - f'(x)|\leq \epsilon$ for all $n\geq N_{\epsilon}$. In particular, $$|g_n(x)| \leq |f'(x)| + \epsilon \leq B + \epsilon $$ for all $n\geq N_{\epsilon}$. So this essentially covers up the second part since this implies that $$ \sup_{n\geq N_{\epsilon}}|g_n(x)| \leq B + \epsilon $$ Note that we cannot just let $\epsilon$ tend to $0$ because in general $N_{\epsilon}$ would then converge to infinity.

For the first part, you should use that $g_n$ is continuous and thus bounded on compact sets such as $[0,1]$. So we could get $|g_n(x)| \leq \max_{x\in[0,1]} |g_n(x)| = M_n < \infty$ and since we are considering only a finite number of these $n=1,2,\dots,N_{\epsilon}$ this will still be bounded. Indeed, $$ \sup_{n\leq N_{\epsilon}} \sup_{x\in[0,1]} |g_n(x)| = \sup_{n\leq N_{\epsilon}} M_n = \max(M_1,M_2,\dots,M_{N_\epsilon}) < \infty. $$

In the end, you get $$ \sup_{n\in\mathbb{N}}\sup_{x\in[0,1]} |g_n(x)| \leq \max(M_1,M_2,\dots,M_{N_\epsilon},B+\epsilon). $$ for any $\epsilon$. This is enough if the goal was for example to show that this supremum is bounded (take for example $\epsilon=1$ and you'll see that the supremum will be finite. However, as Maliesen pointed out, you cannot expect that this $|g_n(x)| \leq B$ but if the goal was to bound it, my line of reasoning is the best you can do.