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How can I formally show that if $f:[0,1]to mathbb{R}$ continuous, $f'(x)$ exists for everything $xin[0,1]$ 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 $f'(x)=lim_{hto 0} frac{f(x+h)-f(x)}{h}$ 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 $x_0in [0,1]$ 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

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].

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 $yin [0,1-x_0]$. Then, $F$ is differentiable with $F'(y)=f(x_0+y)$, in particular $F'(0)=f(x_0)=lim_n (frac{1}{n})^{-1} int_0^{1/n}[f(x_0+x)],dx$.

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 $$ A = {|g_{n}(x)|: ninmathbb{N},xin[0,1]} $$
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 $N_{epsilon}inmathbb{N}$ such that $|g_n(x) - f'(x)|leq epsilon$ for all $ngeq N_{epsilon}$. In particular,
$$|g_n(x)| leq |f'(x)| + epsilon leq B + epsilon $$
for all $ngeq N_{epsilon}$. So this essentially covers up the second part since this implies that
$$ sup_{ngeq N_{epsilon}}|g_n(x)| leq B + epsilon $$
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 $|g_n(x)| leq max_{xin[0,1]} |g_n(x)| = M_n < infty$ and since we are considering only a finite number of these $n=1,2,dots,N_{epsilon}$ this will still be bounded.
Indeed,
$$ sup_{nleq N_{epsilon}} sup_{xin[0,1]} |g_n(x)| = sup_{nleq N_{epsilon}} M_n = max(M_1,M_2,dots,M_{N_epsilon}) < infty. $$

In the end, you get
$$ sup_{ninmathbb{N}}sup_{xin[0,1]} |g_n(x)| leq max(M_1,M_2,dots,M_{N_epsilon},B+epsilon). $$
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.