Why Do the Points from the Mean Value Theorem Depend on Both n and x?

[evinda]

Hello! ;)

Let $f: \mathbb{R} \to \mathbb{R}$ twice differentiable,such that $f''$ is bounded.We set $f_n(x)=n(f(x+\frac{1}{n})-f(x)), x \in \mathbb{R}$.Check the pointwise and uniform convergence of $f_n$.

$\lim_{n \to +\infty} {n(f(x+\frac{1}{n})-f(x))}=\lim_{n \to +\infty} {\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}}=f'(x)$
So, $f_n \to f'$ pointwise.

$f''$ is bounded,so $\exists M>0$ such that $|f''| \leq M, \forall x \in \mathbb{R}$

From the Mean value Theorem,we have:

$$f_n(x)={\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}}=f'(x+a_{n,x}), 0<a_{n,x}<\frac{1}{n}$$

Again,from the Mean value Theorem,we have:

$$\frac{f'(x+a_{n,x})-f'(x)}{a_{n,x}}=f''(x+b_{n,x}) , 0<b_{n,x}<a_{n,x}$$

Therefore, $|f_n(x)-f'(x)|=\frac{|f'(x+a_{n,x})-f'(x)|}{|a_{n,x}|} \cdot |a_{n,x}|=|f''(x+b_{n,x})| \cdot |a_{n,x}| \leq M \cdot |a_{n,x}| \leq \frac{M}{n} \to 0$

The above relation is true for each $x$,so $\sup_{x \in \mathbb{R}} {|f_n(x)-f'(x)|} \leq \frac{M}{n} \to 0$.
So, $f_n \to f'$ uniformly.But..why $a_{n,x}$ and $b_{n,x}$,that we get from the Mean Value Theorem,depend,except from $n$ also from $x$ ?

 

[Evgeny.Makarov]

In general, if a statement of the form
\[
\forall a\,\forall b\,\exists c\;P(a,b,c)
\]
is true, then the $c$, whose existence the statement guarantees, may depend on $a$ and $b$.

The mean value theorem says the following: For all $a,b\in\Bbb R$, if $f:[a,b]\to\Bbb R$ is a function such that bla-bla-bla, then there exists a $c\in (a,b)$ such that $f ' (c) = \frac{f(b) - f(a)}{b - a}$. Therefore, in general $c$ depends on $a$ and $b$ and, of course, on $f$.

In your case, the segment is $[x,x+1/n]$ in the first application of the MVT and $[x,a_{n,x}]$ in the second one. So, the points whose existence is guaranteed by the MVT depend on both ends of the segment.

But..why $a_{n,x}$ and $b_{n,x}$,that we get from the Mean Value Theorem,depend,except from $n$ also from $x$ ?

I would put it like this: "But why do $a_{n,x}$ and $b_{n,x}$, which we get from the Mean Value Theorem, depend on $x$ in addition to $n$?"

 

[evinda]

In general, if a statement of the form
\[
\forall a\,\forall b\,\exists c\;P(a,b,c)
\]
is true, then the $c$, whose existence the statement guarantees, may depend on $a$ and $b$.

The mean value theorem says the following: For all $a,b\in\Bbb R$, if $f:[a,b]\to\Bbb R$ is a function such that bla-bla-bla, then there exists a $c\in (a,b)$ such that $f ' (c) = \frac{f(b) - f(a)}{b - a}$. Therefore, in general $c$ depends on $a$ and $b$ and, of course, on $f$.

In your case, the segment is $[x,x+1/n]$ in the first application of the MVT and $[x,a_{n,x}]$ in the second one. So, the points whose existence is guaranteed by the MVT depend on both ends of the segment.

I would put it like this: "But why $a_{n,x}$ and $b_{n,x}$, which we get from the Mean Value Theorem, depend on $x$ in addition to $n$?"

I understand (Mmm) Thank you very much!