Suppose that $f_n : \mathbb{R} \rightarrow \mathbb{R}$, $n=1,2,\dots$ is a sequence of functions with continuous second derivatives. Further, let $f_n \rightarrow 0$ uniformly as $n \rightarrow \infty$ and let $M$ be a constant such that $|f_n '' (x)| \leq M$ for all $n$ and $x \in \mathbb{R}$.

Show that there exists a constant $K$ such that $|f_n '(x)| \leq K$ for all $n$ and $x \in \mathbb{R}$.

I am stuck on how to relate the above information. I know that given $a,b\in \mathbb{R}$, with $a<b$, we have that for some $c,d\in(a,b)$

$$ f_n(b) = f_n(a) + f_n'(a)(b-a) + \frac{f_n''(c)}{2}(b-a)^2 \leq f_n(a) + f_n'(b-a) + \frac{M}{2}(b-a)^2 $$


$$ f_n'(b) - f_n'(a) = f''(d)(b-a) \leq M(b-a) $$

But I can't see how to combine this with the uniform convergence of $\{f_n\}$.


After reading Hans' comment below, I think the connection comes from $f_n$ being uniformly bounded. Since $f_n \rightarrow 0$ uniformly, there exists a sequence $L_n \rightarrow 0$ where

$$ L_n = \sup_{x} |f_n(x)| $$

Thus, if we let $L$ be the LUB of the $L_n$, we can combine this with Hans' trick below to get

$$ |f_n'(x)| \leq |f_n(x+1) - f_n(x)| + M/2 \leq 2L + M/2 $$

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    $\begingroup$ Drop the subscript $n$ as it is irrelevant. Rewrite the first identity as $f'(a) = (f(b) - f(a))/(b-a) + (b-a)f''(c)/2$. At this point you are still free to choose $a$ and $b$. So choose any $a = x$, set $b = x+1$, and look at the result again. What can you conclude, using your assumptions? $\endgroup$ Jun 6 '13 at 2:37
  • $\begingroup$ @HansEngler That $|f'(x)|\leq |f(x+1) - f(x)| + M/2$. But I think this only gets me half way. I am now thinking that the uniform convergence of $\{f_n\}$ shows that it is also uniformly bounded, so that there exists $L$ such that $|f_n(x)| \leq L$ for all $n$ and $x$. Therefore, $|f_n'(x)| \leq 2L + M/2$ for all $n$ and $x\in \mathbb{R}$. $\endgroup$ Jun 6 '13 at 2:47
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    $\begingroup$ Related question. $\endgroup$
    – 23rd
    Jun 6 '13 at 3:20
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    $\begingroup$ @AnonSubmitter85 - your edit gives a correct solution, and you can conclude more. Using your notation, set $x = a$ (arbitrary) and then choose $b$ such that $b - a = \sqrt{M/(4L_n)}$. The result is now $|f_n'(x)| \le 2 \sqrt{L_n \cdot M}$. This is known as Landau's Inequality. Therefore the $f_n'$ are not only uniformly bounded but go to $0$. $\endgroup$ Jun 6 '13 at 15:16

Suppose that $\|f_n''\|_\infty\le M$. I believe that the only part of $f_n\to0$ uniformly that we need is that for some $N$ and all $n\ge N$, we have $\|f_n\|_\infty\le M$. Let $f=f_n$ for any $n\ge N$.

The Mean Value Theorem says that for any $x,y$ there are $z_1,z_2$ between $x$ and $y$ so that $$ |f'(z_1)(x-y)|=|f(x)-f(y)|\le2M\tag{1} $$ and $$ |f'(x)-f'(y)|=|f''(z_2)(x-y)|\le M|x-y|\tag{2} $$ Consider $f$ on any interval $[k,k+1]$.

By $(1)$, there is a $z_1\in[k,k+1]$ so that $|f'(z_1)|\le2M$.

By $(2)$, for any $x\in[k,k+1]$, $|f'(x)-f'(z_1)|\le M|x-z_1|\le M$.

The Triangle Inequality then gives $$ \begin{align} |f'(x)| &\le|f'(x)-f'(z_1)|+|f'(z_1)|\\ &\le3M\tag{3} \end{align} $$ Since $[k,k+1]$ was arbitrary, we have that, for $n\ge N$, $\|f_n'\|_\infty\le3M$.

Unfortunately, we cannot bound $\|f_1'\|_\infty$ since there is nothing preventing $f_1(x)=x^2$. $\|f_1''\|_\infty=2$; however, the condition $f_n\to0$ only puts limits on the tail of the sequence of functions.


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