Integration and uniform norm

Question

Suppose that $f$ is twice differentiable on $\mathbb{R}$ and $\|f\|_\infty = A$ and $\|f''\|_\infty = C$.

Prove that $\|f'\|_\infty\leq \sqrt{2AC}$.

Hint: $f'(x_0) = b > 0$. Show that $b-C|t| \leq $$f'(x_0+t)$. Integrate from $x_0-\frac {b} {C}$ to $x_0+\frac {b} {C}$.

How can I prove that? I don't even know where to start. I known this uniform norm. Thanks.

Answer 1

Let $f \, : \, \mathbb{R} \, \rightarrow \, \mathbb{R}$ a function which is twice differentiable on $\mathbb{R}$. The inequality

$$ \Vert f' \Vert_{\infty} \leq \sqrt{2 \Vert f \Vert_{\infty} \Vert f'' \Vert_{\infty} } $$

is a particular case of the Landau-Kolomogorov Inequalities. In the hint you gave, $x_{0}$ is any point in $\mathbb{R}$. To prove this inequality, I will need the following result (which I call "the Taylor-Lagrange equality" but I'm not sure that's how people call it in general) :

Let $f \, : \, [a,b] \, \rightarrow \, \mathbb{R}$ a function which is $\mathcal{C}^{n+1}$ on $[a,b]$. There exists $c \in [a,b]$ such that :

$$ f(b) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(b-a)^{k} + \frac{(b-a)^{n+1}}{(n+1)!}f^{(n+1)}(c) $$

Let $h \in ]0,+\infty[$, $x_{0} \in \mathbb{R}$ (since $x_{0}$ is any real number, I'd rather write $x$ instead of $x_{0}$ but I keep $x_{0}$ to be coherent with the hint you gave in your post) and let's apply this result to our function $f$ on $[x_{0},x_{0}+h]$. We know that there exists $c_{1},c_{2} \in [x_{0},x_{0}+h]$ such that :

$$ f(x_{0}+h) = f(x_{0}) + hf'(x_{0}) + \frac{h^{2}}{2!} f''(c_{1}) $$

and

$$ f(x_{0}-h) = f(x_{0}) -hf'(x_{0}) + \frac{h^{2}}{2!} f''(c_{2}) $$

It gives :

$$ f(x_{0}+h) - f(x_{0}-h) - \frac{h^{2}}{2!} \Big( f''(c_{1}) - f''(c_{2}) \Big) = 2h f'(x_{0}) $$

Using triangular inequality, we get :

$$ 2h \vert f'(x_{0}) \vert \leq \vert f(x_{0}+h) \vert + \vert f(x_{0}-h) \vert + \frac{h^{2}}{2!} \Big( \vert f''(c_{1}) \vert + \vert f''(c_{2}) \vert \Big) $$

By definition of $\Vert f \Vert_{\infty}$ and $\Vert f'' \Vert_{\infty}$, we get :

$$ 2h \vert f'(x_{0}) \vert \leq 2 \Vert f \Vert_{\infty} + h^{2} \Vert f'' \Vert_{\infty} $$

which is :

$$ \vert f'(x_{0}) \vert \leq \frac{\Vert f \Vert_{\infty}}{h} + \frac{h}{2} \Vert f'' \Vert_{\infty} \tag{$\star$} $$

We conclude by minimizing the function $h \, \longmapsto \, \frac{\Vert f \Vert_{\infty}}{h} + \frac{h}{2} \Vert f'' \Vert_{\infty}$. Take $h = \sqrt{\frac{2\Vert f \Vert_{\infty}}{\Vert f'' \Vert_{\infty}}}$ and $(\star)$ becomes :

$$ \vert f'(x_{0}) \vert \leq \sqrt{2 \Vert f \Vert_{\infty}\Vert f'' \Vert_{\infty}} \tag{$\diamond$}$$

Since $(\diamond)$ is true for all $x_{0}$ in $\mathbb{R}$, we can conclude that :

$$ \Vert f' \Vert_{\infty} \leq \sqrt{2 \Vert f \Vert_{\infty} \Vert f'' \Vert_{\infty} } $$