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I tried to prove that if $f\in C^2[0,\infty)$ and there exist two positive numbers $M_0$ and $M_2$ such that for all $x\in [0,\infty)$, $|f(x)|\leq M_0$ and $|f''(x)|\leq M_2$ then there exists $M_1\in [0,\infty)$ such that $|f'(x)|\leq M_1.$

If $f\in C^2[0,b]$ then I can prove it using the Taylor expension of $f$ i.e. for $a\in [0,b]$, $$ \begin{align} \exists x_0 \in [0,b],\;\; \ f(0) &= f(a) -af'(a) + \frac{1}{2}a^2f''(x_0) \\ \exists x_1 \in [0,b],\;\;f(b) &= f(a) +(b-a)f'(a) + \frac{1}{2}(b-a)^2f''(x_1) \end{align}.$$ From that, $$f'(a) = \frac{f(b) - f(0)}{b} + \frac{1}{4}[a^2f’’(x_0) - (b-a)^2f’’(x_1)].$$ It follows that $|f'(a)|\leq M_1$ for some $M_1$.

Is it possible to extend that proof so that we will have the same result for $f\in C^2[0,\infty)$?

Jax
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1 Answers1

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From the Extended MVT, also known as Taylor's Theorem, there exists a number $\xi \in (x,x+1)$ such that

$$f(x+1)=f(x)+f'(x)+\frac12 f''(\xi)$$

Solving for $f'(x)$, we find that

$$|f'(x)|=|f(x+1)-f(x)-\frac12 f''(\xi)|\le 2M_0+\frac12M_2=M_1$$

And we are done!

Mark Viola
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