# A proof using Taylor's Theorem

Show that if $f : \mathbb{R} \longrightarrow \mathbb{R}$ is differentiable and $f'(x) \geq x$, then $f(x) \leq f(0) + \frac{1}{2}x^2$ for all $x \leq 0$.

I'm wondering whether this statement can be proved using Taylor's Theorem. Since we're given that $f$ is only one times differentiable, we can use, $f(x) = P_{0}(x_0) + R_1(x),$ $x_0 = 0$ and $x \leq 0$. This gives:

\begin{align} f(x) = f(0) + f'(c)x, \quad\quad c \in [x, 0], \end{align}

Using the assumption, we can write:

\begin{align} f(x) & = f(0) + f'(c)x \\ & \geq f(0) + cx \\ & \geq f(0) + x^2 \\ \end{align}

But using this method, I not only miss out on the factor of $2$ but also get the reverse inequality. Any suggestions?

• it is for x negative, so you get $-\int_x^0$ and the reverse inequality.
– zwim
May 6, 2017 at 17:20
• Oh, right. I see that now. The remainder term, using the form that I have used above, will be negative and not positive. But I'm still not sure where to get the factor of 2 from. May 6, 2017 at 17:22
• a primitive of $t$ is $t^2/2$
– zwim
May 6, 2017 at 17:23
• I'm explicitly trying not to use the integral form of the remainder term. I'm trying to use the form given by the expression: $R_n(x) = \frac{f^{n+1}(c)}{(n+1)!}(x - x_0)^{n+1}$ May 6, 2017 at 17:24
• For $n = 0$, this gives me $R_0(x) = f'(c)x$. Since $n+1$ is odd, the remainder term will be negative, but I'm still not sure how to get the factor of 2 using this approach. May 6, 2017 at 17:26

$$f(x) = f(0) + \int_0^x f'(t) \text{ d}t$$
• Try substituting $f'(t) = t$ into the integral above May 6, 2017 at 17:23