Assume that $f : A \rightarrow \mathbb{R}$ is two times differentiable with $f''(x)=0$ for all $x \in A$, with $A$ an interval.

Show, (not by integration), that $f$ is of the form $f(x) = ax + b$ for some constants $a, b$.

My answer: I will use the Mean Value Theorem:

  1. Take $x,y \in A, x<y$. Applying the Mean Value Theorem on $[x,y]$ gives: $$f''(c)=\frac{f'(y)-f'(x)}{y-x}$$ for some $c \in A$. If $f''(x)=0 $ for all $x \in A$, then $f''(c)=0$, wich means that $f''(c)=\frac{f'(y)-f'(x)}{y-x}= 0$, which implies that $f'(y)-f'(x)=0$. So $f'(y)=f'(x)$. Set $k$ equal to this common value. Because x and y are arbitrary, it follows that $f'(x)=a$ for all $x \in A$

  2. We are now given that $f'(x)=a$ for all $x \in A$. This means that, if we take $x,y \in A, x<y, $, by applying the Mean Value Theorem on [x,y], $$f'(c)=\frac{f(y)-f(x)}{y-x}$$ for some $c \in A$. We know that $f'(x)=a$ for all $x \in A$, which means that $$f'(c)=\frac{f(y)-f(x)}{y-x}=a \implies f(y)-f(x)=a(y-x)$$

What's next?

I could also use the definition of derivative, considering:

$$f'(x)=a = \lim_{x \to c} \frac{f(x)-f(c)}{x-c}$$

How can I show that $f(x) = ax + b$ ?

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    $\begingroup$ Change to $\,y-x\,$ your denominator in (1)-(2) $\endgroup$ – DonAntonio Oct 22 '12 at 17:35
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    $\begingroup$ At the end of point 1., you write that $f'(x)=f'(y)=0$ and then you write that $f'(x)=k$ for all $x\in A$. Am I not seeing why those two statements don't conflict? Did you mean to write $f'(x)-f'(y)=0$? $\endgroup$ – Todd Wilcox Oct 22 '12 at 17:36
  • $\begingroup$ @ToddWilcox Ty. I updated it. Do you guys know how to go further with this proof? :-) $\endgroup$ – Applied mathematician Oct 22 '12 at 17:39
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    $\begingroup$ you can't say $\forall c \in A$. It has to be $\forall c \in ]x,y[$. $\endgroup$ – user31280 Oct 22 '12 at 18:15
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    $\begingroup$ @EmmadKareem No, because this is an exercise from Real Analysis. I have to understand simple things deeply :) $\endgroup$ – Applied mathematician Oct 22 '12 at 18:20

Using the taylor expansion of $\displaystyle f(x+0) = \sum_{k\ge0}x^k\cfrac {f^{(k)}(0)}{k!} = f(0) + xf'(0)+x^2\cfrac{f''(0)}{2!}+ \cdots$ Since $f''(x) = 0$ then every subsequent derivative is $0$. Thus we have $f(x) = f(0) + xf'(0)$ with $a=f'(0)$ and $b=f(0)$.

EDIT: But the Taylor series of a function does not always converge to the function, in this case the Taylor Series is a polynomial. Check this proof to see how such functions equal to their Taylor series.

You see, no integration at all. This is one of the most important uses of Taylor Series, solving differential equations. Check here for futher details.

Lemme restart your work: $A = [c,d]$

Given a function $f(x) \in C^2[c,d] \space|\space f''(x) =0,\space \forall x\in[c,d]$. Using the Mean Value Theorem, since the second derivative is $0$ then $f'(x) = a \in [c,d]$.

Using the theorem again on $f'(x)$, we have that $\forall x\in [c,d], \space \exists b_1\in[c,x]\space |\space f'(b_1)=\cfrac {f(x) - f(c)}{x -c} =a$ and $\exists b_2\in[x,d]\space |\space f'(b_2)=\cfrac {f(d) - f(x)}{d -x} = a$. Since the function is two times derivable then both the right and left derivative should be equal. Thus we have $f(d) - f(x) = a(d-x)$ and $f(x) - f(c) = a(x-c)$ which implies $$\begin{matrix} f(x) = ax +f(d) - ad \\ f(x) = ax + f(c) - ac \end{matrix}$$ Now we have a problem where we have to prove that $f(d) -ad = f(c) -ac$. Assuming it is true then $f(d) -ad = f(c) -ac \space\Rightarrow\space a=\cfrac{f(d)-f(c) }{d-c}$. Since $a= f'(x) = \cfrac{f(d)-f(c) }{d-c}, \space \forall x\in [c,d]$, then our proof is complete and we can conclude that $f(d) -ad = f(c) -ac=b \in [c,d]$ and that $$f(x) = ax +b$$

  • $\begingroup$ Thank you for your help. But I have to use the theory of derivatives, especially the mean value theorem :) $\endgroup$ – Applied mathematician Oct 22 '12 at 18:05
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    $\begingroup$ check now! I've used the Mean Value Theorem $\endgroup$ – user31280 Oct 22 '12 at 19:00
  • $\begingroup$ very good, thanks!!! $\endgroup$ – Applied mathematician Oct 22 '12 at 19:09
  • $\begingroup$ I'm just asking: Is it clear that the Taylor Series really is equal to the function, i.e. that f is analytical? $\endgroup$ – AndreasS Oct 22 '12 at 19:13
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    $\begingroup$ @Hempo: It's probably worth noting that these two proofs are in some sense the same proof expanded differently -- the way you prove Taylor's theorem is by repeatedly applying the MVT to the derivatives of your function... $\endgroup$ – Micah Oct 22 '12 at 20:47

You're effectively done once you have ${\displaystyle {f(y) - f(x) \over y - x} = a}$. For this can be rewritten as $$f(y) - f(x) = a(y - x)$$ or just $$f(x) = ax + ay - f(y)$$ This holds for all $x$ and $y$ in your interval. So you can fix one such $y$, and define $b = ay - f(y)$. Then the above equation is exactly $$f(x) = ax + b$$ Since this holds for all $x$ in the interval you are done.

By the way, if you want to use Taylor series, you can use the form with remainder which says that for fixed $x_0$ in the interior of your interval, for any $x$ in the interval you have $$f(x) = f(x_0) + f'(x_0)(x - x_0) + {1 \over 2}f''(c_x)(x-x_0)^2$$ Here $c_x$ is between $b$ and $x$. Since $f''(c_x) = 0$, this means that for all $x$ you have $$f(x) = f(x_0) + f'(x_0)(x - x_0)$$ This is of the form $ax + b$ as needed.


Just finish it from your last line. Fix any $y_0$. Then, $f(x) = a(x-y_0) + f(y_0)$ for all $x$, i.e. $f(x)$ is a line.


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