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I am working on a practice problem and there is step in the solution that deals with the application of the mean value theorem (MVT) in a Taylor series. The problem is asking for a condition on $f''(x)$ s.t. $\{(x,y)\in\mathbb{R}:y\ge f(x)\}$ is convex if $f:\mathbb{R}\rightarrow\mathbb{R}$ and $f$ is twice differentiable.

Taking the Taylor series up to the second term and applying $y\ge f(x)$ as is given in the problem statement gives $$y\ge f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}$$ but the solution instead proceeds to the following step instructing the reader to use the MVT $$y\ge f(x)=f(a)+f'(a)(x-a)+f''(a^*)\frac{(x-a)^2}{2!},a^*\in(a,x).$$ This seems to be indicating to me that $f''(a)=f''(a^*)$ but since $a^*\in(a,x)$ not $\in[a,x]$ I don't see how this can be true. My understanding of the MVT simply says that if $f$ is differentiable over $(a,b)$ and cts. over $[a,b]$ then there exists a $f'(c)=\frac{f(b)-f(a)}{b-a},c\in(a,b)$.

Am I incorrect in my understanding of this application of the MVT?

WnGatRC456
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    First you should note that your equation $f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}$ is simply not right - it implies that $f$ is a polynomial. – David C. Ullrich Dec 30 '17 at 18:29
  • @DavidC.Ullrich is it true if $x$ is a specific point on $\mathbb{R}$? If it's not true then the solution manual is incorrect because that's how they write it. – WnGatRC456 Dec 30 '17 at 18:33
  • No. And that's not how they write it, at least not according to what you said: They wrote $f''(a^*)$, not $f''(a)$. – David C. Ullrich Dec 30 '17 at 18:36
  • @DavidC.Ullrich could you explain why using $f''(a)$ not $f''(a^)$ makes the expression a polynomial? I was under the assumption that $f''(a)$ and $f''(a^)$ were both just constant values of $f\in [a,x)$. – WnGatRC456 Dec 30 '17 at 18:56
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    Because the value of $a^$ depends on $x$. For every $x$ there exists $a^$ such that right version is true... – David C. Ullrich Dec 30 '17 at 18:59
  • @DavidC.Ullrich, thank you! I guess I had some confusion about to write out Taylor Series. I'll look up and see how it is related to the MVT. – WnGatRC456 Dec 30 '17 at 19:03
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    Perhaps a different letter could have been used instead of $a^{\text{}} $ otherwise it looks so similar to $a$ and creates confusion. Some textbooks even use subscript like $a_{x} $ in place of $a^{\text{}} $ to indicate the dependence on $x$. – Paramanand Singh Dec 31 '17 at 05:44

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It may be relevant to note that there exists a version of Taylor's Theorem that requires the existence of $f^{(k)}$ at only one point. Should be better known than it seems to be, in any case:

Theorem If $f:\mathbb R\to\mathbb C$, $k\ge1$, and $f^{(k)}(a)$ exists then $$f(x)=P_{a,k}(x)+o((x-a)^k)\quad(x\to a).$$

Thanks to Paramanand Singh: It appears that this is known as Taylor's Theorem with Peano's form of the remainder.

Here of course $P_{a,k}(x)=\sum_{j=0}^k\frac{f^{(j)}(a)}{j!}(x-a)^j$.

(Note that the hypothesis implies that $f^{(j)}$ exists in some neighborhood of $a$ for $j<k$.)

The conclusion means by definition that $f(x)=P_{a,k}(x)+E(x)$ where $$\lim_{x\to a}\frac{E(x)}{(x-a)^k}=0.$$

It's enough to prove this:

If $f:\mathbb R\to\mathbb R$, $k\ge1$, and $f(0)=f'(0)=\dots=f^{(k)}(0)=0$ then $f(x)=o(x^k)$ as $x\to0$.

The proof is by induction on $k$. For $k=1$ this is just the definition of the derivative: $$0=f'(0)=\lim_{x\to0}\frac{f(x)}x,$$which says exactly that $f(x)=o(x)$.

Now suppose that $k>1$. Note that $g=f'$ satisfies the hypothesis with $k-1$ in place of $k$, so by induction we have $$f'(x)=o(x^{k-1}).$$

Now MVT shows that if $x\ne0$ (and $x$ is small enough that we're inside the interval about $a$ where $f$ is differentiable) there exists $\xi$ with $|\xi|<|x|$ such that $$\frac{f(x)}x=f'(\xi).$$ So $$f(x)=xf'(\xi)=o(x^k)$$(since $|\xi|\le|x|$).

David C. Ullrich
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