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I was given this question by a friend:

Let $f(x)$ be 2 times differentiable in $R$ and $x_{0}$ a local extremum.

Show that there are $a, b, c \in R$ such that a function $g(x)=f(x+a)+b$ exists. And $g(x)=cx^2+h(x)$ where $\lim_{x\to 0}\frac{h(x)}{x^2}=0$.

I began by writing a first degree Taylor expansion of $f(x)$ around $x_{0}=a$ using Lagrange's remainder:

$$f(x)=f(a)+\frac{f''(\xi)}{2}(x-a)^2$$

From the above we see that $f(x+a)=f(a)+\frac{f''(\xi)}{2}x^2$. Now comes the part I'm not sure of: Can I say that $g(x)=\frac{f''(\xi)}{2}x^2 \implies g(x)=f(x+a)-f(a)$? Is $g(x)$ defined properly? The reason I'm not sure is because the point $\xi$ is different for every $x$.

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A local extremum: extrema is the plural. –  Mariano Suárez-Alvarez Feb 5 '11 at 21:15
    
Your $g(x)$ is defined properly, but does not satisfy the conditions you are requiring, since you are requiring that $g(x)$ be of the form $cx^2 + h(x)$, with $c$ constant. In your expression, the coefficient of $x^2$ in $g$ is not constant, but, as you note, a function of $g(x)$. Instead, try going as far as you are guaranteed to go with the Taylor expansion, and then add a remainder (whatever it may be). Remember that you don't need to be able to write down a formula for $h$ for it to be a function. –  Arturo Magidin Feb 5 '11 at 21:34
    
@Arturo: I see. In that case the only other option I have is: $$f(x+a)=f(a)+\frac{f''(a)}{2}x^2+R_3(x+a)$$ But now I'm faced with the same problem. I can't say $$g(x)=f(x+a)\overbrace{-f(a)-\frac{f''(a)}{2}x^2-R_3(x+a)}^{b}$$ because again my $b$ in this case changes for every $x$ because of the remainder. –  user6679 Feb 5 '11 at 21:45
    
@robert: I haven't thought this through carefully, but have you noticed that you haven't used $x_0$ in any way whatsoever yet? –  Arturo Magidin Feb 5 '11 at 21:48
    
@Arturo: I have, when I wrote $f(x)$ Taylors expansion around $a=x_{0}$ I used the fact that $f'(x_{0})=0$. –  user6679 Feb 5 '11 at 21:50
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1 Answer

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The key here is to look at the condition $$g(x) = cx^2 + h(x)\qquad\mbox{with $\lim\limits_{x\to 0}\frac{h(x)}{x^2} = 0$.}$$

If you think about a potential Taylor expansion of $g(x)$ of degree $2$, this tells you that $h(x)$ has to consists of the degree $3$ and above terms only; that means that the Taylor expansion of $g(x)$ around $0$ cannot have a linear or constant terms, so $g(0)=0$ and $g'(0)=0$.

Now, since we will want $g(x) = f(x+a)+b$, then $g'(x) = f'(x+a)$, so in order to get $g'(0)=0$, we need $f'(a)=0$. The only point where we know $f'(a)=0$ is at $a=x_0$, so we'll pick $a=x_0$. Thus, $g(x) = f(x+x_0) + b$. In order to also get $g(0) = 0$, we need $f(x_0) + b = 0$, so $b=-f(x_0)$. Thus, the function we expect will work is $$g(x) = f(x+x_0) - f(x_0).$$ Now, to see this works, we proceed as you did: taking the Taylor expansion for $f$ around $x_0$ up to degree $2$, we get $$f(x) = f(x_0) + \frac{f''(x_0)}{2}(x-x_0)^2 + R_3(x).$$ So $$g(x) = f(x+x_0) - f(x_0) = \frac{f''(x_0)}{2}x^2 + R_3(x+x_0).$$ Thus, $c=\frac{f''(x_0)}{2}$, and $h(x) = R_3(x+x_0)$. We just need to show that $h(x)/x^2 \to 0$ as $x\to 0$, but this is equivalent to $R_3(x+x_0)/x^2\to 0$ as $x\to 0$, which in turn is equivalent to $$\lim_{x\to x_0}\frac{R_3(x)}{(x-x_0)^2} = 0.$$ This hold precisely because $R_3$ is the degree $3$ remainder at $x_0$.

So, in summary, we set $a=x_0$, $b=-f(x_0)$, $c=\frac{f''(x_0)}{2}$, $g(x) = f(x+x_0) - f(x_0)$, and $h(x) = R_3(f,x_0,x)$ (the Remainder of degree $3$ in the Taylor expansion of $f$ around $x_0$, evaluated at $x$).

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Now that's a great explanation. Many thanks! –  user6679 Feb 5 '11 at 22:47
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