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Let $f : \mathbb R → \mathbb R$ be twice differentiable. Find $D^2 f (x_0)$ in terms of $f ′′(x_0)$.

I do not understand what does $D^2f(x_0)$ mean. If it means second derivative, then shouldn't it be $D^2 f (x_0) = f''(x_0)$?

Thank you all.

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    Are you learning about $Df(x_0)$ and $D^2 f(x_0)$ in the multivariable setting? Are they then asking you to verify that the more general definitions agree with the usual ones for single-variable functions $f$? – 2017-02-27
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    Aha! This is a kind of weird question. Note that $f''(x_{0})$ is a number, while $D^{2} f(x_{0})$ is a bilinear map $\mathbb{R}^{2} \to \mathbb{R}$. They are related in quite a trivial way; can you see what it is? – 2017-02-27
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    It might be easier to explain what $Df(x_{0})$ is in terms of $f'(x_{0})$. Again, they are not *quite* the same! – 2017-02-27
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    @DanielLittlewood I can see $D^2f(x_0)$ is a map, but still I do not know the relation. – 2017-02-27
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    @TedShifrin Yes – 2017-02-27

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As I say in the comments, the distinction between $D f(x_{0})$ and $f'(x_{0})$ is easier to explain, so I'll do that. I this doesn't make the case of $D^{2}f(x_{0})$ clear, then tell me and I'll explain it.

The definition of $f'(x_{0})$ is the following limit, if it exists
$$\lim_{h \to 0} \frac{f(x_{0}+h)-f(x)}{h}$$ The definition of $D f(x_{0})$ is that it is a linear map $\mathbb{R} \to \mathbb{R}$ such that, for any $h \in \mathbb{R}$, $$ f(x+h)-f(x)=Df(x_{0})(h)+o(h)$$ where $o(h)/h \to 0$ as $h \to 0$. By linearity, $Df(x_{0})(h)=hDf(x_{0})(1)$, and it follows that $$f'(x_{0})=\lim_{h \to 0} \left[Df(x_{0})(1)+\frac{o(h)}{h}\right]$$ Hence $f'(x_{0})=Df(x_{0})(1)$. Is that clear?

EDIT: Ok, so what is the definition of $D^{2}f(x_{0})$? It's probably easiest to think of this as a matrix of partial derivatives. In general (given a basis $\left\{e_{i}\right\}$), if $H$ is the Hessian defined by $$H_{ij}=\frac{\partial^{2}f}{\partial x_{i} \partial x_{j}}$$ then $D^{2}f(x_{0})(u)(v)=H_{ij}u_{i}v_{j}$. In particular, $$\frac{\partial^{2}f}{\partial x_{i} \partial x_{j}}=D^{2}f(x_{0})(e_{i},e_{j})$$ So, in the case $\mathbb{R} \to \mathbb{R}$, we have $f''(x_{0})=D^{2}f(x_{0})(1,1)$

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    What you said about $Df(x_0)$ and $f'(x_0)$ is clear to me, but I still do not get the case of $D^2f(x_0)$. Could you please explain that to me as well? – 2017-02-27
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    I hope the edit clears up your problems. Notice that it is delicate to talk about second derivatives in $\mathbb{R}^{n}$: we view the map $Df(x_{0})$ as an element of $L(\mathbb{R}^{n},\mathbb{R}^{m})$. which is itself a finite dimensional normed space, and we differentiate there. This becomes complicated quickly (there is a way to handle it, but it requires some difficult algebra). – 2017-02-27
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    Thanks Daniel, but I do not quite understand what you mean by saying "hessian". Also, how do we know $D^2f(x_0)(u)(v)=H_{ij}u_iv_j$? – 2017-02-27
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    The Hessian is just the matrix of partial derivatives - have you seen the fact that $D^{2}(e_{i},e_{j})=\frac{\partial^{2} f}{\partial x_{i} \partial x_{j}}$? – 2017-02-27