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Let $f : \mathbb R^n → \mathbb R^m$ be a differentiable function such that $f (\mathbf 0) = \mathbf 0$ and $D f (\mathbf x)(\mathbf h) = 2⟨\mathbf x, \mathbf h⟩$ for any $\mathbf x, \mathbf h ∈ \mathbb R^n$. Show that $f (\mathbf x) = ⟨\mathbf x, \mathbf x⟩$.

How should I use the theorem that $f : U ⊆ \mathbb R^n → \mathbb R^m$ with $U$ open and connected. If $D f (\mathbf x) = O$ for every $x ∈ U$, then $f$ is constant to prove this?

Thanks.

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Hint: consider $h(x) = f(x) - \langle x, x \rangle$ and show that $Dh(x) = 0$ for all $x\in \mathbb{R}^n$.

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    Ok I understand that if $\mathbf x=\mathbf 0$, then $h(\mathbf x)=0$, but how can I show the case when $\mathbf x\neq \mathbf 0$? – 2017-02-21
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    Take the derivative of $\langle x, x \rangle$. – 2017-02-21
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Note that $m=1$ in your question.

As a linear map, the matrix of $Df (\mathbf x) $ is $(2x_1,\ldots,2x_n) $. Thus $D_jf (\mathbf x)=2x_j $, and this forces $f (\mathbf x)=x_1^2+\cdots+x_n^2+c $. But $c=0$ by the condition at the origin.