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I'm trying to solve the following problem:

Is there a differentiable function $f:\mathbb R ^3\to \mathbb R$ satisfying for all $a,x\in \mathbb R^3$ $$Df(a)x=a_1x_1+a_2x_3?$$

How to approach such questions? I don't know what to do.

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    Is $\;Df(a)x\;$ the map of the differential of $\;f\;$ at $\;a\;$ in the direction of $\;x\;$ ....or what? – 2017-02-07
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    @DonAntonio it's the value of the derivative of $f$ at $a$, at $x$. – 2017-02-07

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A necessary condition for a vector field to be a gradient is to be curl-free. You can easily show that $$ F(x_1,x_2,x_3)=(x_1,x_3,0) $$ is not curl free. Hence, you can't find such a function $f$.

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    Your welcome. You could mark it as answer indeed! ;) – 2017-02-07
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We need to check if $F(a):=(a_1,0,a_2)$ is a conservative vector field(i.e. if $F(a) = ∇ f(a)$ for some $f$. If it were, then the fundamental theorem of line integrals says that

$$ ∫_{[0,y]} ∇ f(x) · t(x) \ dx = f(y) - f(0)$$ where $[0,y]$ is the line segment joining 0 and $y$, and $f(0) = 0$ without loss of generality. Plugging this in, we see that $$f(x) = ∫_0^1 t x_1^2 + t x_2x_3 \ dx = \frac{1}{2}(x_1^2 + x_2x_3)$$ On inspection, this function does not satisfy $ F = \nabla f$.

If such $f$ existed, one could proceed is as follows. $Df : \Bbb R^3 \times \Bbb R^3 → \Bbb R^3$ is a bilinear form. Can you find $f$ such that $Df(a)x = a_1x_1 + a_2x_2 + a_3x_3 = a · x$? Try differentiating the squared euclidean norm function.

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    In my question $a_2x_3$ appears, not $a_2x_2$.. – 2017-02-07
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    @eleventeen In my answer, I'm trying to guide you. – 2017-02-07
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    I understand you're helping, I just don't understand the help. (The answer to the exercise which I have is "no".) – 2017-02-07
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    @eleventeen you responded to the help in <2 minutes with "you don't understand". Take a break if you need to, but getting the answer given to you may not be the most helpful long term. (Once you have done the above, the labelling issue can be fixed with matrices.) – 2017-02-07
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    I agree. I think the derivative of the norm is $(x,y,z)\mapsto 2(x,y,z)$. – 2017-02-07
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    If a function $f$ is between vector spaces $A→ B$, then since $df(x)$ is the "best linear approximation", the spaces must match; hence you need $df(x) : A → B$. Your derivative is not of this form. – 2017-02-07
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    In that case I think $df(a_1,a_2,a_3):(x,y,z)\mapsto (a_1,a_2,a_3)\cdot 2(x,y,z)$. – 2017-02-07
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    Sorry, I made an error. Let me emphasise that $df$ is not necessarily linear; but $df(x)$ is always linear and between the same vector spaces as $f$. In your case $df$ is linear, and this hints towards the type of function you have (a generalisation of 'if $f':\Bbb R → \Bbb R$ is linear, then f is quadratic') – 2017-02-07
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    @eleventeen thats correct, which if you expand the inner product as a sum, almost answers the hint I gave. Try differentiating other "dot products" than $x·x$, maybe $x·Ax$. – 2017-02-07
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    I think I understand what you meant earlier by "fixed with matrices", and I feel I will be able to figure out the necessary matrix which will have some zeros. I am confused because another student says there's no such differentiable function. His argument looks weird and mentions second order partial derivatives. It sounds wrong, but just to make sure, is it? – 2017-02-07
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    I'm horribly sorry @eleventeen I didn't realise that you were given a "prove or disprove" question. In this case, have you heard of conservative vector fields and their line integrals? – 2017-02-07
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    @eleventeen you can't find such a function... – 2017-02-07
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    @CalvinKhor what goes wrong with the matrix tinkering you suggested? – 2017-02-07
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    @eleventeen the derivative is $x^TAy + x^TA^T y$ ; a factor of two doesn't appear unless the matrix is symmetric. You can't solve this for the elements of $A$. – 2017-02-07