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Is this true that if $\frac{\partial^3 \log{f(x,\alpha)}}{\partial \alpha^3}$ and $\frac{\partial^3 \log{g(x,\alpha)}}{\partial \alpha^3}$ exist $\forall \alpha \in (0,1)$, then $\frac{\partial^3 \log{(f(x,\alpha)+g(x,\alpha))}}{\partial \alpha^3}$ exists $\forall \alpha \in (0,1)$?

Intutively, I think the answer is yes. But, I don't know how to show it. My main question is when does logarithm of a function exists? when does the derivative of a function exists to conclude what it means when the third derivative of log of something exists. Can you please help me understand these and solve the problem above?

Please vote for it up if it is worth it.

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    @zhw any idea how to solve this? – 2017-01-25

2 Answers 2

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I think that that is the case. Also, $\log(f(x))$ exists if $f(x)>0$.

Notice $\frac{\partial^3 ln(f)}{\partial \alpha^3}=\frac{F}{f^3}$, $\frac{\partial^3 ln(g)}{\partial \alpha^3}=\frac{G}{g}$, and $\frac{\partial ln(f+g)}{\partial \alpha}=\frac{H}{(f+g)^3}$, where $F,G,H$ are sums of derivatives of $f,g$, and combinations and multiples of $f$ and $g$'s derivatives, resp. They are discontinuous whenever $f,f',f'',g,g'$, or $g''$ are discontinuous or $f\leq0$, $g\leq0$, or $f+g\leq0$. Since the first two are assumed to be continuous, $f,f',f'',g,g',g''$ are continuous, $f>0$, and $g>0$ in their domain. This implies that $f+g>0$, and therefore it must be that the third equation is also continuous.

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    @ Thanks. There is something wrong with your reasoning. If the third derivative of $\log(f+g)$ exists, then it is equal to $\frac{H}{(f+g)^3}$. You are using it before you prove it. – 2017-01-25
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    Ah lol good point – 2017-01-25
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Answer: Yes.

Explanation: First of all let's compute the second expression:

\begin{equation} \dfrac{\partial}{\partial \alpha}\dfrac{\partial}{\partial \alpha}\dfrac{\partial}{\partial \alpha} \bigg ( \log{\big ( f(x,\alpha) +g(x,\alpha) \big)} \bigg ) \end{equation}

\begin{equation} \dfrac{\partial}{\partial \alpha}\dfrac{\partial}{\partial \alpha} \bigg (\dfrac{ \dfrac{\partial f(x,\alpha)}{\partial \alpha}}{f(x,\alpha) +g(x,\alpha)} + \dfrac{\dfrac {\partial g(x,\alpha)}{\partial \alpha}}{f(x,\alpha) +g(x,\alpha)} \bigg ) \end{equation}

\begin{equation} \dfrac{\partial}{\partial \alpha} \bigg (\dfrac{ \dfrac{\partial^2 f(x,\alpha)}{\partial \alpha^2}[f(x,\alpha) +g(x,\alpha)] - \dfrac{\partial f(x,\alpha)}{\partial \alpha} [\dfrac{\partial f(x,\alpha)}{\partial \alpha}+\dfrac{\partial g(x,\alpha)}{\partial \alpha}] }{[f(x,\alpha) +g(x,\alpha)]^2} + ... ) \end{equation} \begin{equation} ... \end{equation}

And you can see that the final expression will be a combination of addition, subtraction, multiplication and divisions of $f(x,\alpha),g(x,\alpha),\dfrac {\partial f (x,\alpha)}{\partial \alpha},...,\dfrac {\partial^3 f(x,\alpha)}{\partial \alpha^3},\dfrac {\partial^3 g(x,\alpha)}{\partial \alpha^3}$, so if these functions exist on the interval, so have to this expression; the only source of trouble could be a 0 on the denominator.

Now let's consider the two partial derivatives you give at the beginning, and compute the first one (the second is identical changing $f$ by $g$)

\begin{equation} \dfrac{\partial}{\partial \alpha}\dfrac{\partial}{\partial \alpha}\dfrac{\partial}{\partial \alpha} \bigg ( \log{\big ( f(x,\alpha) \big)} \bigg ) \end{equation}

\begin{equation} \dfrac{\partial}{\partial \alpha}\dfrac{\partial}{\partial \alpha} \bigg ( \dfrac{{\dfrac {\partial f (x,\alpha)}{\partial \alpha}}}{f(x,\alpha)} \bigg ) \end{equation}

\begin{equation} \dfrac{\partial}{\partial \alpha} \bigg ( \dfrac{\dfrac {\partial^2 f (x,\alpha)}{\partial \alpha^2} f(x,\alpha)-[{\dfrac {\partial f (x,\alpha)}{\partial \alpha}}]^2}{[f(x,\alpha)]^2} \bigg ) \end{equation} \begin{equation} ... \end{equation}

And it becomes clear that if $\dfrac{\partial^3}{\partial \alpha^3} \bigg ( \log{\big ( f(x,\alpha) \big)} \bigg )$ exists then $f(x,\alpha),\dfrac {\partial f (x,\alpha)}{\partial \alpha},\dfrac {\partial^2 f(x,\alpha)}{\partial \alpha^2},\dfrac {\partial^3 f(x,\alpha)}{\partial \alpha^3}$ must exist as well. Doing an analogous reasoning for $g(x,\alpha)$ we can conclude that $f(x,\alpha),g(x,\alpha),\dfrac {\partial f (x,\alpha)}{\partial \alpha},...,\dfrac {\partial^3 f(x,\alpha)}{\partial \alpha^3},\dfrac {\partial^3 g(x,\alpha)}{\partial \alpha^3}$ exist, and thanks to our previous deduction we can assert that $\dfrac{\partial^3}{\partial \alpha^3} \bigg ( \log{\big ( f(x,\alpha) +g(x,\alpha) \big)} \bigg )$ must exist, excepting the case where the value of the functions leads to a 0 on the denominator, i.e. any $g,f$ and $(x,\alpha_0), \alpha_0 \in (0,1)$ such that make $g(x,\alpha_0) = -f(x,\alpha_0)$. Since that would imply that whether $\log{f(x,\alpha)}$ or $\log{g(x,\alpha)}$ don't exist for some $\alpha_0 \in (0,1)$ and the premise is that the third derivative of both functions exist on that interval which implies both functions exist too, as we saw, there can't be any $\alpha_0 \in (0,1) $ causing $g(x,\alpha_0) = -f(x,\alpha_0)$ and $\bigg ( \log{\big ( f(x,\alpha) +g(x,\alpha) \big)} \bigg )$ exists $\forall \alpha \in (0,1)$

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    Thanks. That very first condition on the top $f=-g$ will never happen since $f>0$ and $g>0$ because third derivative of $\log(f)$ and $\log(g)$ exist. – 2017-01-25
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    True. If it helped you please vote up – 2017-01-25