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Given that say, $f(x)$ is convex for $x>0$. Would the proof that $f(a-x)$ where $a>x$ is convex proceed as shown below. Please verify the correctness and/or alternative approaches that can establish this or that the alternate variable is not convex (or concave or neither)?

I suppose a similar reasoning would establish that $f(bx)$ is convex, where $b>0$. Would it matter whether $b$ is positive or negative (Does not seem to matter, but please confirm this)?


Proof

\begin{eqnarray*} f''\left(x\right)>0\;;\; x>0 \end{eqnarray*} \begin{eqnarray*} \text{Is }f\left(a-x\right)\text{ convex or concave or neither ?} \end{eqnarray*} \begin{eqnarray*} \text{Let }y=a-x \end{eqnarray*} \begin{eqnarray*} \frac{\partial f\left(y\right)}{\partial x} & = & \frac{\partial f\left(y\right)}{\partial y}\frac{\partial\left(a-x\right)}{\partial x}\;;\; a>x\\ & = & \left(-1\right)f'\left(y\right) \end{eqnarray*} \begin{eqnarray*} \frac{\partial^{2}f\left(y\right)}{\partial x^{2}} & = & \left(-1\right)\frac{\partial f'\left(y\right)}{\partial y}\frac{\partial\left(a-x\right)}{\partial x}\\ & = & f''\left(y\right)\\ & > & 0\;\left[\because f''\left(y\right)>0\left|\forall y>0\right.\right] \end{eqnarray*}

End of proof


Alternately (something seems incorrect here with the terminology or substitutions used),

\begin{eqnarray*} f''\left(x\right)>0\;;\; x>0 \end{eqnarray*} \begin{eqnarray*} \frac{\partial f\left(a-x\right)}{\partial x} & = & \frac{\partial f\left(a-x\right)}{\partial\left(a-x\right)}\frac{\partial\left(a-x\right)}{\partial x}\;;\; a>x\\ & = & \left(-1\right)f'\left(a-x\right) \end{eqnarray*} \begin{eqnarray*} \frac{\partial^{2}f\left(a-x\right)}{\partial x^{2}} & = & \left(-1\right)\frac{\partial f'\left(a-x\right)}{\partial\left(a-x\right)}\frac{\partial\left(a-x\right)}{\partial x}\\ & = & f''\left(a-x\right)\\ & > & 0\;\left[\because f''\left(y\right)>0\left|\forall y>0\right.\right] \end{eqnarray*}

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I think your proof is correct.But there is also a more intuitive way to understand the result.

  • First, if $f(x)$ is convex when $x > 0$, then $f(-x)$ is also convex when $x < 0$. To see this, note that $f(-x)$ is a reflection of $f(x)$ across the $y$-axis, thus the graphs of $f(x)$ and $f(-x)$ are symmetric along the $y$-axis, preserving the convexity.

  • Second, if $f(-x)$ is convex when $x < 0$, then $f(a-x)$ is also convex when $x < a$. The graph of $f(a-x)$ can be obtained by shifting the graph of $f(-x) = f(0 - x)$ by $a$ units along the $x$-axis. Thus the graph of $f(a-x)$ when $x < a$ is essentially the same as $f(-x)$ when $x < 0$, ignoring the $x$-coordinate.

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  • $\begingroup$ Thanks for verifying this echo. How about establishing that $f(bx)$ is convex, where $b>0$. Would it matter whether $b$ is positive or negative (Does not seem to matter, but please confirm this)? $\endgroup$ – texmex Jun 14 '15 at 9:27
  • $\begingroup$ And also how about $xf(x)$ which I suppose we cannot establish whether it is convex or concave without knowing $f'(x)$ or the relative magnitudes of the first and second derivatives? $\endgroup$ – texmex Jun 14 '15 at 9:28
  • $\begingroup$ @texmex For $f(bx)$, you should be able to find that its second derivative with respect to $x$ is $b^2\cdot f^{"}(bx)$. So if $b > 0$, then $f(bx)$ is convex when $x > 0$, and if $b < 0$, then $f(bx)$ is convex when $x < 0$. $\endgroup$ – PSPACEhard Jun 14 '15 at 10:20
  • $\begingroup$ @texmex Yes, it depends. If $f(x) = \frac{1}{\sqrt{x}}$ thus convex when $x > 0$, $xf(x) = \sqrt{x}$ is concave however. If $f(x) = \frac{1}{x^2}$ (still convex when $x > 0$), $xf(x) = \frac{1}{x}$ is convex. $\endgroup$ – PSPACEhard Jun 14 '15 at 10:23
  • $\begingroup$ Thanks for your clarifications. So the function I am trying is $f\left(x\right)=\frac{\phi\left(x\right)}{\Phi\left(x\right)}$. Here, $\phi$ and $\mathbf{\Phi}$ are the standard normal PDF and CDF, respectively. So would $xf(x)$ be convex in this case. Please let me know your suggestions and any thoughts. Please assume that f(x) is convex as above. This can be shown separately. $\endgroup$ – texmex Jun 14 '15 at 11:05
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Proof

\begin{eqnarray*} f''\left(x\right)>0\;;\; x>0 \end{eqnarray*} Is $f\left(a-x\right)$ convex or concave or neither ?

\begin{eqnarray*} \text{Let }y=a-x \end{eqnarray*} \begin{eqnarray*} \frac{\partial f\left(y\right)}{\partial x} & = & \frac{\partial f\left(y\right)}{\partial y}\frac{\partial\left(a-x\right)}{\partial x}\;;\; a>x\\ & = & \left(-1\right)f'\left(y\right) \end{eqnarray*} \begin{eqnarray*} \frac{\partial^{2}f\left(y\right)}{\partial x^{2}} & = & \left(-1\right)\frac{\partial f'\left(y\right)}{\partial y}\frac{\partial\left(a-x\right)}{\partial x}\\ & = & f''\left(y\right)\\ & > & 0\;\left[\because f''\left(y\right)>0\left|\forall y>0\right.\right] \end{eqnarray*}

End of proof

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  • $\begingroup$ There are actual horizontal lines in markdown, FYI $\endgroup$ – John Dvorak Jun 14 '15 at 15:00
  • $\begingroup$ Thanks Han and Homegrown .. Appreciate these pointers. $\endgroup$ – texmex Jun 15 '15 at 1:08

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