Proof that Convex Function with alternate variable is convex 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*}
 A: 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.
A: 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
