If $f$ is a quadratic and $f(x)>0\;\forall x$, and $g= f + f' + f''$, prove $g(x)>0\; \forall x$

Solution 1:

Suppose $f(x)=x^2+ax+b$ with $b=f(0)>0$ and $a^2<4b$. Then, $$ g(x)=x^2+(a+2)x+(2+a+b). $$ We note that $g(0)=2+a+b=1+f(1)>0$ and $$ (a+2)^2-4(2+a+b)=a^2+4a+4-8-4a-4b=(a^2-4b)-4<0 $$ so $g$ is never $0$ for real $x$. You now can infer that $g$ is always positive.

The more general case $f(x)=C(x^2+ax+b)$ is the same. It just adds the constant $C>0$ to everything.

Solution 2:

Since $f$ is positive, we can complete the square and write $f(x)=a(x-h)^2+k$ with $a,k > 0$. Then \begin{eqnarray} g(x)&=&a(x-h)^2+k+2a(x-h)+2a\\ &=&a(x-h)(x-h+2)+k+2a\\ &=&a[(x-h+1)^2-1]+k+2a\\ &=&a(x-h+1)^2+k+a \end{eqnarray} is a similarly positive quadratic expression.

ETA: A goofy alternative way of phrasing this would be to say:

• Since $f$ is quadratic, all its higher-order derivatives vanish.
• So, if $D$ is the differential operator, then $f(x)+f'(x)+\frac{f''(x)}{2}=(e^Df)(x)=f(x+1)$.
• So $g(x)=f(x+1)+\frac{f''(x)}{2}$. But if $f$ is quadratic and everywhere positive, then its second derivative must be positive; thus $g$ is everywhere positive.

Solution 3:

Without loss of generality, we can assume that $f(x)= (x-a)^2 + b^2$.

Note that $g$ is also a quadratic polynomial. We find the critical point of $g$.

$$ g'(x) = f'(x) + f''(x)$$

Note also that third derivative of $f$ is zero.

Then we see that the critical point is $x=a-1$, and also it gives the minimum value of $g$.

Plug in the value to $g$, then we have

$$ g(a-1) = f(a-1) + g'(a-1) = f( a-1) > 0.$$

Therefore $g(x)>0$ for all $x\in \mathbb{R}$.