\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
If \(n\) is odd, then \(p\) must be negative when either \(x \gg 0\) or \(x \ll 0\), for a sufficiently large \(\abs *{x}\), since the leading term (term with \(x^n\)) will be sufficiently large at this point. Since \(p(x) > 0\), \(n\) must be even. Furthermore, the leading term coefficient must be positive.
For \(0 \leq k \leq n\), \(p^{(k)} (x)\) is an \(n - k\) degree polynomial. Hence, \(q\) is also a degree \(n\) polynomial, with a positive leading term coefficient. This means when \(\abs *{x}\) is sufficiently large, the leading term will be sufficiently positive and \(q\) will be positive.
We would like to show that \(q(x) - q'(x) = p(x)\), and we have \begin {align*} q(x) - q'(x) & = \sum _{k = 0}^{n} p^{(k)} (x) - \DiffOp {x} \sum _{k = 0}^{n} p^{(k)} (x) \\ & = \sum _{k = 0}^{n} p^{(k)} (x) - \sum _{k = 0}^{n} p^{(k + 1)} (x) \\ & = \sum _{k = 0}^{n} p^{(k)} (x) - \sum _{k = 1}^{n + 1} p^{(k)} (x) \\ & = p^{(0)} (x) - p^{(n + 1)} (x) \\ & = p(x) - 0 \\ & = p(x), \end {align*}
as desired.
If \(q'(x) = 0\) for some \(x\), then \(0 = p(x) - q(x)\), giving \(p(x) = q(x)\) for that point. This means \(p(x)\) and \(q(x)\) will meet at that point, proving precisely \(p(x)\) and \(q(x)\) meet at every stationary point of \(y = q(x)\).
This means \(q\) has all local minimums being positive, since they must be stationary points, situated on \(p\) as well, being positive.
Since \(q\) is an even-degree polynomial, it must also be the case that one of the local minimums is a global minimum, which is positive.
Hence, \(q\) is always positive, and \(q(x) > 0\) for all \(x\).
By differentiating, we have \begin {align*} \DiffFrac {e^{-x} q(x)}{x} & = e^{-x} q'(x) - e^{-x} q(x) \\ & = e^{-x} (q'(x) - q(x)) \\ & = - e^{-x} p(x). \end {align*}
We have \(e^{-x} > 0\) and \(p(x) > 0\) for all \(x\), which means the gradient is always negative, which shows that \(e^{-x} q(x)\) is decreasing.
For sufficiently large \(x\), \(q(x) > 0\), and hence \(e^{-x} q(x) > 0\) for sufficiently large \(x\).
Since this function is decreasing, we can conclude that \(e^{-x} q(x) > 0\) for all \(x\), and since \(e^{-x}\) is always positive, it must be the case that \(q(x) > 0\) for all \(x\).
Let the upper bound of the integral be \(N\). Using integration by parts, we have \begin {align*} \int _{0}^{N} p(x + t) e^{-t} \Diff t & = - \int _{0}^{N} p(x + t) \Diff e^{-t} \\ & = - \left [p(x + t) e^{-t}\right ]_{0}^{N} + \int _{0}^{N} e^{-t} \Diff p(x + t) \\ & = p(x) - p(x + N) e^{-N} + \int _{0}^{N} p'(x + t) e^{-t} \Diff t. \end {align*}
Let \(N \to \infty \), \(e^{-N} p(x + N) \to 0\) since an exponential dominates a polynomial. Hence, \[ \int _{0}^{\infty } p(x + t) e^{-t} \Diff t = p(x) + \int _{0}^{\infty } p^{(1)} (x + t) e^{-t} \Diff t \] as desired.
Repeating this process, we have \begin {align*} \int _{0}^{\infty } p(x + t) e^{-t} \Diff t & = p(x) + \int _{0}^{\infty } p^{(1)} (x + t) e^{-t} \Diff t \\ & = p(x) + p^{(1)} (x) + \int _{0}^{\infty } p^{(2)} (x + t) e^{-t} \Diff t \\ & = \cdots \\ & = p(x) + p^{(1)} (x) + \cdots + p^{(n)} (x) + \int _{0}^{\infty } p^{(n + 1)} (x + t) e^{-t} \Diff t \\ & = \sum _{k = 0}^{n} p^{(k)}(x) + \int _{0}^{\infty } 0 \Diff t \\ & = q(x) + 0 \\ & = q(x), \end {align*}
as desired.
Since the integrand of this integral is positive for all \(t \geq 0\), the integral must evaluate to a positive value, and hence \(q(x) > 0\) for all \(x\) as desired.