Uniform approximation on $[0,\infty)$

Question

Let $f:[0,\infty) \to \mathbb R$ be a continuous function such that $f(x) \to 0$ as $x \to \infty$. If $\epsilon >0$ then there exists a polynomial $p$ such that $|f(x)-e^{-x}p(x)|<\epsilon$ for all $x \in [0,\infty)$. This is an exercise by Serge Lang. I will post an answer after 24 hours to see if there is a solution better than mine. My proof is fairly elementary but somewhat long and I have been shown a shorter proof using a couple of non-trivial theorems.

Answer 1

Let $g(x)=f(-\ln x)$ if $0 < x \leq 1$, $g(0)=0$. Then $g \in C[0,1]$. Let $\epsilon>0$ By Wierstrass Theorem there existse a polynomial $q_1$ such that $|g(x)-q_1(x)| <\epsilon$ for $0 \leq x \leq 1$. Let $q_1 (x)=\sum_{j=0}^{N} a_jx^{j}$ Note that $|a_0|=|g(0)-q_1 (0)|< \epsilon$. If $q(x)=\sum_{j=1}^{N} a_jx^{j}$ then $|g(x)-q(x)|<2\epsilon$ for all $x \in [0,1]$. Going back to $[0,\infty)$ we get $|f(x)-\sum_{j=1}^{N} a_je^{-jx}|<2\epsilon$ for all $x \geq 0$. From this we conclude that it is enough to prove the result for the special functions $e^{-kx}$ where $k$ is a positive integer. We do this by induction on $k$. For $k=1$ we take $p \equiv 1$. Not assume that $|e^{-kx}-e^{-x}p_k(x)| <\epsilon$ for $ x \geq 0$. Changing $x$ to $a_k x$ where $a_k=1+\frac 1 k$ we get $$|e^{-(k+1)x}-e^{-a_k x}r_k(x)|<\epsilon$$ where $r_k x=p_k (a_k x)$. If we show that there is a polynomial $\phi$ such that $$|e^{-x}\phi(x)r_k(x)-e^{-a_kx}r_k(x)|<\epsilon$$ for all $x\geq 0$ the proof would be complete. We claim that $\phi(x)=\sum_{m=0}^{2N} \frac {(-x/k)^{m}} {m!}$ will suffice provided we choose $N$ sufficiently large. We use an elementary fact about alternating series: if $s_n$ is the n-th partial sum of the series $\sum_{m=0}^{\infty} \frac {(-x)^{m}} {m!}$ then $s_{2n-1} \leq e^{-x} \leq s_{2n}$ which implies $|e^{-x}-s_{2N}| \leq \{s_{2N}-{s_{2N-1}}\} =\frac {(-x)^{2N}} {2N!}$. Hence $$|e^{-x}\phi(x)r_k(x)-e^{-a_kx}r_k(x)| \leq M e^{-x/k} \frac {(-x/k)^{2N}} {2N!}$$ where $M=\sup \{e^{-(1-\frac 1 k)x}r_k(x):x \geq 0\}$. The last expression attains its maximum on $[0,\infty)$ at $x=2kN$. It remains only to see that this maximum value, namely, $\frac {Me^{-2N} (2N)^{2N}} {(2N)!} \to 0 $ as $N \to \infty$. This is an immediate consequence of Stirling's Formula.