Ideal not contained in maximal ideal in Banach Algebra $C_0(\mathbb{R})$ Let $A = C_0(\mathbb{R})$ be the (non unital) commutative Banach Algebra (with the uniform norm) of continuous functions vanishing at infinity. Let 
$$ 
I=\lbrace f \in A : \lim_{x \to \infty} xf(x) = 0 \rbrace,
$$
which is a proper ideal of $A$. The goal of the exercise I am trying to solve is to show that $I$ is not contained in a maximal ideal $\mathcal{M}$. Reasoning by contradiction, if we suppose that $g_0 \in A \setminus \mathcal{M}$ with $\vert \vert g_0 \vert \vert =1$, I am given the following hints in order to deduce that $g_0 \in \mathcal{M}$:


*

*There exists $\varphi_1 \in A$ such that $\vert g_0 \vert^{1/4} - \varphi_1 \vert g_0 \vert^{1/2} \in \mathcal{M}$.

*There exist $C$, $\alpha > 0$ such that if $E$ is the compact $E = \lbrace t \in \mathbb{R} : \vert g_0(t) \vert \geq \alpha \rbrace$, then 
$$
C \chi_E + \vert g_0 \vert^{1/4} -\varphi_1\vert g_0 \vert^{1/2} \geq \frac{\vert g_0 \vert^{1/4}}{2}.
$$

*There exists $h \in I$ with $h\geq 0$ and $h \geq 1$ in $E$.

*There exists $f \in \mathcal{M}$ with $f \geq \vert g_0 \vert^{1/4}$.

*Deduce that $g_0 \in \mathcal{M}$.


My thoughts: 


*

*We can prove that $\lbrace \varphi \vert g_0 \vert^{1/2} + \psi: \varphi \in A, \psi \in \mathcal{M} \rbrace$ is an ideal which contains $\mathcal{M}$ and $g_0$, and therefore it is equal to $A$. Then there exists $\varphi_1 \in A$ and $\psi \in \mathcal{M}$ such that 
$$
\vert g_0 \vert^{1/4} = \varphi_1 \vert g_0 \vert^{1/2} + \psi.
$$

*Take $C=\vert \vert \varphi_1 \vert \vert$ and 
$\alpha = \left( \frac{1}{2\vert \vert \varphi_1 \vert \vert} \right)$. If $x \in E$, then 
\begin{align}
C \chi_E(x) + \frac{\vert g_0(x) \vert^{1/4}}{2}  - \varphi_1(x)\vert g_0(x) \vert^{1/2}  = \vert \vert\varphi_1 \vert \vert + \frac{\vert g_0(x) \vert^{1/4} }{2}- \varphi_1(x)\vert g_0(x) \vert^{1/2} \geq \\
   \geq  \vert \vert \varphi_1 \vert \vert + \frac{\vert g_0(x) \vert^{1/4} }{2} - \vert\vert\varphi_1\vert\vert =  \frac{\vert g_0(x) \vert^{1/4} }{2} \geq 0.
\end{align}
If $x \neq E$,
    \begin{align}
   C \chi_E(x) + \frac{\vert g_0(x) \vert^{1/4} }{2} - \varphi_1(x)\vert g_0(x) \vert^{1/2} = \frac{\vert g_0(x) \vert^{1/4} }{2}- \varphi_1(x)\vert g_0(x) \vert^{1/2} = \\
   =\vert g_0(x) \vert^{1/4} \left(\frac{1}{2}-\varphi_1(x)\vert g_0(x) \vert^{1/4}\right) \geq \vert g_0(x) \vert^{1/4} \left(\frac{1}{2}-\varphi_1(x)\frac{1}{2\vert\vert\varphi_1\vert\vert}\right) \geq 0.
  \end{align}

*Let $M > 0$ such that $E \subset [-M,M]$ and take any continuous function satisfying $h=1$ in $[-M,M]$ and $h=0$ in $\mathbb{R} \setminus [-M-1,M+1]$.

*We have that $f=2 \left( Ch + \vert g_0 \vert ^{1/4} - \varphi_1 \vert g_0 \vert^{1/2} \right) \in \mathcal{M}$ and by (2) it satisfies $f \geq \vert g_0 \vert^{1/4}$.


However, I can not see how to deduce (5) from the previous ones. I would really appreciate any suggestions.
 A: 

*Define a function $h$ as follows: $h(x) = g_0(x) / f(x)$ if $g_0(x) \neq 0$, and $h(x) = 0$ otherwise.


Note that $h$ is well-defined, as $g_0(x) \neq 0$ implies that $f(x) \geq |g_0(x)|^{1/4} > 0$.
We show that $h$ is an element of $A$.

Firstly, we verify that $h$ is a continuous function. The continuity is clear at points $x$ with $g_0(x) \neq 0$ (which implies $f(x)\neq 0$). Now take $x_0\in \Bbb R$ such that $g(x_0) = 0$ and we want to show that $h$ is continuous at $x_0$.
By continuity of $g_0$, for any $\epsilon > 0$, there exists an open neighborhood $I$ of $x_0$ such that $|g_0(x)| < \epsilon^{4/3}$ for all $x\in I$.
Now for every $x\in I$, we have either $g_0(x) = 0$, in which case $|h(x)| = 0$, or $g_0(x) \neq 0$, in which case $|h(x)| = |g_0(x)| / |f(x)| \leq |g_0(x)|^{3/4} < \epsilon$. This proves the continuity at $x_0$.
Secondly, we prove that $\lim_{x\rightarrow\infty}h(x) = 0$. This is equivalent to saying that for any $\epsilon > 0$, the set $\{x\in\Bbb R: |h(x)| \geq \epsilon\}$ is bounded. 
But this is clear, since $|h(x)| \geq \epsilon$ implies that $g_0(x) \neq 0$ and hence $\epsilon \leq |h(x)| = |g_0(x)| / |f(x)| \leq |g_0(x)|^{3/4}$, or $|g_0(x)| \geq \epsilon^{4/3}$.
Therefore the above set is in fact contained in the set $\{x\in \Bbb R: |g(x)| \geq \epsilon^{4/3}\}$, which is bounded, since $g$ is an element of $A$.

We have finished the proof that $h$ is an element of $A$.
Now it only remains to remark that $g_0 = f\cdot h$.
