Ideal not contained in maximal ideal in Banach Algebra $C_0(\mathbb{R})$

Question

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 $g_0 \in \mathcal{M}$:

1. There exists $\varphi_1 \in A$ such that $\vert g_0 \vert^{1/4} - \varphi_1 \vert g_0 \vert^{1/2} \in \mathcal{M}$.
2. 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 $$ C \chi_E + \vert g_0 \vert^{1/4} -\varphi_1\vert g_0 \vert^{1/2} \geq \frac{\vert g_0 \vert^{1/4}}{2}. $$
3. There exists $h \in I$ with $h\geq 0$ and $h \geq 1$ in $E$.
4. There exists $f \in \mathcal{M}$ with $f \geq \vert g_0 \vert^{1/4}$.
5. Deduce that $g_0 \in \mathcal{M}$.

My thoughts:

1. 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 $$ \vert g_0 \vert^{1/4} = \varphi_1 \vert g_0 \vert^{1/2} + \psi. $$
2. 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}
3. Let $M > 0$ such that $E \subset [-M,M]$ and take any continuous function satisfying $h=1$ in $[-M,M]$ and $h=0$ in $\mathbb{R} \setminus [-M-1,M+1]$.
4. 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.

Answer 1

5. 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 $A$.

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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 $A$.

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We have finished the proof that $h$ is an element of $A$.

Now it only remains to remark that $g_0 = f\cdot h$.