A Banach algebra with a proper dense two-sided ideal.

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1. C_{c}(R) = {f ∈ C_{0}(R); supp(f) = the closure of {x ∈ R; f(x) ≠ 0} is compact } is a dense ideal of C_{0}(R). Note that the function f defined by

f(x) = \{ \begin{matrix}\frac{1}{1+x} & x ≥ 0 \\ \frac{1}{1−x} & x < 0 \end{matrix}

belongs to C_{0}(R) − C_{c}(R).

2. A = {f ∈ C([0, 1]); f(0) = 0} is a closed subalgebra of C([0, 1]) not containing the constant function 1. So A is a non-unital Banach algebra. Let f_{∘}(t) = t , t ∈ [0, 1]. I = {f_{∘}g; g ∈ C[0, 1]} is a proper ideal of A (since if

h(t) = \{ \begin{matrix}tsin\frac{1}{t} & t ∈ (0, 1] \\ 0 & t = 0 \end{matrix}

and for some g ∈ C[0, 1], tg(t) = h(t) whenever t ∈ [0, 1] then lim_{t→0} sin \frac{1}{t} = g(0), a contradiction). By the Stone-Weierstrass theorem, each f ∈ A is the uniform limit of a sequence (p_{n}) of polynomials with p_{n}(0) = 0. Moreover t ⟼ \frac{p_{n}(t)}{t} belongs to C[0, 1] and t\frac{p_{n}(t)}{t} ⟶ f(t) uniformly on [0, 1]. So f belongs to the closure of I. Hence I is dense in A.