Suppose f and g are continuous at p_0 in D. Show that fg is continuous at p_0. (see below for full question)

Let f and g be real valued functions D ⊂ R^n. Consider the functions fg defined on D. Suppose f and g are continuous at p_0 in D. Show that fg is continuous at p_0.

Let ε > 0 be given.

Note that ||(fg)(x) - (fg)(p₀)||
= ||f(x) g(x) - f(p₀) g(p₀)||
≤ ||f(x) g(x) - f(p₀) g(x)|| + ||f(p₀) g(x) - f(p₀) g(p₀)||
= ||f(x) - f(p₀)|| ||g(x)|| + ||f(p₀)|| ||g(x) - g(p₀)||

Since f and g are continuous at p₀, there exist δ₁, δ₂, δ₃ > 0 such that
(i) For all x in D such that ||x - p₀|| < δ₁, we have ||f(x) - f(p₀)|| < ε/(2(1 + ||g(p₀)||))
(ii) For all x in D such that ||x - p₀|| < δ₂, we have ||g(x) - g(p₀)|| < ε/(2(1 + ||f(p₀)||))
(iii) For all x in D such that ||x - p₀|| < δ₃, we have ||g(x) - g(p₀)|| < 1.

Letting δ = min{δ₁, δ₂, δ₃}: For all x in D such that ||x - p₀|| < δ, we have
||f(x) - f(p₀)|| < ε/(2(1 + ||g(p₀)||)), ||g(x) - g(p₀)|| < ε/(2(1 + ||f(p₀)||)), and ||g(x) - g(p₀)|| < 1.

By (iii), we can bound ||g(x)||:
||g(x)|| ≤ ||g(x) - g(p₀)|| + ||g(p₀)||
......... < 1 + ||g(p₀)||.
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Finally, for all x in D such that ||x - p₀|| < δ, we have
||(fg)(x) - (fg)(p₀)||
≤ ||f(x) - f(p₀)|| ||g(x)|| + ||f(p₀)|| ||g(x) - g(p₀)||
< [ε/(2(1 + ||g(p₀)||))] * (1 + ||g(p₀)||) + ||f(p₀)|| [ε/(2(1 + ||f(p₀)||))]
< ε/2 + ε/2
= ε.

I hope this helps!