Continuity under algebraic operations

We create new functions from old, using +, -, {\cdot}, /, {\circ}, and see how this affects continuity.

Theorem.

 If f, g: {\bf R}^n {\rightarrow} {\bf R} are continuous at x = x, then so is f + g.

Proof. Let

 h = f + g.

We want to show that for every \epsilon > 0, there is a \delta >0 such that

 | x - a | < \delta, x \in D(h) => | h(x) - h(a) | < \epsilon.

Let's write the definition of continuity twice - for f and g. As it turns our choosing \frac{\epsilon}{2} is preferable in order to get from these to the one above.

 (1) there is a \delta_1 > 0 such that 
      | x - a | < \delta_1, x \in D(f) => | f(x) - f(a) | < \frac{\epsilon}{2},
 (2) there is a \delta_2 > 0 such that 
      | x - a | < \delta_2, x \in D(g)  => | g(x) - g(a) | < \frac{\epsilon}{2}.

Let

 \delta = min( \delta_1, \delta_2 ). 

Then

 \delta_1, \delta_2 \leq \delta.

Hence we can re-write the above two definitions as follows:

 there is \delta > 0 such that 
 | x - a | < \delta, x \in D(f) \cap D(g) = D(h) => 
 | f(x) - f(a) | < \frac{\epsilon}{2}, and 
 | g(x) - g(a) | < \frac{\epsilon}{2}.

To get to h, use the two inequalities, as follows:

 | h(x) - h(a) | = | ( f(x) + g(x) ) - ( f(a) + g(a) ) | 
                 = | ( f(x) - f(a) ) + ( g(x) - g(a) ) |
                 \leq | f(x) - f(a) | + | g(x) - g(a) |     Triangle Inequality
                 < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. QED

Theorem (Sum).

 If f: {\bf R}^n {\rightarrow} {\bf R} is continuous at a and c is a constant, then c{\cdot}f is continuous at a.

Hint:

 | cf(x) - cf(a) | = | c | | f(x) - f(a) |

with f continuous, and we have to show that

 | c | | f(x) - f(a) | < \epsilon.

Choose \delta = \frac{\epsilon}{|c|}.

Theorem (Constant multiple).

 Let f, g: {\bf R}^n {\rightarrow} {\bf R} be continuous at a. Then f{\cdot}g is continuous at a.

Theorem (Ratio). Let f, g: {\bf R}^n {\rightarrow} {\bf R} be continuous at a. Then \frac{f}{g} is continuous at a, provided g(a) ≠ 0.

Theorem (Composition).

 Let f: {\bf R}^n {\rightarrow} {\bf R}, g: {\bf R} {\rightarrow} {\bf R}, where f is continuous at a and g is continuous at f(a). 
 Then h = g {\circ}f is continuous at a.

Read the proof.

Exercise. Prove f( x, y ) = x + y is continuous, from the definition.

Exercise. Prove f(u) = || u || (the norm) is continuous, two ways:

• from the definition: use || u - a || < \delta ⇒ | ||u|| - ||a|| | < \epsilon;
• from the theorems: use f(u) = ( x_1^2 + ... + x_n^2 )^{\frac{1}{2}}.

Theorem. Affine functions are continuous.

Proof. Recall

 f(x) = Ax + b, 
 Ax = < v, x > = v_1 x_1 + ... + v_n x_n

Now, the definition:

 || x - a || < \delta ⇒ | ( Ax + b ) - ( Aa + b ) | < \epsilon,

where

 | ( Ax + b ) - ( Aa + b ) | = | Ax - Aa | = | A ( x - a ) | = | < v, x - a > | 
                             = || v || {\cdot} || x - a || < \epsilon.

Then choose \delta = \epsilon / || v || if v ≠ 0. If v = 0, then f is constant. QED