Question

A function is called “piecewise linear” if it is (i) continuous and (ii) its graph consists of finitely many linear segments. Prove that a continuous function on an interval [a, b] is the uniform limit of a sequence of piecewise linear functions. 

Answer 1

Answer:

as given,

f is piecewise linear so,

f is continuous on [a, b], it is uniformly continuous.

Thus, given e > 0, we may partition [a, b] by points $x_0 = a, x_1, . . . , x_n = b,$

so that for each j,

$|f(x) − f(y)| < e$ if x and y belong to $[x_{j-1}, x_j]$

Let g be the function which is linear on each interval [x_j-1, x_j] and is such that g(xj) = f(xj ) for each j.

Then it is easy to see that $|f(x) − g(x)| < 2e$ for all $x ∈ [a, b]$ .

We conclude that f can be uniformly approximated by piecewise-linear functions, so it suffices to show that any piecewise-linear function can be approximated by polynomials.

Now, any piecewise-linear function g can be written as

$g(x)=b+\sum_{j=1}^{n}c_j|x-a_j|$

or suitable constants $n, a_j , c_j$ and b.

By the Lemma, |x| can be approximated by polynomials.

Translating these polynomials by aj shows that we can approximate $|x−a_j |$ , and then taking a linear combination shows that we can approximate the piecewise-linear function g,