$\forall \epsilon >0 \exists$ a piecewise linear function $g$% such that for a continous function $f$

Question

How would I show the following Edit

A function is called piecewise linear if it is (1)Continuous (2)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 piece wise linear function.

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$\forall \epsilon >0 \exists$ a piecewise linear function $g$ such that for a continous function $f$

$|g(x)-f(x)|<\epsilon$ $\forall x \in E$

I am not sure how I would do this problem.

Answer 1

Let $f:[a,b]\to\mathbb R$ be continuous. For $n\in\mathbb N$ and $h_n = (b-a) / n$ define $x_{n,k} = a + kh_n$, $0\le k \le n$, and the piecewise linear function $g_n:[a,b]\to\mathbb R$ by $$ g_n(x_{n,k}) = f(x_{n,k}). $$

Claim: $g_n \to f$ uniformly.

Hints:

1. For every $\epsilon > 0$ there is a $\delta > 0$ such that $|f(x) -f(y)| < \epsilon$ holds for every $x,y\in[a,b]$ with $|x-y|<\delta$.
2. There is a $N\in \mathbb N$ such that for $x_{n,k+1} - x_{n,k} = h_n = (b-a)/n < \delta$ for every $n\ge N$ and $1\le k \le n$.
3. Let $x\in [x_{n,k}, x_{n,k+1}]$. Wlog assume $f(x_{n,k}) \le f(x_{n,k+1})$. Then, we have $$ f(x_{n,k}) = g_n(x_{n,k}) \le g_n(x) \le g_n(x_{n,k+1}) \le f(x_{n,k+1}). $$

Could you bound $g_n(x) - f(x)$?