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.
I am not sure how to do this problem.
If I have $[a,b]$ then you can make a partion which
$a=x_0<x_1<.....<x_n=b$ then the function r is linear on the interval $[x_i,x_{i+1}]$
So then is f a continuous function on $[a,b]$ and $\forall \epsilon >0$
I would have to show $\exists \delta>0$ st $|x-y|<\delta$ then $|r(x)-f(y)|<\epsilon$
for all $x,y \in [a,b]$ But I am not sure how.
Let $\gamma>0$. By the uniform continuity of $f(x)$ on $[a,b]$ there exists $\delta>0$ such that $|f(x)-f(y)|<\gamma$ for any $x,y\in [a,b]$ with $|x-y|<\delta$. Let $\{x_0,x_1,\dotsc,x_n\}$ be a partition of $[a,b]$, i.e. $a=x_0<x_1\dotso<x_n=b$ such that $|x_i-x_{i-1}|<\delta$ for all $i=1,2,\dotsc,n$ (it is easy to construct such a partition). Let $y_i=f(x_i)$ for $i=0,1,2,\dotsc, n$. Then it follows that $|y_i-y_{i-1}|<\gamma$ for any $i=1,2,\dotsc,n$. Let $\phi_n(x)$ be the piece-wise linear function defined as the straight lines connecting the points $(a,f(a))$ to $(x_1,f(x_1))$, $(x_1,f(x_1))$ to $(x_2,f(x_2))$ and so forth (up to $n$). The difference of $|f(x)-\phi_n(x)|$ for any $x\in [a,b]$ can be made arbitrarily small by considering $$|f(x)-\phi_n(x)|\leq |f(x)-f(x_i)|+|f(x_i)-\phi_n(x)|,$$ where $x\in [x_{i-1},x_i]$ for some $1\leq i \leq n$ (this follows by the partition of $[a,b]$ every element is in some subinterval), if I am not mistaken. Uniform convergence of $\phi_n$ to $f$ would then follow.