Approximation by piecewise linear functions

A function $f$, continuous on the closed interval $[a ; b]$, is called piecewise linear if there exists a partition of the closed interval $[a ; b]$ by points $x_{0}, x_{1}, \ldots , x_{n}$ such that

and the function $f$ is linear on each closed interval $\left[x_{i-1} ; x_{i}\right], i = 1,2, \ldots , n$ (i.e., $f(\mathbf{x}) = a_{i} x+b_{i}$). Prove that every function $F$, continuous on the closed interval $[a ; b]$, can be approximated to any degree of accuracy by a piecewise linear function, i.e., for any $\varepsilon > 0$ there exists a piecewise linear function $f$ such that for all $x \in [a ; b]$ the inequality holds

For the function $F$, choose a piecewise linear function $f$ so that for all $x \in [a ; b]$ the inequality $\left|F(x)-f(x)\right| < 0,1$ holds:

Find a piecewise linear function $f$ on the closed interval $[0 ; 100]$ such that for all $x \in [0 ; 100]$ the inequality holds

A function $f$, continuous on $\mathbb {R}$, is called piecewise linear on $\mathbb {R}$ if there exist numbers $a$ and $b, a < b$, such that the function $f$ is linear on the intervals $(-\infty ; a],[b ;+\infty )$ and piecewise linear on $[a ; b]$. Prove that, whatever the number $\varepsilon$ may be, there is no function $f$, piecewise linear on $\mathbb {R}$, for which $\left|x^{2}-f(x)\right| < 1$ for all $x \in \mathbb {R}$.

Prove that every piecewise linear function can be given by the formula

Find the formula indicated in the preceding problem for a function $f$, piecewise linear on a closed interval, if:

Prove that a function piecewise linear on $\mathbb {R}$ is uniformly continuous on $\mathbb {R}$.

Let

Prove that for no value of $a$ does there exist a function $g$, piecewise linear on $[-1 / \pi ; 1 / \pi ]$, such that $\left|f(x)-g(x)\right| < 1 / 2$ for any $x \in [-1 / \pi ; 1 / \pi ]$.

Formulate in positive form the statement that a function, defined on a closed interval, cannot be approximated to any degree of accuracy by a piecewise linear function.

Prove that if a function $f$ is defined, but discontinuous, on the closed interval $[a ; b]$, then it cannot be approximated to any degree of accuracy by a piecewise linear function.

Prove that if a function $f$ on an interval can be approximated to any degree of accuracy by a piecewise linear function, then $f$ is uniformly continuous on this interval.

Prove that if the function $f$ on the interval $[c ;+\infty )$ can be approximated to any degree of accuracy by a piecewise linear function, then $f$ has an asymptote as $x \rightarrow +\infty$ (a function is piecewise linear on $[c ;+\infty )$ if it is continuous on $[c ;+\infty )$, linear on $[d ;+\infty )$ for some $d > c$, and piecewise linear on $[c ; d]$).

Prove that every continuous function convex on $\left[a, b\right]$ is the limit of a uniformly convergent decreasing sequence of:

A continuous function $\varphi$ on $[a, b]$ is called piecewise linear provided there is a partition $a = x_{0} < x_{1} < \cdots < x_{n} = b$ of $[a, b]$ for which $\varphi$ is linear on each interval $\left[x_{i}, x_{i+1}\right]$. Let $f$ be a continuous function on $[a, b]$ and $\epsilon$ a positive number. Show that there is a piecewise linear function $\varphi$ on $[a, b]$ with $\left|f(x)-\varphi (x)\right| < \epsilon$ for all $x \in [a, b]$.