Solution 1

Since ${\displaystyle f\!\,}$ is absolutely continuous for all ${\displaystyle [a,b]\subset R^{1}\!\,}$,

${\displaystyle \int _{a}^{b}f^{\prime }(x)dx=f(b)-f(a)\!\,}$

Hence

${\displaystyle \int _{-\infty }^{\infty }f^{\prime }(x)dx=\lim _{a,b\rightarrow \infty }\int _{a}^{b}f^{\prime }(x)=\lim _{a,b\rightarrow \infty }[f(b)-f(a)]\!\,}$

Since ${\displaystyle f^{\prime }\!\,}$ is integrable i.e. ${\displaystyle f^{\prime }\in L^{1}(-\infty ,\infty )\!\,}$, ${\displaystyle \lim _{b\rightarrow \infty }f(b)\!\,}$ and ${\displaystyle \lim _{a\rightarrow \infty }f(a)\!\,}$ exist.

Assume for the sake of contradiction that

${\displaystyle \lim _{b\rightarrow \infty }|f(b)|=\delta >0\!\,}$

Then there exists ${\displaystyle M\!\,}$ such that for all ${\displaystyle x>M\!\,}$

${\displaystyle |f(x)|>{\frac {\delta }{2}}\!\,}$

since ${\displaystyle f\!\,}$ is continuous. (At some point, ${\displaystyle |f|\!\,}$ will either monotonically increase or decrease to ${\displaystyle \delta \!\,}$.) This implies

${\displaystyle {\begin{aligned}\int _{-\infty }^{\infty }|f(x)|dx&\geq \int _{M}^{\infty }|f(x)|dx\\&\geq \int _{M}^{\infty }{\frac {\delta }{2}}dx\\&=\infty \end{aligned}}}$

which contradicts the hypothesis that ${\displaystyle f\!\,}$ is integrable i.e. ${\displaystyle f\in L^{1}(-\infty ,\infty )\!\,}$. Hence,

${\displaystyle \lim _{b\rightarrow \infty }f(b)=0\!\,}$

Using the same reasoning as above,

${\displaystyle \lim _{a\rightarrow -\infty }f(a)=0\!\,}$

Hence,

${\displaystyle \int _{-\infty }^{\infty }f^{\prime }(x)dx=\lim _{b\rightarrow \infty }f(b)-\lim _{a\rightarrow -\infty }f(a)=0\!\,}$

Alternate Solution

Suppose ${\displaystyle \int _{-\infty }^{\infty }f'=c\neq 0}$ (without loss of generality, ${\displaystyle c>0}$). Then for small positive ${\displaystyle \epsilon }$, there exists some real ${\displaystyle M}$ such that for all ${\displaystyle m>M}$ we have ${\displaystyle c-\epsilon <\int _{-m}^{m}f'<c+\epsilon }$. By the fundamental theorem of calculus, this gives

${\displaystyle f(-m)+c-\epsilon <f(m)<f(-m)+c+\epsilon }$ for all ${\displaystyle m>M}$.

Since ${\displaystyle f}$ is integrable, this means that for any small positive ${\displaystyle \delta }$, there exists an ${\displaystyle N}$ such that for all ${\displaystyle n>N}$, we have ${\displaystyle \int _{-\infty }^{-n}+\int _{n}^{\infty }f<\delta }$. But by the above estimate,

${\displaystyle \int _{-\infty }^{-n}f+\int _{n}^{\infty }f>\int _{-\infty }^{-n}f+\int _{-\infty }^{-n}f+(c-\epsilon )=\int _{-\infty }^{-n}2f+\int _{-\infty }^{-n}(c-\epsilon )=\infty }$

This contradicts the integrability of ${\displaystyle f}$. Therefore, we must have ${\displaystyle c=0}$.

Solution 3a

By definition of norm,

${\displaystyle \|f\|_{p}=\left(\int |f(x)|^{p}\,\mathrm {d} x\right)^{\frac {1}{p}}\!\,}$

Since ${\displaystyle \|f_{n}\|_{p}\leq M\!\,}$,

${\displaystyle \|f_{n}\|_{p}^{p}\leq M^{p}\!\,}$

By Fatou's Lemma,

${\displaystyle {\begin{aligned}\|f\|_{p}^{p}&=\int _{0}^{1}|f(x)|^{p}dx\\&=\int _{0}^{1}{\underset {n}{\lim \inf }}|f_{n}(x)|^{p}dx\\&\leq {\underset {n}{\lim \inf }}\int _{0}^{1}|f_{n}(x)|^{p}dx\\&\leq M^{p}\end{aligned}}}$

which implies, by taking the ${\displaystyle p\!\,}$th root,

${\displaystyle \|f\|_{p}\leq M\!\,}$

Solution 3b

By Holder's Inequality, for all ${\displaystyle A\subset [0,1]\!\,}$ that are measurable,

${\displaystyle {\begin{aligned}\int _{A}|(f(x)-f_{n}(x))\cdot 1|dx&\leq \left(\int _{A}|f(x)-f_{n}(x)|^{p}dx\right)^{\frac {1}{p}}\cdot \left(\int _{A}1^{q}dx\right)^{\frac {1}{q}}\\&\leq \left(\int _{A}|2f(x)|^{p}dx\right)^{\frac {1}{p}}\cdot \left(\int _{A}1^{q}dx\right)^{\frac {1}{q}}\\&\leq 2\left(\int _{A}|f(x)|^{p}dx\right)^{\frac {1}{p}}\cdot \left(\int _{A}1^{q}dx\right)^{\frac {1}{q}}\\&\leq 2M\cdot (m(A))^{\frac {1}{q}}\end{aligned}}}$

where ${\displaystyle {\frac {1}{p}}+{\frac {1}{q}}=1\!\,}$

Hence, ${\displaystyle |f(x)-f_{n}(x)|\leq 2M(m(A))^{\frac {1}{q}}\!\,}$

The Vitali Convergence Theorem then implies

${\displaystyle \lim _{n\rightarrow \infty }\int _{0}^{1}|f(x)-f_{n}(x)|dx=\int _{0}^{1}\lim _{n\rightarrow \infty }|f(x)-f_{n}(x)|dx=0\!\,}$

Solution 5

${\displaystyle {\begin{aligned}\int _{R}|f(x)|dx&=\int _{|x|>1}|xf(x|\cdot {\frac {1}{|x|}}dx+\int _{|x|<1}|f(x)|\cdot 1dx\\&\leq \left(\int _{|x|>1}|xf(x)|^{2}dx\right)^{\frac {1}{2}}\cdot \left(\int _{|x|>1}{\frac {1}{|x|^{2}}}dx\right)^{\frac {1}{2}}+\left(\int _{|x|<1}|f(x)|^{2}dx\right)^{\frac {1}{2}}\cdot \left(\int _{|x|<1}1^{2}dx\right)^{\frac {1}{2}}\end{aligned}}}$