Suppose that is a continuous real-valued function with domain and that is absolutely continuous on every finite interval .
Prove: If and are both integrable on , then
![{\displaystyle \int _{-\infty }^{\infty }f^{\prime }=0\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0e94d80d5ee0590824c8b63a26a2a21836e8b90)
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Since
is absolutely continuous for all
,
![{\displaystyle \int _{a}^{b}f^{\prime }(x)dx=f(b)-f(a)\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ab57ccc8a80f85d6a0e6a71c30acbd6467530eb)
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)]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd1088a71901ca07c9f42c09a480536778799b37)
Since
is integrable i.e.
,
and
exist.
Assume for the sake of contradiction that
![{\displaystyle \lim _{b\rightarrow \infty }|f(b)|=\delta >0\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a734ba53bbc105cfbf4bdaaf4d6b548106392c3)
Then there exists
such that for all
![{\displaystyle |f(x)|>{\frac {\delta }{2}}\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a443697fbc35372d25713686450dc76be8dc2fd)
since
is continuous. (At some point,
will either monotonically increase or decrease to
.) This implies
which contradicts the hypothesis that
is integrable i.e.
. Hence,
![{\displaystyle \lim _{b\rightarrow \infty }f(b)=0\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75c652fb0e5ef07877d66042b9b36a3fbea1c5c4)
Using the same reasoning as above,
![{\displaystyle \lim _{a\rightarrow -\infty }f(a)=0\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/155f0e8005e459cd0a1d2e50f4b3a4d6574db21f)
Hence,
![{\displaystyle \int _{-\infty }^{\infty }f^{\prime }(x)dx=\lim _{b\rightarrow \infty }f(b)-\lim _{a\rightarrow -\infty }f(a)=0\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10b4a02bc8d2723144ecf15a1bf3166b13358c57)
Suppose
(without loss of generality,
). Then for small positive
, there exists some real
such that for all
we have
. By the fundamental theorem of calculus, this gives
for all
.
Since
is integrable, this means that for any small positive
, there exists an
such that for all
, we have
. But by the above estimate,
This contradicts the integrability of
. Therefore, we must have
.
Suppose that is a sequence of real valued measurable functions defined on the interval and suppose that for almost every . Let and and suppose that for all
(a) Prove that .
(b)Prove that as
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By definition of norm,
![{\displaystyle \|f\|_{p}=\left(\int |f(x)|^{p}\,\mathrm {d} x\right)^{\frac {1}{p}}\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75b55c956065900c28e09855849b20997270a4ae)
Since
,
![{\displaystyle \|f_{n}\|_{p}^{p}\leq M^{p}\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ae6cee67b810f2b2b216bcdfd0558149a522238)
By Fatou's Lemma,
which implies, by taking the
th root,
![{\displaystyle \|f\|_{p}\leq M\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c729fec2cb8e31f1cf25fa3f6adc17981f35a4dc)
By Holder's Inequality, for all
that are measurable,
where
Hence,
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\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/743827043c279356da7a2415f0e6d8be942d0b2b)
Suppose . Prove that and that
![{\displaystyle \|f\|_{1}\leq {\sqrt {2}}(\|f\|_{2}+\|xf\|_{2})\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ab4ed011d22b57c802d3043caf1b9cf771ee5cd)
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