Convergence locally uniformly VS $L^1$ convegence for probability density functions
$begingroup$
Let $f_1, f_2, ldots$ and $f$ be probability density functions on $(0, infty)$ - so $int_{(0,infty)}f_n(x)dx=1$, $int_{(0,infty)}f(x)dx=1$. Assume that for every $x in (0, infty)$ there exists a neighborhood $N_x$ of $x$ such that
$$
sup_{y in N_x}|f_n(y)-f(y)|to 0,
$$
i.e. $f_n$ converges to $f$ locally uniformly. This can also be thought in terms of uniform convergence on compact sets. Now, is it possible to also conclude that $Vert f_n - f Vert_1 to 0$? My feeling is that the answer is no, unless additional assumptions are included, but I'm not managing to construct a counterexample.
probability convergence
asked Dec 12 '18 at 16:46
Jack LondonJack London
33318
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add a comment |
$begingroup$
Let $f_1, f_2, ldots$ and $f$ be probability density functions on $(0, infty)$ - so $int_{(0,infty)}f_n(x)dx=1$, $int_{(0,infty)}f(x)dx=1$. Assume that for every $x in (0, infty)$ there exists a neighborhood $N_x$ of $x$ such that
$$
sup_{y in N_x}|f_n(y)-f(y)|to 0,
$$
i.e. $f_n$ converges to $f$ locally uniformly. This can also be thought in terms of uniform convergence on compact sets. Now, is it possible to also conclude that $Vert f_n - f Vert_1 to 0$? My feeling is that the answer is no, unless additional assumptions are included, but I'm not managing to construct a counterexample.
probability convergence
asked Dec 12 '18 at 16:46
Jack LondonJack London
33318
$endgroup$
add a comment |
$begingroup$
Let $f_1, f_2, ldots$ and $f$ be probability density functions on $(0, infty)$ - so $int_{(0,infty)}f_n(x)dx=1$, $int_{(0,infty)}f(x)dx=1$. Assume that for every $x in (0, infty)$ there exists a neighborhood $N_x$ of $x$ such that
$$
sup_{y in N_x}|f_n(y)-f(y)|to 0,
$$
i.e. $f_n$ converges to $f$ locally uniformly. This can also be thought in terms of uniform convergence on compact sets. Now, is it possible to also conclude that $Vert f_n - f Vert_1 to 0$? My feeling is that the answer is no, unless additional assumptions are included, but I'm not managing to construct a counterexample.
probability convergence
asked Dec 12 '18 at 16:46
Jack LondonJack London
33318
$endgroup$
Let $f_1, f_2, ldots$ and $f$ be probability density functions on $(0, infty)$ - so $int_{(0,infty)}f_n(x)dx=1$, $int_{(0,infty)}f(x)dx=1$. Assume that for every $x in (0, infty)$ there exists a neighborhood $N_x$ of $x$ such that
$$
sup_{y in N_x}|f_n(y)-f(y)|to 0,
$$
i.e. $f_n$ converges to $f$ locally uniformly. This can also be thought in terms of uniform convergence on compact sets. Now, is it possible to also conclude that $Vert f_n - f Vert_1 to 0$? My feeling is that the answer is no, unless additional assumptions are included, but I'm not managing to construct a counterexample.
probability convergence
probability convergence
asked Dec 12 '18 at 16:46
Jack LondonJack London
33318
asked Dec 12 '18 at 16:46
Jack LondonJack London
33318
asked Dec 12 '18 at 16:46
Jack LondonJack London
33318
asked Dec 12 '18 at 16:46
Jack LondonJack London
33318
asked Dec 12 '18 at 16:46
Jack LondonJack London
33318
33318
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
Yes, it is possible to conclude that the convergence is in $L^1$ norm.
Given any $varepsilon>0$, there exists a compact set $K$ such that
$$
int_{(0,infty)backslash K} f(x), dx<varepsilon/4
$$
since $fin L^1(0,infty)$. On the other hand, uniform convergence on $K$, a bounded set, implies $L^1(K)$ convergence, hence $exists NinBbb N$ such that
$$
int_{ K} |f(x)-f_n(x)| , dx <varepsilon/4
$$
for all $nge N$. This also implies that for each $nge N$, we have
$$begin{align}
int_{ K} |f_n(x) |, dx
&ge
int_{ K} |f(x)|, dx
-int_{ K} |f(x)-f_n(x)| , dx \
&ge (1-varepsilon/4) - varepsilon/4 \
&= 1 - varepsilon/2,
end{align}$$
hence
$$
int_{(0,infty)backslash K} |f_n(x) |, dx le varepsilon/2.
$$
This means that for all $nge N$,
$$begin{align}
int_{(0,infty)} |f(x)-f_n(x)| , dx
&le
int_{K} |f(x)-f_n(x)| , dx + int_{(0,infty)backslash K} |f(x)-f_n(x)| , dx \
&le varepsilon/4 + int_{(0,infty)backslash K} |f(x)| , dx
+int_{(0,infty)backslash K} |f_n(x)| , dx \
&le varepsilon/4 + varepsilon/4 + varepsilon/2 \
&le varepsilon,
end{align}$$
which proves convergence in $L^1(0,infty)$.
answered Dec 12 '18 at 18:03
BigbearZzzBigbearZzz
8,79421652
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, it is possible to conclude that the convergence is in $L^1$ norm.
Given any $varepsilon>0$, there exists a compact set $K$ such that
$$
int_{(0,infty)backslash K} f(x), dx<varepsilon/4
$$
since $fin L^1(0,infty)$. On the other hand, uniform convergence on $K$, a bounded set, implies $L^1(K)$ convergence, hence $exists NinBbb N$ such that
$$
int_{ K} |f(x)-f_n(x)| , dx <varepsilon/4
$$
for all $nge N$. This also implies that for each $nge N$, we have
$$begin{align}
int_{ K} |f_n(x) |, dx
&ge
int_{ K} |f(x)|, dx
-int_{ K} |f(x)-f_n(x)| , dx \
&ge (1-varepsilon/4) - varepsilon/4 \
&= 1 - varepsilon/2,
end{align}$$
hence
$$
int_{(0,infty)backslash K} |f_n(x) |, dx le varepsilon/2.
$$
This means that for all $nge N$,
$$begin{align}
int_{(0,infty)} |f(x)-f_n(x)| , dx
&le
int_{K} |f(x)-f_n(x)| , dx + int_{(0,infty)backslash K} |f(x)-f_n(x)| , dx \
&le varepsilon/4 + int_{(0,infty)backslash K} |f(x)| , dx
+int_{(0,infty)backslash K} |f_n(x)| , dx \
&le varepsilon/4 + varepsilon/4 + varepsilon/2 \
&le varepsilon,
end{align}$$
which proves convergence in $L^1(0,infty)$.
answered Dec 12 '18 at 18:03
BigbearZzzBigbearZzz
8,79421652
$endgroup$
add a comment |
$begingroup$
Yes, it is possible to conclude that the convergence is in $L^1$ norm.
Given any $varepsilon>0$, there exists a compact set $K$ such that
$$
int_{(0,infty)backslash K} f(x), dx<varepsilon/4
$$
since $fin L^1(0,infty)$. On the other hand, uniform convergence on $K$, a bounded set, implies $L^1(K)$ convergence, hence $exists NinBbb N$ such that
$$
int_{ K} |f(x)-f_n(x)| , dx <varepsilon/4
$$
for all $nge N$. This also implies that for each $nge N$, we have
$$begin{align}
int_{ K} |f_n(x) |, dx
&ge
int_{ K} |f(x)|, dx
-int_{ K} |f(x)-f_n(x)| , dx \
&ge (1-varepsilon/4) - varepsilon/4 \
&= 1 - varepsilon/2,
end{align}$$
hence
$$
int_{(0,infty)backslash K} |f_n(x) |, dx le varepsilon/2.
$$
This means that for all $nge N$,
$$begin{align}
int_{(0,infty)} |f(x)-f_n(x)| , dx
&le
int_{K} |f(x)-f_n(x)| , dx + int_{(0,infty)backslash K} |f(x)-f_n(x)| , dx \
&le varepsilon/4 + int_{(0,infty)backslash K} |f(x)| , dx
+int_{(0,infty)backslash K} |f_n(x)| , dx \
&le varepsilon/4 + varepsilon/4 + varepsilon/2 \
&le varepsilon,
end{align}$$
which proves convergence in $L^1(0,infty)$.
answered Dec 12 '18 at 18:03
BigbearZzzBigbearZzz
8,79421652
$endgroup$
add a comment |
$begingroup$
Yes, it is possible to conclude that the convergence is in $L^1$ norm.
Given any $varepsilon>0$, there exists a compact set $K$ such that
$$
int_{(0,infty)backslash K} f(x), dx<varepsilon/4
$$
since $fin L^1(0,infty)$. On the other hand, uniform convergence on $K$, a bounded set, implies $L^1(K)$ convergence, hence $exists NinBbb N$ such that
$$
int_{ K} |f(x)-f_n(x)| , dx <varepsilon/4
$$
for all $nge N$. This also implies that for each $nge N$, we have
$$begin{align}
int_{ K} |f_n(x) |, dx
&ge
int_{ K} |f(x)|, dx
-int_{ K} |f(x)-f_n(x)| , dx \
&ge (1-varepsilon/4) - varepsilon/4 \
&= 1 - varepsilon/2,
end{align}$$
hence
$$
int_{(0,infty)backslash K} |f_n(x) |, dx le varepsilon/2.
$$
This means that for all $nge N$,
$$begin{align}
int_{(0,infty)} |f(x)-f_n(x)| , dx
&le
int_{K} |f(x)-f_n(x)| , dx + int_{(0,infty)backslash K} |f(x)-f_n(x)| , dx \
&le varepsilon/4 + int_{(0,infty)backslash K} |f(x)| , dx
+int_{(0,infty)backslash K} |f_n(x)| , dx \
&le varepsilon/4 + varepsilon/4 + varepsilon/2 \
&le varepsilon,
end{align}$$
which proves convergence in $L^1(0,infty)$.
answered Dec 12 '18 at 18:03
BigbearZzzBigbearZzz
8,79421652
$endgroup$
Yes, it is possible to conclude that the convergence is in $L^1$ norm.
Given any $varepsilon>0$, there exists a compact set $K$ such that
$$
int_{(0,infty)backslash K} f(x), dx<varepsilon/4
$$
since $fin L^1(0,infty)$. On the other hand, uniform convergence on $K$, a bounded set, implies $L^1(K)$ convergence, hence $exists NinBbb N$ such that
$$
int_{ K} |f(x)-f_n(x)| , dx <varepsilon/4
$$
for all $nge N$. This also implies that for each $nge N$, we have
$$begin{align}
int_{ K} |f_n(x) |, dx
&ge
int_{ K} |f(x)|, dx
-int_{ K} |f(x)-f_n(x)| , dx \
&ge (1-varepsilon/4) - varepsilon/4 \
&= 1 - varepsilon/2,
end{align}$$
hence
$$
int_{(0,infty)backslash K} |f_n(x) |, dx le varepsilon/2.
$$
This means that for all $nge N$,
$$begin{align}
int_{(0,infty)} |f(x)-f_n(x)| , dx
&le
int_{K} |f(x)-f_n(x)| , dx + int_{(0,infty)backslash K} |f(x)-f_n(x)| , dx \
&le varepsilon/4 + int_{(0,infty)backslash K} |f(x)| , dx
+int_{(0,infty)backslash K} |f_n(x)| , dx \
&le varepsilon/4 + varepsilon/4 + varepsilon/2 \
&le varepsilon,
end{align}$$
which proves convergence in $L^1(0,infty)$.
answered Dec 12 '18 at 18:03
BigbearZzzBigbearZzz
8,79421652
answered Dec 12 '18 at 18:03
BigbearZzzBigbearZzz
8,79421652
answered Dec 12 '18 at 18:03
BigbearZzzBigbearZzz
8,79421652
answered Dec 12 '18 at 18:03
BigbearZzzBigbearZzz
8,79421652
8,79421652
add a comment |
add a comment |
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