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Show that $displaystylelim_x rightarrow 0^+ f(x) = f(0)$
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Show that if $f$ is bounded function on $[0;1]$ and satisfying
$$f(ax) = b f(x)$$
with $0 leq x leq frac1a$ and $a,b > 1$ then
$$displaystylelim_x rightarrow 0^+ f(x) = f(0)$$
Solution:
There is exist $M geq 0$ such that $|f(x)| geq M$ with $x in (0;1)$. From $f(ax) = b f(x)$, $x in [0;frac1a]$ implies $f(a^2 x) = b^2 f(x)$ with $x in [0;frac1a^2]$. Use induction on $n$, we have
$$f(a^n x) = b^n f(x),quad with quad x in left[0;frac1a^nright], n in mathbbN$$
So,
$$|f(x)| leq M frac1b^n, quad with quad x in left[0; frac1a^nright], n in mathbbN qquad (*)$$
Other way, from $f(ax) = bf(x)$ implies $f(0)=0$. Associate with $(*)$ we have Q.E.D.
I do not understand why we conclude that $|f(x)| leq M frac1b^n$ after use the induction. Could you explain this to me. Thanks all!
real-analysis limits
asked Jul 25 at 6:55
Minh
727
add a comment |Â
up vote
1
down vote
favorite
Show that if $f$ is bounded function on $[0;1]$ and satisfying
$$f(ax) = b f(x)$$
with $0 leq x leq frac1a$ and $a,b > 1$ then
$$displaystylelim_x rightarrow 0^+ f(x) = f(0)$$
Solution:
There is exist $M geq 0$ such that $|f(x)| geq M$ with $x in (0;1)$. From $f(ax) = b f(x)$, $x in [0;frac1a]$ implies $f(a^2 x) = b^2 f(x)$ with $x in [0;frac1a^2]$. Use induction on $n$, we have
$$f(a^n x) = b^n f(x),quad with quad x in left[0;frac1a^nright], n in mathbbN$$
So,
$$|f(x)| leq M frac1b^n, quad with quad x in left[0; frac1a^nright], n in mathbbN qquad (*)$$
Other way, from $f(ax) = bf(x)$ implies $f(0)=0$. Associate with $(*)$ we have Q.E.D.
I do not understand why we conclude that $|f(x)| leq M frac1b^n$ after use the induction. Could you explain this to me. Thanks all!
real-analysis limits
asked Jul 25 at 6:55
Minh
727
1
There exists $M ge 0$ such that $lvert f(x) rvert colorredle M$ for $x in [0,1]$.
– user539887
Jul 25 at 7:07
I have a similar idea with you. This solution have some error.
– Minh
Jul 25 at 7:12
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that if $f$ is bounded function on $[0;1]$ and satisfying
$$f(ax) = b f(x)$$
with $0 leq x leq frac1a$ and $a,b > 1$ then
$$displaystylelim_x rightarrow 0^+ f(x) = f(0)$$
Solution:
There is exist $M geq 0$ such that $|f(x)| geq M$ with $x in (0;1)$. From $f(ax) = b f(x)$, $x in [0;frac1a]$ implies $f(a^2 x) = b^2 f(x)$ with $x in [0;frac1a^2]$. Use induction on $n$, we have
$$f(a^n x) = b^n f(x),quad with quad x in left[0;frac1a^nright], n in mathbbN$$
So,
$$|f(x)| leq M frac1b^n, quad with quad x in left[0; frac1a^nright], n in mathbbN qquad (*)$$
Other way, from $f(ax) = bf(x)$ implies $f(0)=0$. Associate with $(*)$ we have Q.E.D.
I do not understand why we conclude that $|f(x)| leq M frac1b^n$ after use the induction. Could you explain this to me. Thanks all!
real-analysis limits
asked Jul 25 at 6:55
Minh
727
Show that if $f$ is bounded function on $[0;1]$ and satisfying
$$f(ax) = b f(x)$$
with $0 leq x leq frac1a$ and $a,b > 1$ then
$$displaystylelim_x rightarrow 0^+ f(x) = f(0)$$
Solution:
There is exist $M geq 0$ such that $|f(x)| geq M$ with $x in (0;1)$. From $f(ax) = b f(x)$, $x in [0;frac1a]$ implies $f(a^2 x) = b^2 f(x)$ with $x in [0;frac1a^2]$. Use induction on $n$, we have
$$f(a^n x) = b^n f(x),quad with quad x in left[0;frac1a^nright], n in mathbbN$$
So,
$$|f(x)| leq M frac1b^n, quad with quad x in left[0; frac1a^nright], n in mathbbN qquad (*)$$
Other way, from $f(ax) = bf(x)$ implies $f(0)=0$. Associate with $(*)$ we have Q.E.D.
I do not understand why we conclude that $|f(x)| leq M frac1b^n$ after use the induction. Could you explain this to me. Thanks all!
real-analysis limits
asked Jul 25 at 6:55
Minh
727
asked Jul 25 at 6:55
Minh
727
asked Jul 25 at 6:55
Minh
727
asked Jul 25 at 6:55
Minh
727
727
1
There exists $M ge 0$ such that $lvert f(x) rvert colorredle M$ for $x in [0,1]$.
– user539887
Jul 25 at 7:07
I have a similar idea with you. This solution have some error.
– Minh
Jul 25 at 7:12
add a comment |Â
1
There exists $M ge 0$ such that $lvert f(x) rvert colorredle M$ for $x in [0,1]$.
– user539887
Jul 25 at 7:07
I have a similar idea with you. This solution have some error.
– Minh
Jul 25 at 7:12
1
1
There exists $M ge 0$ such that $lvert f(x) rvert colorredle M$ for $x in [0,1]$.
– user539887
Jul 25 at 7:07
There exists $M ge 0$ such that $lvert f(x) rvert colorredle M$ for $x in [0,1]$.
– user539887
Jul 25 at 7:07
I have a similar idea with you. This solution have some error.
– Minh
Jul 25 at 7:12
I have a similar idea with you. This solution have some error.
– Minh
Jul 25 at 7:12
add a comment |Â
1 Answer
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1
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As you said, $f(a^nx) = b^n f(x)$ with $x in [0, 1/a^n]$. So you can also write $f(x) = fracf(a^nx)b^n$, and then by upper bounding $f(a^nx) leq M$ you get your result, i.e. :
$$|f(x)| = left|fracf(a^nx)b^nright| = fracf(a^nx)rightb^n leq fracMb^n$$
answered Jul 25 at 7:14
ippiki-ookami
308216
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
As you said, $f(a^nx) = b^n f(x)$ with $x in [0, 1/a^n]$. So you can also write $f(x) = fracf(a^nx)b^n$, and then by upper bounding $f(a^nx) leq M$ you get your result, i.e. :
$$|f(x)| = left|fracf(a^nx)b^nright| = fracf(a^nx)rightb^n leq fracMb^n$$
answered Jul 25 at 7:14
ippiki-ookami
308216
add a comment |Â
up vote
1
down vote
As you said, $f(a^nx) = b^n f(x)$ with $x in [0, 1/a^n]$. So you can also write $f(x) = fracf(a^nx)b^n$, and then by upper bounding $f(a^nx) leq M$ you get your result, i.e. :
$$|f(x)| = left|fracf(a^nx)b^nright| = fracf(a^nx)rightb^n leq fracMb^n$$
answered Jul 25 at 7:14
ippiki-ookami
308216
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As you said, $f(a^nx) = b^n f(x)$ with $x in [0, 1/a^n]$. So you can also write $f(x) = fracf(a^nx)b^n$, and then by upper bounding $f(a^nx) leq M$ you get your result, i.e. :
$$|f(x)| = left|fracf(a^nx)b^nright| = fracf(a^nx)rightb^n leq fracMb^n$$
answered Jul 25 at 7:14
ippiki-ookami
308216
As you said, $f(a^nx) = b^n f(x)$ with $x in [0, 1/a^n]$. So you can also write $f(x) = fracf(a^nx)b^n$, and then by upper bounding $f(a^nx) leq M$ you get your result, i.e. :
$$|f(x)| = left|fracf(a^nx)b^nright| = fracf(a^nx)rightb^n leq fracMb^n$$
answered Jul 25 at 7:14
ippiki-ookami
308216
answered Jul 25 at 7:14
ippiki-ookami
308216
answered Jul 25 at 7:14
ippiki-ookami
308216
answered Jul 25 at 7:14
ippiki-ookami
308216
308216
add a comment |Â
add a comment |Â
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1
There exists $M ge 0$ such that $lvert f(x) rvert colorredle M$ for $x in [0,1]$.
– user539887
Jul 25 at 7:07
I have a similar idea with you. This solution have some error.
– Minh
Jul 25 at 7:12