Pointwise limit of the sequence of continuously differentiable functions defined inductively.
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Let $f_1 : [-1, 1] rightarrow mathbb{R};: f_1(0) = 0 $ be a continuously differentiable function and $lambda > 1$. Consider the sequence of functions defined inductively by $f_k(x) := lambda f_{k-1}left(dfrac{x}{lambda}right);: k geq 2;: xin [-1, 1]$. Find the pointwise limit of the sequence of functions $(f_n)$.
I figured out that $f_k(0)=0 :forall: kimplies f_krightarrow 0 $ when $ x=0$. Also $f_k$ is continuously differentiable $forall :k$.
$Bigg($Since $quaddisplaystylelim_{hrightarrow0}dfrac{f_2(x+h)-f_2(x)}{h}=lambdalim_{hrightarrow0}dfrac{f_1left(frac{x+h}{lambda}right)-f_1left(frac{x}{lambda}right)}{h}.: $Since$ f_1 $is continuously differentiable we see $ f_2 $ is differentiable and its derivative is continuous and hence $f_k$. $Bigg)$.
Further since $f_1(0)=0$. So I assumed $f_1(x)=xcdot g(x)$, where $g(x)$ is continuously differentiable and $g(0)neq0$. Assume $xneq0,:f_2(x)=lambdacdotdfrac{x}{lambda}gleft(dfrac{x}{lambda}right)=xcdot gleft(dfrac{x}{lambda}right)implies f_3(x)=xcdot gleft(dfrac{x}{lambda^2}right)$
$implies f_k(x)=xcdot gleft(dfrac{x}{lambda^{k-1}}right)$.
So taking limit as $krightarrow infty$, we get $displaystyle lim_{krightarrowinfty}f_k(x)=xcdot g(0)$. Now I am not sure how to proceed or whether there is something wrong here. Please provide hints or suggestions.
$rule{17cm}{1pt}$
To sum up (using the arguments made by $textbf{zhw}$ and $textbf{Paul Frost}$)
When $displaystyle xneq0, lim_{nrightarrowinfty}f_{n+1}(x) = lim_{nrightarrowinfty}x,frac{f_1(x/lambda^n)}{x/lambda^n} = lim_{nrightarrowinfty}x,frac{f_1(x/lambda^n)-f_1(0)}{x/lambda^n}=x:f'(0).$ Thus
$$f_nrightarrowbegin{cases}0&x=0\x:f'_1(0) &xneq0end{cases}=f $$
real-analysis derivatives recurrence-relations pointwise-convergence
asked Nov 19 at 17:59
Yadati Kiran
1,243417
add a comment |
up vote
0
down vote
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Let $f_1 : [-1, 1] rightarrow mathbb{R};: f_1(0) = 0 $ be a continuously differentiable function and $lambda > 1$. Consider the sequence of functions defined inductively by $f_k(x) := lambda f_{k-1}left(dfrac{x}{lambda}right);: k geq 2;: xin [-1, 1]$. Find the pointwise limit of the sequence of functions $(f_n)$.
I figured out that $f_k(0)=0 :forall: kimplies f_krightarrow 0 $ when $ x=0$. Also $f_k$ is continuously differentiable $forall :k$.
$Bigg($Since $quaddisplaystylelim_{hrightarrow0}dfrac{f_2(x+h)-f_2(x)}{h}=lambdalim_{hrightarrow0}dfrac{f_1left(frac{x+h}{lambda}right)-f_1left(frac{x}{lambda}right)}{h}.: $Since$ f_1 $is continuously differentiable we see $ f_2 $ is differentiable and its derivative is continuous and hence $f_k$. $Bigg)$.
Further since $f_1(0)=0$. So I assumed $f_1(x)=xcdot g(x)$, where $g(x)$ is continuously differentiable and $g(0)neq0$. Assume $xneq0,:f_2(x)=lambdacdotdfrac{x}{lambda}gleft(dfrac{x}{lambda}right)=xcdot gleft(dfrac{x}{lambda}right)implies f_3(x)=xcdot gleft(dfrac{x}{lambda^2}right)$
$implies f_k(x)=xcdot gleft(dfrac{x}{lambda^{k-1}}right)$.
So taking limit as $krightarrow infty$, we get $displaystyle lim_{krightarrowinfty}f_k(x)=xcdot g(0)$. Now I am not sure how to proceed or whether there is something wrong here. Please provide hints or suggestions.
$rule{17cm}{1pt}$
To sum up (using the arguments made by $textbf{zhw}$ and $textbf{Paul Frost}$)
When $displaystyle xneq0, lim_{nrightarrowinfty}f_{n+1}(x) = lim_{nrightarrowinfty}x,frac{f_1(x/lambda^n)}{x/lambda^n} = lim_{nrightarrowinfty}x,frac{f_1(x/lambda^n)-f_1(0)}{x/lambda^n}=x:f'(0).$ Thus
$$f_nrightarrowbegin{cases}0&x=0\x:f'_1(0) &xneq0end{cases}=f $$
real-analysis derivatives recurrence-relations pointwise-convergence
asked Nov 19 at 17:59
Yadati Kiran
1,243417
You do not need $g$ continuously differentiable. This in general not true (also $g(0) ne 0$ is in general not true). Your approach shows that it suffices to know that $g$ is continuous at $x = 0$.
– Paul Frost
Nov 19 at 22:48
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f_1 : [-1, 1] rightarrow mathbb{R};: f_1(0) = 0 $ be a continuously differentiable function and $lambda > 1$. Consider the sequence of functions defined inductively by $f_k(x) := lambda f_{k-1}left(dfrac{x}{lambda}right);: k geq 2;: xin [-1, 1]$. Find the pointwise limit of the sequence of functions $(f_n)$.
I figured out that $f_k(0)=0 :forall: kimplies f_krightarrow 0 $ when $ x=0$. Also $f_k$ is continuously differentiable $forall :k$.
$Bigg($Since $quaddisplaystylelim_{hrightarrow0}dfrac{f_2(x+h)-f_2(x)}{h}=lambdalim_{hrightarrow0}dfrac{f_1left(frac{x+h}{lambda}right)-f_1left(frac{x}{lambda}right)}{h}.: $Since$ f_1 $is continuously differentiable we see $ f_2 $ is differentiable and its derivative is continuous and hence $f_k$. $Bigg)$.
Further since $f_1(0)=0$. So I assumed $f_1(x)=xcdot g(x)$, where $g(x)$ is continuously differentiable and $g(0)neq0$. Assume $xneq0,:f_2(x)=lambdacdotdfrac{x}{lambda}gleft(dfrac{x}{lambda}right)=xcdot gleft(dfrac{x}{lambda}right)implies f_3(x)=xcdot gleft(dfrac{x}{lambda^2}right)$
$implies f_k(x)=xcdot gleft(dfrac{x}{lambda^{k-1}}right)$.
So taking limit as $krightarrow infty$, we get $displaystyle lim_{krightarrowinfty}f_k(x)=xcdot g(0)$. Now I am not sure how to proceed or whether there is something wrong here. Please provide hints or suggestions.
$rule{17cm}{1pt}$
To sum up (using the arguments made by $textbf{zhw}$ and $textbf{Paul Frost}$)
When $displaystyle xneq0, lim_{nrightarrowinfty}f_{n+1}(x) = lim_{nrightarrowinfty}x,frac{f_1(x/lambda^n)}{x/lambda^n} = lim_{nrightarrowinfty}x,frac{f_1(x/lambda^n)-f_1(0)}{x/lambda^n}=x:f'(0).$ Thus
$$f_nrightarrowbegin{cases}0&x=0\x:f'_1(0) &xneq0end{cases}=f $$
real-analysis derivatives recurrence-relations pointwise-convergence
asked Nov 19 at 17:59
Yadati Kiran
1,243417
Let $f_1 : [-1, 1] rightarrow mathbb{R};: f_1(0) = 0 $ be a continuously differentiable function and $lambda > 1$. Consider the sequence of functions defined inductively by $f_k(x) := lambda f_{k-1}left(dfrac{x}{lambda}right);: k geq 2;: xin [-1, 1]$. Find the pointwise limit of the sequence of functions $(f_n)$.
I figured out that $f_k(0)=0 :forall: kimplies f_krightarrow 0 $ when $ x=0$. Also $f_k$ is continuously differentiable $forall :k$.
$Bigg($Since $quaddisplaystylelim_{hrightarrow0}dfrac{f_2(x+h)-f_2(x)}{h}=lambdalim_{hrightarrow0}dfrac{f_1left(frac{x+h}{lambda}right)-f_1left(frac{x}{lambda}right)}{h}.: $Since$ f_1 $is continuously differentiable we see $ f_2 $ is differentiable and its derivative is continuous and hence $f_k$. $Bigg)$.
Further since $f_1(0)=0$. So I assumed $f_1(x)=xcdot g(x)$, where $g(x)$ is continuously differentiable and $g(0)neq0$. Assume $xneq0,:f_2(x)=lambdacdotdfrac{x}{lambda}gleft(dfrac{x}{lambda}right)=xcdot gleft(dfrac{x}{lambda}right)implies f_3(x)=xcdot gleft(dfrac{x}{lambda^2}right)$
$implies f_k(x)=xcdot gleft(dfrac{x}{lambda^{k-1}}right)$.
So taking limit as $krightarrow infty$, we get $displaystyle lim_{krightarrowinfty}f_k(x)=xcdot g(0)$. Now I am not sure how to proceed or whether there is something wrong here. Please provide hints or suggestions.
$rule{17cm}{1pt}$
To sum up (using the arguments made by $textbf{zhw}$ and $textbf{Paul Frost}$)
When $displaystyle xneq0, lim_{nrightarrowinfty}f_{n+1}(x) = lim_{nrightarrowinfty}x,frac{f_1(x/lambda^n)}{x/lambda^n} = lim_{nrightarrowinfty}x,frac{f_1(x/lambda^n)-f_1(0)}{x/lambda^n}=x:f'(0).$ Thus
$$f_nrightarrowbegin{cases}0&x=0\x:f'_1(0) &xneq0end{cases}=f $$
real-analysis derivatives recurrence-relations pointwise-convergence
real-analysis derivatives recurrence-relations pointwise-convergence
asked Nov 19 at 17:59
Yadati Kiran
1,243417
asked Nov 19 at 17:59
Yadati Kiran
1,243417
edited Nov 20 at 8:19
asked Nov 19 at 17:59
Yadati Kiran
1,243417
asked Nov 19 at 17:59
Yadati Kiran
1,243417
asked Nov 19 at 17:59
Yadati Kiran
1,243417
1,243417
You do not need $g$ continuously differentiable. This in general not true (also $g(0) ne 0$ is in general not true). Your approach shows that it suffices to know that $g$ is continuous at $x = 0$.
– Paul Frost
Nov 19 at 22:48
add a comment |
You do not need $g$ continuously differentiable. This in general not true (also $g(0) ne 0$ is in general not true). Your approach shows that it suffices to know that $g$ is continuous at $x = 0$.
– Paul Frost
Nov 19 at 22:48
You do not need $g$ continuously differentiable. This in general not true (also $g(0) ne 0$ is in general not true). Your approach shows that it suffices to know that $g$ is continuous at $x = 0$.
– Paul Frost
Nov 19 at 22:48
You do not need $g$ continuously differentiable. This in general not true (also $g(0) ne 0$ is in general not true). Your approach shows that it suffices to know that $g$ is continuous at $x = 0$.
– Paul Frost
Nov 19 at 22:48
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
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It looks to me like you have the right answer, but I don't think defining $g$ is necessary. Additionally, I think we need only assume $f_1'(0)$ exists.
Hint: First verify $f_{n+1}(x) = lambda^n f_1(x/lambda^n),$ which you can do by induction. Suppose $xin [-1,1]setminus{0}.$ Then
$$tag 1 f_{n+1}(x) = x,frac{f_1(x/lambda^n)}{x/lambda^n} = x,frac{f_1(x/lambda^n)-f_1(0)}{x/lambda^n}.$$
answered Nov 19 at 19:51
zhw.
70.7k43075
1
@YadatiKiran Your approach is very simliar to zhw's. For $x ne 0$ define $g(x) = frac{f_1(x)}{x} = frac{f_1(x) - f_1(0)}{x - 0}$. This is a continuous function on $[-1,1] setminus { 0 }$. But $f_1$ is differentiable at $x = 0$, hence $g$ has a unique continuous extension to $[-1,1]$ by taking $g(0) = f'_1(0)$.
– Paul Frost
Nov 19 at 22:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It looks to me like you have the right answer, but I don't think defining $g$ is necessary. Additionally, I think we need only assume $f_1'(0)$ exists.
Hint: First verify $f_{n+1}(x) = lambda^n f_1(x/lambda^n),$ which you can do by induction. Suppose $xin [-1,1]setminus{0}.$ Then
$$tag 1 f_{n+1}(x) = x,frac{f_1(x/lambda^n)}{x/lambda^n} = x,frac{f_1(x/lambda^n)-f_1(0)}{x/lambda^n}.$$
answered Nov 19 at 19:51
zhw.
70.7k43075
1
@YadatiKiran Your approach is very simliar to zhw's. For $x ne 0$ define $g(x) = frac{f_1(x)}{x} = frac{f_1(x) - f_1(0)}{x - 0}$. This is a continuous function on $[-1,1] setminus { 0 }$. But $f_1$ is differentiable at $x = 0$, hence $g$ has a unique continuous extension to $[-1,1]$ by taking $g(0) = f'_1(0)$.
– Paul Frost
Nov 19 at 22:44
add a comment |
up vote
1
down vote
accepted
It looks to me like you have the right answer, but I don't think defining $g$ is necessary. Additionally, I think we need only assume $f_1'(0)$ exists.
Hint: First verify $f_{n+1}(x) = lambda^n f_1(x/lambda^n),$ which you can do by induction. Suppose $xin [-1,1]setminus{0}.$ Then
$$tag 1 f_{n+1}(x) = x,frac{f_1(x/lambda^n)}{x/lambda^n} = x,frac{f_1(x/lambda^n)-f_1(0)}{x/lambda^n}.$$
answered Nov 19 at 19:51
zhw.
70.7k43075
1
@YadatiKiran Your approach is very simliar to zhw's. For $x ne 0$ define $g(x) = frac{f_1(x)}{x} = frac{f_1(x) - f_1(0)}{x - 0}$. This is a continuous function on $[-1,1] setminus { 0 }$. But $f_1$ is differentiable at $x = 0$, hence $g$ has a unique continuous extension to $[-1,1]$ by taking $g(0) = f'_1(0)$.
– Paul Frost
Nov 19 at 22:44
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It looks to me like you have the right answer, but I don't think defining $g$ is necessary. Additionally, I think we need only assume $f_1'(0)$ exists.
Hint: First verify $f_{n+1}(x) = lambda^n f_1(x/lambda^n),$ which you can do by induction. Suppose $xin [-1,1]setminus{0}.$ Then
$$tag 1 f_{n+1}(x) = x,frac{f_1(x/lambda^n)}{x/lambda^n} = x,frac{f_1(x/lambda^n)-f_1(0)}{x/lambda^n}.$$
answered Nov 19 at 19:51
zhw.
70.7k43075
It looks to me like you have the right answer, but I don't think defining $g$ is necessary. Additionally, I think we need only assume $f_1'(0)$ exists.
Hint: First verify $f_{n+1}(x) = lambda^n f_1(x/lambda^n),$ which you can do by induction. Suppose $xin [-1,1]setminus{0}.$ Then
$$tag 1 f_{n+1}(x) = x,frac{f_1(x/lambda^n)}{x/lambda^n} = x,frac{f_1(x/lambda^n)-f_1(0)}{x/lambda^n}.$$
answered Nov 19 at 19:51
zhw.
70.7k43075
edited Nov 19 at 19:57
answered Nov 19 at 19:51
zhw.
70.7k43075
answered Nov 19 at 19:51
zhw.
70.7k43075
answered Nov 19 at 19:51
zhw.
70.7k43075
70.7k43075
1
@YadatiKiran Your approach is very simliar to zhw's. For $x ne 0$ define $g(x) = frac{f_1(x)}{x} = frac{f_1(x) - f_1(0)}{x - 0}$. This is a continuous function on $[-1,1] setminus { 0 }$. But $f_1$ is differentiable at $x = 0$, hence $g$ has a unique continuous extension to $[-1,1]$ by taking $g(0) = f'_1(0)$.
– Paul Frost
Nov 19 at 22:44
add a comment |
1
@YadatiKiran Your approach is very simliar to zhw's. For $x ne 0$ define $g(x) = frac{f_1(x)}{x} = frac{f_1(x) - f_1(0)}{x - 0}$. This is a continuous function on $[-1,1] setminus { 0 }$. But $f_1$ is differentiable at $x = 0$, hence $g$ has a unique continuous extension to $[-1,1]$ by taking $g(0) = f'_1(0)$.
– Paul Frost
Nov 19 at 22:44
1
1
@YadatiKiran Your approach is very simliar to zhw's. For $x ne 0$ define $g(x) = frac{f_1(x)}{x} = frac{f_1(x) - f_1(0)}{x - 0}$. This is a continuous function on $[-1,1] setminus { 0 }$. But $f_1$ is differentiable at $x = 0$, hence $g$ has a unique continuous extension to $[-1,1]$ by taking $g(0) = f'_1(0)$.
– Paul Frost
Nov 19 at 22:44
@YadatiKiran Your approach is very simliar to zhw's. For $x ne 0$ define $g(x) = frac{f_1(x)}{x} = frac{f_1(x) - f_1(0)}{x - 0}$. This is a continuous function on $[-1,1] setminus { 0 }$. But $f_1$ is differentiable at $x = 0$, hence $g$ has a unique continuous extension to $[-1,1]$ by taking $g(0) = f'_1(0)$.
– Paul Frost
Nov 19 at 22:44
add a comment |
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You do not need $g$ continuously differentiable. This in general not true (also $g(0) ne 0$ is in general not true). Your approach shows that it suffices to know that $g$ is continuous at $x = 0$.
– Paul Frost
Nov 19 at 22:48