Reason for the integer case and the rational case to be solved differently
$begingroup$
Assume $f$ is continuous,$f(0)=1$ , and $f(m+n+1)=f(m)+f(n)$
for all real $m, n$.
Show that $f(x) = 1 + x$ for all real numbers $x$.
This is referenced from Terence Tao’s solving mathematical problems and in the exercise he provided a hint;
first prove this for integer $x$, then for rational $x$, then finally for real $x.$
The questions are as follows:
Why would there be a separate case to be considered for this question? Wouldn’t a direct method of solving suffice? Is there another way of approaching the question?
Any help would be much appreciated.
real-analysis functions continuity functional-equations
edited Dec 17 '18 at 16:13
user10354138
7,4322925
asked Dec 17 '18 at 15:38
299792458299792458
527
$endgroup$
add a comment |
$begingroup$
Assume $f$ is continuous,$f(0)=1$ , and $f(m+n+1)=f(m)+f(n)$
for all real $m, n$.
Show that $f(x) = 1 + x$ for all real numbers $x$.
This is referenced from Terence Tao’s solving mathematical problems and in the exercise he provided a hint;
first prove this for integer $x$, then for rational $x$, then finally for real $x.$
The questions are as follows:
Why would there be a separate case to be considered for this question? Wouldn’t a direct method of solving suffice? Is there another way of approaching the question?
Any help would be much appreciated.
real-analysis functions continuity functional-equations
edited Dec 17 '18 at 16:13
user10354138
7,4322925
asked Dec 17 '18 at 15:38
299792458299792458
527
$endgroup$
$begingroup$
A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
$endgroup$
– the_fox
Dec 17 '18 at 16:19
$begingroup$
the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
$endgroup$
– Conrad
Dec 17 '18 at 16:24
$begingroup$
Maybe exists a direct method. But surely the solution using the hint is easier.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 17:36
$begingroup$
But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
$endgroup$
– 299792458
Dec 17 '18 at 17:40
$begingroup$
Because the integer case is used in the proof of rational case?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 20:17
add a comment |
$begingroup$
Assume $f$ is continuous,$f(0)=1$ , and $f(m+n+1)=f(m)+f(n)$
for all real $m, n$.
Show that $f(x) = 1 + x$ for all real numbers $x$.
This is referenced from Terence Tao’s solving mathematical problems and in the exercise he provided a hint;
first prove this for integer $x$, then for rational $x$, then finally for real $x.$
The questions are as follows:
Why would there be a separate case to be considered for this question? Wouldn’t a direct method of solving suffice? Is there another way of approaching the question?
Any help would be much appreciated.
real-analysis functions continuity functional-equations
edited Dec 17 '18 at 16:13
user10354138
7,4322925
asked Dec 17 '18 at 15:38
299792458299792458
527
$endgroup$
Assume $f$ is continuous,$f(0)=1$ , and $f(m+n+1)=f(m)+f(n)$
for all real $m, n$.
Show that $f(x) = 1 + x$ for all real numbers $x$.
This is referenced from Terence Tao’s solving mathematical problems and in the exercise he provided a hint;
first prove this for integer $x$, then for rational $x$, then finally for real $x.$
The questions are as follows:
Why would there be a separate case to be considered for this question? Wouldn’t a direct method of solving suffice? Is there another way of approaching the question?
Any help would be much appreciated.
real-analysis functions continuity functional-equations
real-analysis functions continuity functional-equations
edited Dec 17 '18 at 16:13
user10354138
7,4322925
asked Dec 17 '18 at 15:38
299792458299792458
527
edited Dec 17 '18 at 16:13
user10354138
7,4322925
asked Dec 17 '18 at 15:38
299792458299792458
527
edited Dec 17 '18 at 16:13
user10354138
7,4322925
edited Dec 17 '18 at 16:13
user10354138
7,4322925
edited Dec 17 '18 at 16:13
user10354138
7,4322925
7,4322925
asked Dec 17 '18 at 15:38
299792458299792458
527
asked Dec 17 '18 at 15:38
299792458299792458
527
asked Dec 17 '18 at 15:38
299792458299792458
527
527
$begingroup$
A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
$endgroup$
– the_fox
Dec 17 '18 at 16:19
$begingroup$
the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
$endgroup$
– Conrad
Dec 17 '18 at 16:24
$begingroup$
Maybe exists a direct method. But surely the solution using the hint is easier.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 17:36
$begingroup$
But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
$endgroup$
– 299792458
Dec 17 '18 at 17:40
$begingroup$
Because the integer case is used in the proof of rational case?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 20:17
add a comment |
$begingroup$
A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
$endgroup$
– the_fox
Dec 17 '18 at 16:19
$begingroup$
the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
$endgroup$
– Conrad
Dec 17 '18 at 16:24
$begingroup$
Maybe exists a direct method. But surely the solution using the hint is easier.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 17:36
$begingroup$
But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
$endgroup$
– 299792458
Dec 17 '18 at 17:40
$begingroup$
Because the integer case is used in the proof of rational case?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 20:17
$begingroup$
A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
$endgroup$
– the_fox
Dec 17 '18 at 16:19
$begingroup$
A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
$endgroup$
– the_fox
Dec 17 '18 at 16:19
$begingroup$
the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
$endgroup$
– Conrad
Dec 17 '18 at 16:24
$begingroup$
the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
$endgroup$
– Conrad
Dec 17 '18 at 16:24
$begingroup$
Maybe exists a direct method. But surely the solution using the hint is easier.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 17:36
$begingroup$
Maybe exists a direct method. But surely the solution using the hint is easier.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 17:36
$begingroup$
But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
$endgroup$
– 299792458
Dec 17 '18 at 17:40
$begingroup$
But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
$endgroup$
– 299792458
Dec 17 '18 at 17:40
$begingroup$
Because the integer case is used in the proof of rational case?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 20:17
$begingroup$
Because the integer case is used in the proof of rational case?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 17 '18 at 20:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.
Continous Additive functions are linear
Coming back to your Orignal Problem
Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function.
Keep it aside for a while.
By the functional equation given $f(y)=f({color{red}{0}}+{color{blue}{y-1}}+1)=f({color{red}{0}})+f({color{blue}{y-1}})$
Hence $f(y)=1+f(y-1)$
Further consider $f(x+y)$
$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}}+{color{blue}{y-1}}+1)$$
$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}})+f({color{blue}{y-1}})$$
Hence we have $f(x+y)=f(x)+f(y)-1$
This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$
Hence $g(x+y)=g(x)+g(y)$
Hence $g(x)=g(1)x$
Which implies $f(x)-1=g(1)x$
Hence $f(x)=1+(f(x)-1)x$
Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 times 1=2$
We conclude $f(x)=1+x$
You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above
answered Dec 18 '18 at 4:26
Rakesh BhattRakesh Bhatt
967214
$endgroup$
$begingroup$
But why $2f(0)=2.1$?
$endgroup$
– 299792458
Dec 19 '18 at 10:22
$begingroup$
@299792458 because $f(0)=1$ I'd given in the question.
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 3:38
$begingroup$
In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
$endgroup$
– 299792458
Dec 23 '18 at 4:18
1
$begingroup$
You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
$endgroup$
– Milo Brandt
Dec 23 '18 at 5:29
$begingroup$
@MiloBrandt May Be. I will inform him about this
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 5:48
|
show 2 more comments
$begingroup$
You need the integer case to prove the rational case. How else do you get a handle on $f(frac 12)?$ If we take $n=0$ we are told $f(m+1)=f(m)+1$ from which we can derive that $f(x)=x+1$ for the integers.
Once we have that, we can substitute in $m=n=frac 12$ and get $f(2)=2f(frac 12)$. Since we know $f(2)=3$ because we solved the integer case already, we get $f(frac 12)=frac 32$. We can continue this approach to get the function on all the dyadic rationals. Of course this does not prove that there is not a direct approach
Since the dyadics are dense, we can then extend to all the reals. This is where continuity comes in. Clearly the integers do not suffice for this because they are not dense. As Conrad comments, if you remove the continuity restriction there are other solutions. You need a dense set for continuity to work on. That shows you need to compute it on a dense set before you cover all the reals.
answered Dec 18 '18 at 4:42
Ross MillikanRoss Millikan
299k24200374
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.
Continous Additive functions are linear
Coming back to your Orignal Problem
Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function.
Keep it aside for a while.
By the functional equation given $f(y)=f({color{red}{0}}+{color{blue}{y-1}}+1)=f({color{red}{0}})+f({color{blue}{y-1}})$
Hence $f(y)=1+f(y-1)$
Further consider $f(x+y)$
$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}}+{color{blue}{y-1}}+1)$$
$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}})+f({color{blue}{y-1}})$$
Hence we have $f(x+y)=f(x)+f(y)-1$
This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$
Hence $g(x+y)=g(x)+g(y)$
Hence $g(x)=g(1)x$
Which implies $f(x)-1=g(1)x$
Hence $f(x)=1+(f(x)-1)x$
Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 times 1=2$
We conclude $f(x)=1+x$
You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above
answered Dec 18 '18 at 4:26
Rakesh BhattRakesh Bhatt
967214
$endgroup$
$begingroup$
But why $2f(0)=2.1$?
$endgroup$
– 299792458
Dec 19 '18 at 10:22
$begingroup$
@299792458 because $f(0)=1$ I'd given in the question.
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 3:38
$begingroup$
In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
$endgroup$
– 299792458
Dec 23 '18 at 4:18
1
$begingroup$
You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
$endgroup$
– Milo Brandt
Dec 23 '18 at 5:29
$begingroup$
@MiloBrandt May Be. I will inform him about this
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 5:48
|
show 2 more comments
$begingroup$
Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.
Continous Additive functions are linear
Coming back to your Orignal Problem
Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function.
Keep it aside for a while.
By the functional equation given $f(y)=f({color{red}{0}}+{color{blue}{y-1}}+1)=f({color{red}{0}})+f({color{blue}{y-1}})$
Hence $f(y)=1+f(y-1)$
Further consider $f(x+y)$
$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}}+{color{blue}{y-1}}+1)$$
$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}})+f({color{blue}{y-1}})$$
Hence we have $f(x+y)=f(x)+f(y)-1$
This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$
Hence $g(x+y)=g(x)+g(y)$
Hence $g(x)=g(1)x$
Which implies $f(x)-1=g(1)x$
Hence $f(x)=1+(f(x)-1)x$
Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 times 1=2$
We conclude $f(x)=1+x$
You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above
answered Dec 18 '18 at 4:26
Rakesh BhattRakesh Bhatt
967214
$endgroup$
$begingroup$
But why $2f(0)=2.1$?
$endgroup$
– 299792458
Dec 19 '18 at 10:22
$begingroup$
@299792458 because $f(0)=1$ I'd given in the question.
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 3:38
$begingroup$
In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
$endgroup$
– 299792458
Dec 23 '18 at 4:18
1
$begingroup$
You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
$endgroup$
– Milo Brandt
Dec 23 '18 at 5:29
$begingroup$
@MiloBrandt May Be. I will inform him about this
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 5:48
|
show 2 more comments
$begingroup$
Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.
Continous Additive functions are linear
Coming back to your Orignal Problem
Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function.
Keep it aside for a while.
By the functional equation given $f(y)=f({color{red}{0}}+{color{blue}{y-1}}+1)=f({color{red}{0}})+f({color{blue}{y-1}})$
Hence $f(y)=1+f(y-1)$
Further consider $f(x+y)$
$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}}+{color{blue}{y-1}}+1)$$
$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}})+f({color{blue}{y-1}})$$
Hence we have $f(x+y)=f(x)+f(y)-1$
This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$
Hence $g(x+y)=g(x)+g(y)$
Hence $g(x)=g(1)x$
Which implies $f(x)-1=g(1)x$
Hence $f(x)=1+(f(x)-1)x$
Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 times 1=2$
We conclude $f(x)=1+x$
You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above
answered Dec 18 '18 at 4:26
Rakesh BhattRakesh Bhatt
967214
$endgroup$
Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.
Continous Additive functions are linear
Coming back to your Orignal Problem
Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function.
Keep it aside for a while.
By the functional equation given $f(y)=f({color{red}{0}}+{color{blue}{y-1}}+1)=f({color{red}{0}})+f({color{blue}{y-1}})$
Hence $f(y)=1+f(y-1)$
Further consider $f(x+y)$
$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}}+{color{blue}{y-1}}+1)$$
$$f({color{red}{x}}+{color{blue}{y}})=f({color{red}{x}})+f({color{blue}{y-1}})$$
Hence we have $f(x+y)=f(x)+f(y)-1$
This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$
Hence $g(x+y)=g(x)+g(y)$
Hence $g(x)=g(1)x$
Which implies $f(x)-1=g(1)x$
Hence $f(x)=1+(f(x)-1)x$
Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 times 1=2$
We conclude $f(x)=1+x$
You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above
answered Dec 18 '18 at 4:26
Rakesh BhattRakesh Bhatt
967214
edited Dec 23 '18 at 5:47
answered Dec 18 '18 at 4:26
Rakesh BhattRakesh Bhatt
967214
answered Dec 18 '18 at 4:26
Rakesh BhattRakesh Bhatt
967214
answered Dec 18 '18 at 4:26
Rakesh BhattRakesh Bhatt
967214
967214
$begingroup$
But why $2f(0)=2.1$?
$endgroup$
– 299792458
Dec 19 '18 at 10:22
$begingroup$
@299792458 because $f(0)=1$ I'd given in the question.
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 3:38
$begingroup$
In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
$endgroup$
– 299792458
Dec 23 '18 at 4:18
1
$begingroup$
You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
$endgroup$
– Milo Brandt
Dec 23 '18 at 5:29
$begingroup$
@MiloBrandt May Be. I will inform him about this
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 5:48
|
show 2 more comments
$begingroup$
But why $2f(0)=2.1$?
$endgroup$
– 299792458
Dec 19 '18 at 10:22
$begingroup$
@299792458 because $f(0)=1$ I'd given in the question.
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 3:38
$begingroup$
In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
$endgroup$
– 299792458
Dec 23 '18 at 4:18
1
$begingroup$
You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
$endgroup$
– Milo Brandt
Dec 23 '18 at 5:29
$begingroup$
@MiloBrandt May Be. I will inform him about this
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 5:48
$begingroup$
But why $2f(0)=2.1$?
$endgroup$
– 299792458
Dec 19 '18 at 10:22
$begingroup$
But why $2f(0)=2.1$?
$endgroup$
– 299792458
Dec 19 '18 at 10:22
$begingroup$
@299792458 because $f(0)=1$ I'd given in the question.
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 3:38
$begingroup$
@299792458 because $f(0)=1$ I'd given in the question.
$endgroup$
– Rakesh Bhatt
Dec 23 '18 at 3:38
$begingroup$
In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
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– 299792458
Dec 23 '18 at 4:18
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In that case, should it directly be $2f(0) =2$? Why would there be a transition step to 2.1?
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– 299792458
Dec 23 '18 at 4:18
1
1
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You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
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– Milo Brandt
Dec 23 '18 at 5:29
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You can make this method even faster: Define $g(x)=f(x-1)$. Then, $g(x)+g(y)=f(x-1)+f(y-1)=f(x+y-1)=g(x+y)$. (Also, I think the confusion of the OP is that $2.1$ looks like a decimal for $frac{21}{10}$ rather than $2cdot 1$)
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– Milo Brandt
Dec 23 '18 at 5:29
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@MiloBrandt May Be. I will inform him about this
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– Rakesh Bhatt
Dec 23 '18 at 5:48
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@MiloBrandt May Be. I will inform him about this
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– Rakesh Bhatt
Dec 23 '18 at 5:48
|
show 2 more comments
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You need the integer case to prove the rational case. How else do you get a handle on $f(frac 12)?$ If we take $n=0$ we are told $f(m+1)=f(m)+1$ from which we can derive that $f(x)=x+1$ for the integers.
Once we have that, we can substitute in $m=n=frac 12$ and get $f(2)=2f(frac 12)$. Since we know $f(2)=3$ because we solved the integer case already, we get $f(frac 12)=frac 32$. We can continue this approach to get the function on all the dyadic rationals. Of course this does not prove that there is not a direct approach
Since the dyadics are dense, we can then extend to all the reals. This is where continuity comes in. Clearly the integers do not suffice for this because they are not dense. As Conrad comments, if you remove the continuity restriction there are other solutions. You need a dense set for continuity to work on. That shows you need to compute it on a dense set before you cover all the reals.
answered Dec 18 '18 at 4:42
Ross MillikanRoss Millikan
299k24200374
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add a comment |
$begingroup$
You need the integer case to prove the rational case. How else do you get a handle on $f(frac 12)?$ If we take $n=0$ we are told $f(m+1)=f(m)+1$ from which we can derive that $f(x)=x+1$ for the integers.
Once we have that, we can substitute in $m=n=frac 12$ and get $f(2)=2f(frac 12)$. Since we know $f(2)=3$ because we solved the integer case already, we get $f(frac 12)=frac 32$. We can continue this approach to get the function on all the dyadic rationals. Of course this does not prove that there is not a direct approach
Since the dyadics are dense, we can then extend to all the reals. This is where continuity comes in. Clearly the integers do not suffice for this because they are not dense. As Conrad comments, if you remove the continuity restriction there are other solutions. You need a dense set for continuity to work on. That shows you need to compute it on a dense set before you cover all the reals.
answered Dec 18 '18 at 4:42
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
You need the integer case to prove the rational case. How else do you get a handle on $f(frac 12)?$ If we take $n=0$ we are told $f(m+1)=f(m)+1$ from which we can derive that $f(x)=x+1$ for the integers.
Once we have that, we can substitute in $m=n=frac 12$ and get $f(2)=2f(frac 12)$. Since we know $f(2)=3$ because we solved the integer case already, we get $f(frac 12)=frac 32$. We can continue this approach to get the function on all the dyadic rationals. Of course this does not prove that there is not a direct approach
Since the dyadics are dense, we can then extend to all the reals. This is where continuity comes in. Clearly the integers do not suffice for this because they are not dense. As Conrad comments, if you remove the continuity restriction there are other solutions. You need a dense set for continuity to work on. That shows you need to compute it on a dense set before you cover all the reals.
answered Dec 18 '18 at 4:42
Ross MillikanRoss Millikan
299k24200374
$endgroup$
You need the integer case to prove the rational case. How else do you get a handle on $f(frac 12)?$ If we take $n=0$ we are told $f(m+1)=f(m)+1$ from which we can derive that $f(x)=x+1$ for the integers.
Once we have that, we can substitute in $m=n=frac 12$ and get $f(2)=2f(frac 12)$. Since we know $f(2)=3$ because we solved the integer case already, we get $f(frac 12)=frac 32$. We can continue this approach to get the function on all the dyadic rationals. Of course this does not prove that there is not a direct approach
Since the dyadics are dense, we can then extend to all the reals. This is where continuity comes in. Clearly the integers do not suffice for this because they are not dense. As Conrad comments, if you remove the continuity restriction there are other solutions. You need a dense set for continuity to work on. That shows you need to compute it on a dense set before you cover all the reals.
answered Dec 18 '18 at 4:42
Ross MillikanRoss Millikan
299k24200374
answered Dec 18 '18 at 4:42
Ross MillikanRoss Millikan
299k24200374
answered Dec 18 '18 at 4:42
Ross MillikanRoss Millikan
299k24200374
answered Dec 18 '18 at 4:42
Ross MillikanRoss Millikan
299k24200374
299k24200374
add a comment |
add a comment |
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A direct method would suffice, of course, provided that you can find one! Do you have any ideas?
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– the_fox
Dec 17 '18 at 16:19
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the problem with trying to find a direct method is that you need to use continuity in an essential way since there are standard counterexamples for such an f without some regularity conditions (local boundness at one point is the weakest I know), as g+1 satisfies the above for any additive function g which is the identity on the rationals
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– Conrad
Dec 17 '18 at 16:24
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Maybe exists a direct method. But surely the solution using the hint is easier.
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– Martín-Blas Pérez Pinilla
Dec 17 '18 at 17:36
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But even with the solution proposed by Tao, I still do not see why there must be a rational case before an integer one?
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– 299792458
Dec 17 '18 at 17:40
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Because the integer case is used in the proof of rational case?
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– Martín-Blas Pérez Pinilla
Dec 17 '18 at 20:17