Prove: $f$ is a constant function, as $forall a in mathbb{R} exists lim_{x to a}f(x) and forall n :...
$begingroup$
Given $f: mathbb{R} to mathbb{R}$ , for every $a in mathbb{R}$ there exists $lim_{x to a}f(x)$.
Also. for every $n in mathbb{N}: frac{1}{n}$ is a period of $f$ such that:
$$forall x in mathbb{R}, forall n in mathbb{N}: f(x) = f(x+ frac{1}{n})$$
Provedisprove: $f$ is a constant function.
my attempt:
I understand that you need to divide the proof for two cases, the first for $x in mathbb{R/Q}$ and the other for $x in mathbb{Q}$.
I can't understand how to formal this ideas.
real-analysis calculus limits functions
edited Dec 16 '18 at 12:50
Rebellos
14.5k31246
asked Dec 16 '18 at 12:43
JnevenJneven
783320
$endgroup$
add a comment |
$begingroup$
Given $f: mathbb{R} to mathbb{R}$ , for every $a in mathbb{R}$ there exists $lim_{x to a}f(x)$.
Also. for every $n in mathbb{N}: frac{1}{n}$ is a period of $f$ such that:
$$forall x in mathbb{R}, forall n in mathbb{N}: f(x) = f(x+ frac{1}{n})$$
Provedisprove: $f$ is a constant function.
my attempt:
I understand that you need to divide the proof for two cases, the first for $x in mathbb{R/Q}$ and the other for $x in mathbb{Q}$.
I can't understand how to formal this ideas.
real-analysis calculus limits functions
edited Dec 16 '18 at 12:50
Rebellos
14.5k31246
asked Dec 16 '18 at 12:43
JnevenJneven
783320
$endgroup$
add a comment |
$begingroup$
Given $f: mathbb{R} to mathbb{R}$ , for every $a in mathbb{R}$ there exists $lim_{x to a}f(x)$.
Also. for every $n in mathbb{N}: frac{1}{n}$ is a period of $f$ such that:
$$forall x in mathbb{R}, forall n in mathbb{N}: f(x) = f(x+ frac{1}{n})$$
Provedisprove: $f$ is a constant function.
my attempt:
I understand that you need to divide the proof for two cases, the first for $x in mathbb{R/Q}$ and the other for $x in mathbb{Q}$.
I can't understand how to formal this ideas.
real-analysis calculus limits functions
edited Dec 16 '18 at 12:50
Rebellos
14.5k31246
asked Dec 16 '18 at 12:43
JnevenJneven
783320
$endgroup$
Given $f: mathbb{R} to mathbb{R}$ , for every $a in mathbb{R}$ there exists $lim_{x to a}f(x)$.
Also. for every $n in mathbb{N}: frac{1}{n}$ is a period of $f$ such that:
$$forall x in mathbb{R}, forall n in mathbb{N}: f(x) = f(x+ frac{1}{n})$$
Provedisprove: $f$ is a constant function.
my attempt:
I understand that you need to divide the proof for two cases, the first for $x in mathbb{R/Q}$ and the other for $x in mathbb{Q}$.
I can't understand how to formal this ideas.
real-analysis calculus limits functions
real-analysis calculus limits functions
edited Dec 16 '18 at 12:50
Rebellos
14.5k31246
asked Dec 16 '18 at 12:43
JnevenJneven
783320
edited Dec 16 '18 at 12:50
Rebellos
14.5k31246
asked Dec 16 '18 at 12:43
JnevenJneven
783320
edited Dec 16 '18 at 12:50
Rebellos
14.5k31246
edited Dec 16 '18 at 12:50
Rebellos
14.5k31246
edited Dec 16 '18 at 12:50
Rebellos
14.5k31246
14.5k31246
asked Dec 16 '18 at 12:43
JnevenJneven
783320
asked Dec 16 '18 at 12:43
JnevenJneven
783320
asked Dec 16 '18 at 12:43
JnevenJneven
783320
783320
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Simply observe that
$$
f(a) = lim_{ntoinfty} f(a+frac{1}{n}) = lim_{xto a} f(x).$$ Therefore, $f$ is continuous at every $ainmathbb{R}$. Observe that by the assumption, it holds that
$$
f(0) = f(frac{j}{n}),quad forall ninmathbb{N},jin mathbb{Z},
$$ and hence that
$$
f(q) = f(0),quadforall qinmathbb{Q}.
$$ By the continuity of $f$, it follows that
$$f(x) = f(0),quadforall xinmathbb{R}.$$
answered Dec 16 '18 at 12:48
SongSong
8,699625
$endgroup$
$begingroup$
Oops, I misread it. But about $f(a) = f(a=frac{1}{n})$, it follows directly from the assumption.
$endgroup$
– Song
Dec 16 '18 at 12:53
$begingroup$
@Rebellos The first equality is obvious from the period 1/n. Fixing continuous f on a dense set determines the whole function.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 16 '18 at 12:55
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Simply observe that
$$
f(a) = lim_{ntoinfty} f(a+frac{1}{n}) = lim_{xto a} f(x).$$ Therefore, $f$ is continuous at every $ainmathbb{R}$. Observe that by the assumption, it holds that
$$
f(0) = f(frac{j}{n}),quad forall ninmathbb{N},jin mathbb{Z},
$$ and hence that
$$
f(q) = f(0),quadforall qinmathbb{Q}.
$$ By the continuity of $f$, it follows that
$$f(x) = f(0),quadforall xinmathbb{R}.$$
answered Dec 16 '18 at 12:48
SongSong
8,699625
$endgroup$
$begingroup$
Oops, I misread it. But about $f(a) = f(a=frac{1}{n})$, it follows directly from the assumption.
$endgroup$
– Song
Dec 16 '18 at 12:53
$begingroup$
@Rebellos The first equality is obvious from the period 1/n. Fixing continuous f on a dense set determines the whole function.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 16 '18 at 12:55
add a comment |
$begingroup$
Simply observe that
$$
f(a) = lim_{ntoinfty} f(a+frac{1}{n}) = lim_{xto a} f(x).$$ Therefore, $f$ is continuous at every $ainmathbb{R}$. Observe that by the assumption, it holds that
$$
f(0) = f(frac{j}{n}),quad forall ninmathbb{N},jin mathbb{Z},
$$ and hence that
$$
f(q) = f(0),quadforall qinmathbb{Q}.
$$ By the continuity of $f$, it follows that
$$f(x) = f(0),quadforall xinmathbb{R}.$$
answered Dec 16 '18 at 12:48
SongSong
8,699625
$endgroup$
$begingroup$
Oops, I misread it. But about $f(a) = f(a=frac{1}{n})$, it follows directly from the assumption.
$endgroup$
– Song
Dec 16 '18 at 12:53
$begingroup$
@Rebellos The first equality is obvious from the period 1/n. Fixing continuous f on a dense set determines the whole function.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 16 '18 at 12:55
add a comment |
$begingroup$
Simply observe that
$$
f(a) = lim_{ntoinfty} f(a+frac{1}{n}) = lim_{xto a} f(x).$$ Therefore, $f$ is continuous at every $ainmathbb{R}$. Observe that by the assumption, it holds that
$$
f(0) = f(frac{j}{n}),quad forall ninmathbb{N},jin mathbb{Z},
$$ and hence that
$$
f(q) = f(0),quadforall qinmathbb{Q}.
$$ By the continuity of $f$, it follows that
$$f(x) = f(0),quadforall xinmathbb{R}.$$
answered Dec 16 '18 at 12:48
SongSong
8,699625
$endgroup$
Simply observe that
$$
f(a) = lim_{ntoinfty} f(a+frac{1}{n}) = lim_{xto a} f(x).$$ Therefore, $f$ is continuous at every $ainmathbb{R}$. Observe that by the assumption, it holds that
$$
f(0) = f(frac{j}{n}),quad forall ninmathbb{N},jin mathbb{Z},
$$ and hence that
$$
f(q) = f(0),quadforall qinmathbb{Q}.
$$ By the continuity of $f$, it follows that
$$f(x) = f(0),quadforall xinmathbb{R}.$$
answered Dec 16 '18 at 12:48
SongSong
8,699625
edited Dec 16 '18 at 13:27
answered Dec 16 '18 at 12:48
SongSong
8,699625
answered Dec 16 '18 at 12:48
SongSong
8,699625
answered Dec 16 '18 at 12:48
SongSong
8,699625
8,699625
$begingroup$
Oops, I misread it. But about $f(a) = f(a=frac{1}{n})$, it follows directly from the assumption.
$endgroup$
– Song
Dec 16 '18 at 12:53
$begingroup$
@Rebellos The first equality is obvious from the period 1/n. Fixing continuous f on a dense set determines the whole function.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 16 '18 at 12:55
add a comment |
$begingroup$
Oops, I misread it. But about $f(a) = f(a=frac{1}{n})$, it follows directly from the assumption.
$endgroup$
– Song
Dec 16 '18 at 12:53
$begingroup$
@Rebellos The first equality is obvious from the period 1/n. Fixing continuous f on a dense set determines the whole function.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 16 '18 at 12:55
$begingroup$
Oops, I misread it. But about $f(a) = f(a=frac{1}{n})$, it follows directly from the assumption.
$endgroup$
– Song
Dec 16 '18 at 12:53
$begingroup$
Oops, I misread it. But about $f(a) = f(a=frac{1}{n})$, it follows directly from the assumption.
$endgroup$
– Song
Dec 16 '18 at 12:53
$begingroup$
@Rebellos The first equality is obvious from the period 1/n. Fixing continuous f on a dense set determines the whole function.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 16 '18 at 12:55
$begingroup$
@Rebellos The first equality is obvious from the period 1/n. Fixing continuous f on a dense set determines the whole function.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 16 '18 at 12:55
add a comment |
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