How to show that if $f^2(x)$ is uniformly continous function then f is uniformly continous
$begingroup$
$f:Rto [0,infty)$ is function such that $f^2(x)$ is uniformly continuous on R then I have to show that f is uniformly continuous ?
My attempt :
$|f^2(x)-f^2(y)|<epsilon$ for $|x-y|<delta$
then
$|f(x)-f(y)<epsilon/|f(x)+f(y)|$ for $|x-y|<delta$
My problem is that how to control above difference as f may be 0 at both x and y
SO how to show above is uniformly continuous
Any help will be appreciated
real-analysis continuity uniform-continuity
edited Dec 1 '18 at 7:36
twnly
542110
asked Dec 1 '18 at 6:47
MathLoverMathLover
49310
$endgroup$
add a comment |
$begingroup$
$f:Rto [0,infty)$ is function such that $f^2(x)$ is uniformly continuous on R then I have to show that f is uniformly continuous ?
My attempt :
$|f^2(x)-f^2(y)|<epsilon$ for $|x-y|<delta$
then
$|f(x)-f(y)<epsilon/|f(x)+f(y)|$ for $|x-y|<delta$
My problem is that how to control above difference as f may be 0 at both x and y
SO how to show above is uniformly continuous
Any help will be appreciated
real-analysis continuity uniform-continuity
edited Dec 1 '18 at 7:36
twnly
542110
asked Dec 1 '18 at 6:47
MathLoverMathLover
49310
$endgroup$
2
$begingroup$
if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
$endgroup$
– mathworker21
Dec 1 '18 at 7:02
$begingroup$
It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Dec 1 '18 at 8:27
add a comment |
$begingroup$
$f:Rto [0,infty)$ is function such that $f^2(x)$ is uniformly continuous on R then I have to show that f is uniformly continuous ?
My attempt :
$|f^2(x)-f^2(y)|<epsilon$ for $|x-y|<delta$
then
$|f(x)-f(y)<epsilon/|f(x)+f(y)|$ for $|x-y|<delta$
My problem is that how to control above difference as f may be 0 at both x and y
SO how to show above is uniformly continuous
Any help will be appreciated
real-analysis continuity uniform-continuity
edited Dec 1 '18 at 7:36
twnly
542110
asked Dec 1 '18 at 6:47
MathLoverMathLover
49310
$endgroup$
$f:Rto [0,infty)$ is function such that $f^2(x)$ is uniformly continuous on R then I have to show that f is uniformly continuous ?
My attempt :
$|f^2(x)-f^2(y)|<epsilon$ for $|x-y|<delta$
then
$|f(x)-f(y)<epsilon/|f(x)+f(y)|$ for $|x-y|<delta$
My problem is that how to control above difference as f may be 0 at both x and y
SO how to show above is uniformly continuous
Any help will be appreciated
real-analysis continuity uniform-continuity
real-analysis continuity uniform-continuity
edited Dec 1 '18 at 7:36
twnly
542110
asked Dec 1 '18 at 6:47
MathLoverMathLover
49310
edited Dec 1 '18 at 7:36
twnly
542110
asked Dec 1 '18 at 6:47
MathLoverMathLover
49310
edited Dec 1 '18 at 7:36
twnly
542110
edited Dec 1 '18 at 7:36
twnly
542110
edited Dec 1 '18 at 7:36
twnly
542110
542110
asked Dec 1 '18 at 6:47
MathLoverMathLover
49310
asked Dec 1 '18 at 6:47
MathLoverMathLover
49310
asked Dec 1 '18 at 6:47
MathLoverMathLover
49310
49310
2
$begingroup$
if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
$endgroup$
– mathworker21
Dec 1 '18 at 7:02
$begingroup$
It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Dec 1 '18 at 8:27
add a comment |
2
$begingroup$
if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
$endgroup$
– mathworker21
Dec 1 '18 at 7:02
$begingroup$
It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Dec 1 '18 at 8:27
2
2
$begingroup$
if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
$endgroup$
– mathworker21
Dec 1 '18 at 7:02
$begingroup$
if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
$endgroup$
– mathworker21
Dec 1 '18 at 7:02
$begingroup$
It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Dec 1 '18 at 8:27
$begingroup$
It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
$endgroup$
– Martin R
Dec 1 '18 at 8:27
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Building off of Martin R's comment; If you prove that the square root function is uniformly continuous, and that the composition of two uniformly continuous functions is again uniformly continuous, then the result follows by taking $sqrt.$ composed with $f^{2}$.
$sqrt x$ is uniformly continuous
composition of two uniformly continuous functions.
answered Dec 1 '18 at 8:19
simplemathsimplemath
363
$endgroup$
add a comment |
$begingroup$
To follow up on Martin R's comment:
Theorem: If $h$ is uniformly continuous on an interval $I$, and $g$ is uniformly continuous on $h(I)$, $gcirc h$ is uniformly continuous on $I$.
Proof: Just chain the definitions. For any $epsilon>0$, there is some $delta>0$ such that $|g(y_1)-g(y_2)| < epsilon$ whenever $|y_1 - y_2| < delta$ for $y_1,y_2in h(I)$. Then, for any $delta > 0$, there is some $gamma > 0$ such that $|h(x_1)-h(x_2)| < delta$ whenever $|x_1-x_2| < gamma$ for $x_1,x_2in I$. Take $y_1=h(x_1),y_2=h(x_2)$, and we get that $|g(h(x_1))-g(h(x_2))| <epsilon$ whenever $|x_1-x_2| <gamma$ for $x_1,x_2in I$. That's the definition of uniform continuity, and we're done.
Next, we show that $g(y)=sqrt{y}$ is uniformly continuous on $[0,infty)$. With the slope increasing as we approach zero, the worst cases will be there - so we can write down explicitly that $|sqrt{y_1}-sqrt{y_2}|le sqrt{|y_1-y_2|}$, with equality only if one of the $y_i$ is zero. Invert that to get what we need; if $|y_1-y_2| < epsilon^2$, $|sqrt{y_1}-sqrt{y_2}| <epsilon$.
Finally, the original problem has $f=gcirc h$, where $g$ is the square root function on $[0,infty)$ and $h(x)=f^2(x)$ on $mathbb{R}$. Since $h$ is uniformly continuous on its domain $mathbb{R}$ and $g$ is uniformly continuous on $h(mathbb{R})$, the composition is uniformly continuous on its domain $mathbb{R}$. Do note here that $f$ was chosen to be positive; we need that detail for $f$ to actually be the composite function.
answered Dec 1 '18 at 8:55
jmerryjmerry
3,447413
$endgroup$
add a comment |
$begingroup$
I think the result holds for all $fin C(mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted.
answered Dec 1 '18 at 9:31
MiGangMiGang
636
$endgroup$
$begingroup$
Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
$endgroup$
– Martin R
Dec 1 '18 at 10:08
$begingroup$
@MartinR I mean you can remove the non-negative condition provided f is continuous.
$endgroup$
– MiGang
Dec 1 '18 at 12:05
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
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$begingroup$
Building off of Martin R's comment; If you prove that the square root function is uniformly continuous, and that the composition of two uniformly continuous functions is again uniformly continuous, then the result follows by taking $sqrt.$ composed with $f^{2}$.
$sqrt x$ is uniformly continuous
composition of two uniformly continuous functions.
answered Dec 1 '18 at 8:19
simplemathsimplemath
363
$endgroup$
add a comment |
$begingroup$
Building off of Martin R's comment; If you prove that the square root function is uniformly continuous, and that the composition of two uniformly continuous functions is again uniformly continuous, then the result follows by taking $sqrt.$ composed with $f^{2}$.
$sqrt x$ is uniformly continuous
composition of two uniformly continuous functions.
answered Dec 1 '18 at 8:19
simplemathsimplemath
363
$endgroup$
add a comment |
$begingroup$
Building off of Martin R's comment; If you prove that the square root function is uniformly continuous, and that the composition of two uniformly continuous functions is again uniformly continuous, then the result follows by taking $sqrt.$ composed with $f^{2}$.
$sqrt x$ is uniformly continuous
composition of two uniformly continuous functions.
answered Dec 1 '18 at 8:19
simplemathsimplemath
363
$endgroup$
Building off of Martin R's comment; If you prove that the square root function is uniformly continuous, and that the composition of two uniformly continuous functions is again uniformly continuous, then the result follows by taking $sqrt.$ composed with $f^{2}$.
$sqrt x$ is uniformly continuous
composition of two uniformly continuous functions.
answered Dec 1 '18 at 8:19
simplemathsimplemath
363
edited Dec 1 '18 at 8:59
answered Dec 1 '18 at 8:19
simplemathsimplemath
363
answered Dec 1 '18 at 8:19
simplemathsimplemath
363
answered Dec 1 '18 at 8:19
simplemathsimplemath
363
363
add a comment |
add a comment |
$begingroup$
To follow up on Martin R's comment:
Theorem: If $h$ is uniformly continuous on an interval $I$, and $g$ is uniformly continuous on $h(I)$, $gcirc h$ is uniformly continuous on $I$.
Proof: Just chain the definitions. For any $epsilon>0$, there is some $delta>0$ such that $|g(y_1)-g(y_2)| < epsilon$ whenever $|y_1 - y_2| < delta$ for $y_1,y_2in h(I)$. Then, for any $delta > 0$, there is some $gamma > 0$ such that $|h(x_1)-h(x_2)| < delta$ whenever $|x_1-x_2| < gamma$ for $x_1,x_2in I$. Take $y_1=h(x_1),y_2=h(x_2)$, and we get that $|g(h(x_1))-g(h(x_2))| <epsilon$ whenever $|x_1-x_2| <gamma$ for $x_1,x_2in I$. That's the definition of uniform continuity, and we're done.
Next, we show that $g(y)=sqrt{y}$ is uniformly continuous on $[0,infty)$. With the slope increasing as we approach zero, the worst cases will be there - so we can write down explicitly that $|sqrt{y_1}-sqrt{y_2}|le sqrt{|y_1-y_2|}$, with equality only if one of the $y_i$ is zero. Invert that to get what we need; if $|y_1-y_2| < epsilon^2$, $|sqrt{y_1}-sqrt{y_2}| <epsilon$.
Finally, the original problem has $f=gcirc h$, where $g$ is the square root function on $[0,infty)$ and $h(x)=f^2(x)$ on $mathbb{R}$. Since $h$ is uniformly continuous on its domain $mathbb{R}$ and $g$ is uniformly continuous on $h(mathbb{R})$, the composition is uniformly continuous on its domain $mathbb{R}$. Do note here that $f$ was chosen to be positive; we need that detail for $f$ to actually be the composite function.
answered Dec 1 '18 at 8:55
jmerryjmerry
3,447413
$endgroup$
add a comment |
$begingroup$
To follow up on Martin R's comment:
Theorem: If $h$ is uniformly continuous on an interval $I$, and $g$ is uniformly continuous on $h(I)$, $gcirc h$ is uniformly continuous on $I$.
Proof: Just chain the definitions. For any $epsilon>0$, there is some $delta>0$ such that $|g(y_1)-g(y_2)| < epsilon$ whenever $|y_1 - y_2| < delta$ for $y_1,y_2in h(I)$. Then, for any $delta > 0$, there is some $gamma > 0$ such that $|h(x_1)-h(x_2)| < delta$ whenever $|x_1-x_2| < gamma$ for $x_1,x_2in I$. Take $y_1=h(x_1),y_2=h(x_2)$, and we get that $|g(h(x_1))-g(h(x_2))| <epsilon$ whenever $|x_1-x_2| <gamma$ for $x_1,x_2in I$. That's the definition of uniform continuity, and we're done.
Next, we show that $g(y)=sqrt{y}$ is uniformly continuous on $[0,infty)$. With the slope increasing as we approach zero, the worst cases will be there - so we can write down explicitly that $|sqrt{y_1}-sqrt{y_2}|le sqrt{|y_1-y_2|}$, with equality only if one of the $y_i$ is zero. Invert that to get what we need; if $|y_1-y_2| < epsilon^2$, $|sqrt{y_1}-sqrt{y_2}| <epsilon$.
Finally, the original problem has $f=gcirc h$, where $g$ is the square root function on $[0,infty)$ and $h(x)=f^2(x)$ on $mathbb{R}$. Since $h$ is uniformly continuous on its domain $mathbb{R}$ and $g$ is uniformly continuous on $h(mathbb{R})$, the composition is uniformly continuous on its domain $mathbb{R}$. Do note here that $f$ was chosen to be positive; we need that detail for $f$ to actually be the composite function.
answered Dec 1 '18 at 8:55
jmerryjmerry
3,447413
$endgroup$
add a comment |
$begingroup$
To follow up on Martin R's comment:
Theorem: If $h$ is uniformly continuous on an interval $I$, and $g$ is uniformly continuous on $h(I)$, $gcirc h$ is uniformly continuous on $I$.
Proof: Just chain the definitions. For any $epsilon>0$, there is some $delta>0$ such that $|g(y_1)-g(y_2)| < epsilon$ whenever $|y_1 - y_2| < delta$ for $y_1,y_2in h(I)$. Then, for any $delta > 0$, there is some $gamma > 0$ such that $|h(x_1)-h(x_2)| < delta$ whenever $|x_1-x_2| < gamma$ for $x_1,x_2in I$. Take $y_1=h(x_1),y_2=h(x_2)$, and we get that $|g(h(x_1))-g(h(x_2))| <epsilon$ whenever $|x_1-x_2| <gamma$ for $x_1,x_2in I$. That's the definition of uniform continuity, and we're done.
Next, we show that $g(y)=sqrt{y}$ is uniformly continuous on $[0,infty)$. With the slope increasing as we approach zero, the worst cases will be there - so we can write down explicitly that $|sqrt{y_1}-sqrt{y_2}|le sqrt{|y_1-y_2|}$, with equality only if one of the $y_i$ is zero. Invert that to get what we need; if $|y_1-y_2| < epsilon^2$, $|sqrt{y_1}-sqrt{y_2}| <epsilon$.
Finally, the original problem has $f=gcirc h$, where $g$ is the square root function on $[0,infty)$ and $h(x)=f^2(x)$ on $mathbb{R}$. Since $h$ is uniformly continuous on its domain $mathbb{R}$ and $g$ is uniformly continuous on $h(mathbb{R})$, the composition is uniformly continuous on its domain $mathbb{R}$. Do note here that $f$ was chosen to be positive; we need that detail for $f$ to actually be the composite function.
answered Dec 1 '18 at 8:55
jmerryjmerry
3,447413
$endgroup$
To follow up on Martin R's comment:
Theorem: If $h$ is uniformly continuous on an interval $I$, and $g$ is uniformly continuous on $h(I)$, $gcirc h$ is uniformly continuous on $I$.
Proof: Just chain the definitions. For any $epsilon>0$, there is some $delta>0$ such that $|g(y_1)-g(y_2)| < epsilon$ whenever $|y_1 - y_2| < delta$ for $y_1,y_2in h(I)$. Then, for any $delta > 0$, there is some $gamma > 0$ such that $|h(x_1)-h(x_2)| < delta$ whenever $|x_1-x_2| < gamma$ for $x_1,x_2in I$. Take $y_1=h(x_1),y_2=h(x_2)$, and we get that $|g(h(x_1))-g(h(x_2))| <epsilon$ whenever $|x_1-x_2| <gamma$ for $x_1,x_2in I$. That's the definition of uniform continuity, and we're done.
Next, we show that $g(y)=sqrt{y}$ is uniformly continuous on $[0,infty)$. With the slope increasing as we approach zero, the worst cases will be there - so we can write down explicitly that $|sqrt{y_1}-sqrt{y_2}|le sqrt{|y_1-y_2|}$, with equality only if one of the $y_i$ is zero. Invert that to get what we need; if $|y_1-y_2| < epsilon^2$, $|sqrt{y_1}-sqrt{y_2}| <epsilon$.
Finally, the original problem has $f=gcirc h$, where $g$ is the square root function on $[0,infty)$ and $h(x)=f^2(x)$ on $mathbb{R}$. Since $h$ is uniformly continuous on its domain $mathbb{R}$ and $g$ is uniformly continuous on $h(mathbb{R})$, the composition is uniformly continuous on its domain $mathbb{R}$. Do note here that $f$ was chosen to be positive; we need that detail for $f$ to actually be the composite function.
answered Dec 1 '18 at 8:55
jmerryjmerry
3,447413
answered Dec 1 '18 at 8:55
jmerryjmerry
3,447413
answered Dec 1 '18 at 8:55
jmerryjmerry
3,447413
answered Dec 1 '18 at 8:55
jmerryjmerry
3,447413
3,447413
add a comment |
add a comment |
$begingroup$
I think the result holds for all $fin C(mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted.
answered Dec 1 '18 at 9:31
MiGangMiGang
636
$endgroup$
$begingroup$
Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
$endgroup$
– Martin R
Dec 1 '18 at 10:08
$begingroup$
@MartinR I mean you can remove the non-negative condition provided f is continuous.
$endgroup$
– MiGang
Dec 1 '18 at 12:05
add a comment |
$begingroup$
I think the result holds for all $fin C(mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted.
answered Dec 1 '18 at 9:31
MiGangMiGang
636
$endgroup$
$begingroup$
Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
$endgroup$
– Martin R
Dec 1 '18 at 10:08
$begingroup$
@MartinR I mean you can remove the non-negative condition provided f is continuous.
$endgroup$
– MiGang
Dec 1 '18 at 12:05
add a comment |
$begingroup$
I think the result holds for all $fin C(mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted.
answered Dec 1 '18 at 9:31
MiGangMiGang
636
$endgroup$
I think the result holds for all $fin C(mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted.
answered Dec 1 '18 at 9:31
MiGangMiGang
636
edited Dec 1 '18 at 9:43
answered Dec 1 '18 at 9:31
MiGangMiGang
636
answered Dec 1 '18 at 9:31
MiGangMiGang
636
answered Dec 1 '18 at 9:31
MiGangMiGang
636
636
$begingroup$
Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
$endgroup$
– Martin R
Dec 1 '18 at 10:08
$begingroup$
@MartinR I mean you can remove the non-negative condition provided f is continuous.
$endgroup$
– MiGang
Dec 1 '18 at 12:05
add a comment |
$begingroup$
Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
$endgroup$
– Martin R
Dec 1 '18 at 10:08
$begingroup$
@MartinR I mean you can remove the non-negative condition provided f is continuous.
$endgroup$
– MiGang
Dec 1 '18 at 12:05
$begingroup$
Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
$endgroup$
– Martin R
Dec 1 '18 at 10:08
$begingroup$
Sorry, I don't get how this is related to the question. $f$ is non-negative, and the question is about uniform continuity, not about zeros.
$endgroup$
– Martin R
Dec 1 '18 at 10:08
$begingroup$
@MartinR I mean you can remove the non-negative condition provided f is continuous.
$endgroup$
– MiGang
Dec 1 '18 at 12:05
$begingroup$
@MartinR I mean you can remove the non-negative condition provided f is continuous.
$endgroup$
– MiGang
Dec 1 '18 at 12:05
add a comment |
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if $|f(x)+f(y)| = f(x)+f(y)$ is small, then both $f(x)$ and $f(y)$ are small, which gives $|f(x)-f(y)|$ small. If $f(x)+f(y)$ is large, then $frac{epsilon}{|f(x)+f(y)|}$ is small.
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– mathworker21
Dec 1 '18 at 7:02
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It suffices to prove that the square root function is uniformly continuous, see for example math.stackexchange.com/q/569928/42969.
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– Martin R
Dec 1 '18 at 8:27