Uniformly continuous derivative implies existence of limitHow to show that a uniformly continuous function is...
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Uniformly continuous derivative implies existence of limit
How to show that a uniformly continuous function is bounded?Simple Construction of a Uniformly Continuous Real Valued Function With No Derivative Anywhere In The Domain?Bounded derivative implies uniform continuity- does the domain need to be an open interval?Prove $f$ is uniformly continuous iff $ lim_{xto infty}f(x)=0$The product of uniformly continuous functions is not necessarily uniformly continuousIs $f$ uniformly continuous?Continuous function goes to zero at $pm infty$, show it is uniformly continuousDifficult limit problem involving sine and tangent$f$ is uniformly continuous if and only if the limits exist in $mathbb{R}$Relationship with uniformly continuous function and its derivative.
$begingroup$
Let $f in C^1([0, +infty))$. Suppose that $lim_{x rightarrow +infty} f(x)=L$ and $f'$ is uniformly continuous.
Show that $$lim_{x rightarrow +infty} f'(x) + f(x)=L$$
I tried to apply L'Hospital's Rule to $frac{e^xf(x)}{e^x}$ since $frac{d}{dx}e^xf(x)=e^x(f'(x)+f(x))$. It seems alright but I didn't use the uniform continuity of $f'$ and it doesn't work for the function $f(x)=frac{sin(x^2)}{x}$ whose derivative is $f'(x)=2cos(x^2)-frac{sin(x^2)}{x^2}$ since $lim_{x rightarrow +infty} f'(x)$ doesn't exist.
Any ideas? Thanks in advance.
real-analysis
asked 3 hours ago
lzralbulzralbu
697512
$endgroup$
add a comment |
$begingroup$
Let $f in C^1([0, +infty))$. Suppose that $lim_{x rightarrow +infty} f(x)=L$ and $f'$ is uniformly continuous.
Show that $$lim_{x rightarrow +infty} f'(x) + f(x)=L$$
I tried to apply L'Hospital's Rule to $frac{e^xf(x)}{e^x}$ since $frac{d}{dx}e^xf(x)=e^x(f'(x)+f(x))$. It seems alright but I didn't use the uniform continuity of $f'$ and it doesn't work for the function $f(x)=frac{sin(x^2)}{x}$ whose derivative is $f'(x)=2cos(x^2)-frac{sin(x^2)}{x^2}$ since $lim_{x rightarrow +infty} f'(x)$ doesn't exist.
Any ideas? Thanks in advance.
real-analysis
asked 3 hours ago
lzralbulzralbu
697512
$endgroup$
$begingroup$
The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
$endgroup$
– RRL
2 hours ago
add a comment |
$begingroup$
Let $f in C^1([0, +infty))$. Suppose that $lim_{x rightarrow +infty} f(x)=L$ and $f'$ is uniformly continuous.
Show that $$lim_{x rightarrow +infty} f'(x) + f(x)=L$$
I tried to apply L'Hospital's Rule to $frac{e^xf(x)}{e^x}$ since $frac{d}{dx}e^xf(x)=e^x(f'(x)+f(x))$. It seems alright but I didn't use the uniform continuity of $f'$ and it doesn't work for the function $f(x)=frac{sin(x^2)}{x}$ whose derivative is $f'(x)=2cos(x^2)-frac{sin(x^2)}{x^2}$ since $lim_{x rightarrow +infty} f'(x)$ doesn't exist.
Any ideas? Thanks in advance.
real-analysis
asked 3 hours ago
lzralbulzralbu
697512
$endgroup$
Let $f in C^1([0, +infty))$. Suppose that $lim_{x rightarrow +infty} f(x)=L$ and $f'$ is uniformly continuous.
Show that $$lim_{x rightarrow +infty} f'(x) + f(x)=L$$
I tried to apply L'Hospital's Rule to $frac{e^xf(x)}{e^x}$ since $frac{d}{dx}e^xf(x)=e^x(f'(x)+f(x))$. It seems alright but I didn't use the uniform continuity of $f'$ and it doesn't work for the function $f(x)=frac{sin(x^2)}{x}$ whose derivative is $f'(x)=2cos(x^2)-frac{sin(x^2)}{x^2}$ since $lim_{x rightarrow +infty} f'(x)$ doesn't exist.
Any ideas? Thanks in advance.
real-analysis
real-analysis
asked 3 hours ago
lzralbulzralbu
697512
asked 3 hours ago
lzralbulzralbu
697512
asked 3 hours ago
lzralbulzralbu
697512
asked 3 hours ago
lzralbulzralbu
697512
asked 3 hours ago
lzralbulzralbu
697512
697512
$begingroup$
The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
$endgroup$
– RRL
2 hours ago
add a comment |
$begingroup$
The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
$endgroup$
– RRL
2 hours ago
$begingroup$
The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
$endgroup$
– RRL
2 hours ago
$begingroup$
The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
$endgroup$
– RRL
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have $lim_{x to infty} f'(x) = 0$ because,
$$int_0^x f'(t) , dt = f(x) - f(0), \int_0^infty f'(t) , dt = lim_{x to infty}f(x) - f(0) = L - f(0) quad (text{convergent})$$
and $f'$ is uniformly continuous.
To prove this assume that $lim_{x to infty}f'(x) =0$ does not hold and arrive at contradiction with the fact that the integral of $f'$ is convergent.
If $lim_{x to infty} f'(x) = 0$ does not hold then there exists $epsilon_0 > 0$ and a sequence $x_n to infty$ such that $|f'(x_n)| geqslant epsilon_0$ for all $n$. Next apply uniform continuity.
Assume WLOG that $f'(x_n) geqslant epsilon_0$.
There exists by uniform continuity $delta > 0$ such that $|f'(t) - f'(x_n)| < epsilon_0/2 implies f'(t) > epsilon_0/2$ for all $t in [x_n - delta,x_n + delta],$ and
$$ int_{x_n - delta}^{x_n + delta} f'(t) , dt > epsilondelta$$
This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.
answered 2 hours ago
RRLRRL
54.1k52675
$endgroup$
$begingroup$
I can help you further, but first let me know if these hints makes it obvious to you now.
$endgroup$
– RRL
2 hours ago
$begingroup$
I still can't see how to use uniform continuity. Could you, please, explain it further?
$endgroup$
– lzralbu
2 hours ago
$begingroup$
I shall do so...
$endgroup$
– RRL
2 hours ago
$begingroup$
What about the example given in the question?
$endgroup$
– Jens Schwaiger
1 hour ago
$begingroup$
@JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
$endgroup$
– RRL
1 hour ago
|
show 1 more comment
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1 Answer
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1 Answer
1
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$begingroup$
We have $lim_{x to infty} f'(x) = 0$ because,
$$int_0^x f'(t) , dt = f(x) - f(0), \int_0^infty f'(t) , dt = lim_{x to infty}f(x) - f(0) = L - f(0) quad (text{convergent})$$
and $f'$ is uniformly continuous.
To prove this assume that $lim_{x to infty}f'(x) =0$ does not hold and arrive at contradiction with the fact that the integral of $f'$ is convergent.
If $lim_{x to infty} f'(x) = 0$ does not hold then there exists $epsilon_0 > 0$ and a sequence $x_n to infty$ such that $|f'(x_n)| geqslant epsilon_0$ for all $n$. Next apply uniform continuity.
Assume WLOG that $f'(x_n) geqslant epsilon_0$.
There exists by uniform continuity $delta > 0$ such that $|f'(t) - f'(x_n)| < epsilon_0/2 implies f'(t) > epsilon_0/2$ for all $t in [x_n - delta,x_n + delta],$ and
$$ int_{x_n - delta}^{x_n + delta} f'(t) , dt > epsilondelta$$
This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.
answered 2 hours ago
RRLRRL
54.1k52675
$endgroup$
$begingroup$
I can help you further, but first let me know if these hints makes it obvious to you now.
$endgroup$
– RRL
2 hours ago
$begingroup$
I still can't see how to use uniform continuity. Could you, please, explain it further?
$endgroup$
– lzralbu
2 hours ago
$begingroup$
I shall do so...
$endgroup$
– RRL
2 hours ago
$begingroup$
What about the example given in the question?
$endgroup$
– Jens Schwaiger
1 hour ago
$begingroup$
@JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
$endgroup$
– RRL
1 hour ago
|
show 1 more comment
$begingroup$
We have $lim_{x to infty} f'(x) = 0$ because,
$$int_0^x f'(t) , dt = f(x) - f(0), \int_0^infty f'(t) , dt = lim_{x to infty}f(x) - f(0) = L - f(0) quad (text{convergent})$$
and $f'$ is uniformly continuous.
To prove this assume that $lim_{x to infty}f'(x) =0$ does not hold and arrive at contradiction with the fact that the integral of $f'$ is convergent.
If $lim_{x to infty} f'(x) = 0$ does not hold then there exists $epsilon_0 > 0$ and a sequence $x_n to infty$ such that $|f'(x_n)| geqslant epsilon_0$ for all $n$. Next apply uniform continuity.
Assume WLOG that $f'(x_n) geqslant epsilon_0$.
There exists by uniform continuity $delta > 0$ such that $|f'(t) - f'(x_n)| < epsilon_0/2 implies f'(t) > epsilon_0/2$ for all $t in [x_n - delta,x_n + delta],$ and
$$ int_{x_n - delta}^{x_n + delta} f'(t) , dt > epsilondelta$$
This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.
answered 2 hours ago
RRLRRL
54.1k52675
$endgroup$
$begingroup$
I can help you further, but first let me know if these hints makes it obvious to you now.
$endgroup$
– RRL
2 hours ago
$begingroup$
I still can't see how to use uniform continuity. Could you, please, explain it further?
$endgroup$
– lzralbu
2 hours ago
$begingroup$
I shall do so...
$endgroup$
– RRL
2 hours ago
$begingroup$
What about the example given in the question?
$endgroup$
– Jens Schwaiger
1 hour ago
$begingroup$
@JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
$endgroup$
– RRL
1 hour ago
|
show 1 more comment
$begingroup$
We have $lim_{x to infty} f'(x) = 0$ because,
$$int_0^x f'(t) , dt = f(x) - f(0), \int_0^infty f'(t) , dt = lim_{x to infty}f(x) - f(0) = L - f(0) quad (text{convergent})$$
and $f'$ is uniformly continuous.
To prove this assume that $lim_{x to infty}f'(x) =0$ does not hold and arrive at contradiction with the fact that the integral of $f'$ is convergent.
If $lim_{x to infty} f'(x) = 0$ does not hold then there exists $epsilon_0 > 0$ and a sequence $x_n to infty$ such that $|f'(x_n)| geqslant epsilon_0$ for all $n$. Next apply uniform continuity.
Assume WLOG that $f'(x_n) geqslant epsilon_0$.
There exists by uniform continuity $delta > 0$ such that $|f'(t) - f'(x_n)| < epsilon_0/2 implies f'(t) > epsilon_0/2$ for all $t in [x_n - delta,x_n + delta],$ and
$$ int_{x_n - delta}^{x_n + delta} f'(t) , dt > epsilondelta$$
This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.
answered 2 hours ago
RRLRRL
54.1k52675
$endgroup$
We have $lim_{x to infty} f'(x) = 0$ because,
$$int_0^x f'(t) , dt = f(x) - f(0), \int_0^infty f'(t) , dt = lim_{x to infty}f(x) - f(0) = L - f(0) quad (text{convergent})$$
and $f'$ is uniformly continuous.
To prove this assume that $lim_{x to infty}f'(x) =0$ does not hold and arrive at contradiction with the fact that the integral of $f'$ is convergent.
If $lim_{x to infty} f'(x) = 0$ does not hold then there exists $epsilon_0 > 0$ and a sequence $x_n to infty$ such that $|f'(x_n)| geqslant epsilon_0$ for all $n$. Next apply uniform continuity.
Assume WLOG that $f'(x_n) geqslant epsilon_0$.
There exists by uniform continuity $delta > 0$ such that $|f'(t) - f'(x_n)| < epsilon_0/2 implies f'(t) > epsilon_0/2$ for all $t in [x_n - delta,x_n + delta],$ and
$$ int_{x_n - delta}^{x_n + delta} f'(t) , dt > epsilondelta$$
This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.
answered 2 hours ago
RRLRRL
54.1k52675
edited 2 hours ago
answered 2 hours ago
RRLRRL
54.1k52675
answered 2 hours ago
RRLRRL
54.1k52675
answered 2 hours ago
RRLRRL
54.1k52675
54.1k52675
$begingroup$
I can help you further, but first let me know if these hints makes it obvious to you now.
$endgroup$
– RRL
2 hours ago
$begingroup$
I still can't see how to use uniform continuity. Could you, please, explain it further?
$endgroup$
– lzralbu
2 hours ago
$begingroup$
I shall do so...
$endgroup$
– RRL
2 hours ago
$begingroup$
What about the example given in the question?
$endgroup$
– Jens Schwaiger
1 hour ago
$begingroup$
@JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
$endgroup$
– RRL
1 hour ago
|
show 1 more comment
$begingroup$
I can help you further, but first let me know if these hints makes it obvious to you now.
$endgroup$
– RRL
2 hours ago
$begingroup$
I still can't see how to use uniform continuity. Could you, please, explain it further?
$endgroup$
– lzralbu
2 hours ago
$begingroup$
I shall do so...
$endgroup$
– RRL
2 hours ago
$begingroup$
What about the example given in the question?
$endgroup$
– Jens Schwaiger
1 hour ago
$begingroup$
@JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
$endgroup$
– RRL
1 hour ago
$begingroup$
I can help you further, but first let me know if these hints makes it obvious to you now.
$endgroup$
– RRL
2 hours ago
$begingroup$
I can help you further, but first let me know if these hints makes it obvious to you now.
$endgroup$
– RRL
2 hours ago
$begingroup$
I still can't see how to use uniform continuity. Could you, please, explain it further?
$endgroup$
– lzralbu
2 hours ago
$begingroup$
I still can't see how to use uniform continuity. Could you, please, explain it further?
$endgroup$
– lzralbu
2 hours ago
$begingroup$
I shall do so...
$endgroup$
– RRL
2 hours ago
$begingroup$
I shall do so...
$endgroup$
– RRL
2 hours ago
$begingroup$
What about the example given in the question?
$endgroup$
– Jens Schwaiger
1 hour ago
$begingroup$
What about the example given in the question?
$endgroup$
– Jens Schwaiger
1 hour ago
$begingroup$
@JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
$endgroup$
– RRL
1 hour ago
$begingroup$
@JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
$endgroup$
– RRL
1 hour ago
|
show 1 more comment
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$begingroup$
The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
$endgroup$
– RRL
2 hours ago