Is the Limit of an Integrable Function Equal to Its Pointwise Limit?

In summary, the conversation discusses a solution to the exercise where $f$ is integrable on each closed subinterval of $[0,+\infty)$ and $\lim_{x \to +\infty} f(x)=l$. The solution involves choosing a value $c_1$ and showing that for $x\geq\max\{c_1,c_2\}$, the difference between $\frac{1}{x} \int_0^x f$ and $l$ is less than or equal to $\epsilon$. This is explained by the fact that $\int_{0}^{c_1} |f-l|$ is a fixed number and $x$ increases unboundedly. Additionally,
  • #1
evinda
Gold Member
MHB
3,836
0
Hey again! :rolleyes: (Blush)

I am looking at the following exercise:
Let $f:[0,+\infty) \to \mathbb{R} $ integrable on each closed subinterval of $[0,+\infty)$.We suppose that $\lim_{x \to +\infty} f(x)=l$.Show that $\lim_{x \to +\infty} \frac{1}{x} \int_{0}^{x} f = l$.

The solution is the following:
Let $\epsilon>0$.As $\lim_{x \to +\infty} f(x)=l$, $\exists c_1>0$ such that $|f(x)-l|<\epsilon, \forall x \geq c_1$.
Also,$\exists c_2>0$ such that $\frac{1}{x} \int_{0}^{c_1} |f-l|< \epsilon, \forall x \geq c_2$
Therefore, $ \forall x \geq \max\{c_1,c_2\}$ we have $|\frac{1}{x} \int_0^x f - l|=\frac{1}{x} |\int_0^x (f-l)|\leq \frac{1}{x} \int_0^{c_1} |f-l|+\frac{1}{x} \int_{c_1}^x |f-l| \leq \epsilon + \frac{x-c_1}{x} \epsilon \leq \epsilon$

Could you explain me the red part?? How do we conclude to that?? :confused:
 
Physics news on Phys.org
  • #2
Re: Why is it like that?

evinda said:
Let $f:[0,+\infty) \to \mathbb{R} $ integrable on each closed subinterval of $[0,+\infty)$...

Also,$\exists c_2>0$ such that $\frac{1}{x} \int_{0}^{c_1} |f-l|< \epsilon, \forall x \geq c_2$ ...

Could you explain me the red part??
This is because once we have chosen $c_1$, $\int_{0}^{c_1} |f-l|$ is a fixed number, which exists because $f$, and therefore $|f-l|$, is integrable. Meanwhile, $x$ increases unboundedly.
 
  • #3
Re: Why is it like that?

Evgeny.Makarov said:
This is because once we have chosen $c_1$, $\int_{0}^{c_1} |f-l|$ is a fixed number, which exists because $f$, and therefore $|f-l|$, is integrable. Meanwhile, $x$ increases unboundedly.

A ok!Thanks a lot! :)
 
  • #4
evinda said:
Therefore, $ \forall x \geq \max\{c_1,c_2\}$ we have $|\frac{1}{x} \int_0^x f - l|=\frac{1}{x} |\int_0^x (f-l)|\leq \frac{1}{x} \int_0^{c_1} |f-l|+\frac{1}{x} \int_{c_1}^x |f-l| \leq \epsilon + \frac{x-c_1}{x} \epsilon \leq \epsilon$

Could you also explain me why it stands that :

$$ \epsilon + \frac{x-c_1}{x} \epsilon \leq \epsilon $$

? (Thinking) (Thinking)
 
  • #5
evinda said:
Could you also explain me why it stands that :

$$ \epsilon + \frac{x-c_1}{x} \epsilon \leq \epsilon $$

?
Well, obviously $\epsilon$ plus a positive number cannot be $\le\epsilon$, but it is $\le2\epsilon$.
 
  • #6
Evgeny.Makarov said:
Well, obviously $\epsilon$ plus a positive number cannot be $\le\epsilon$, but it is $\le2\epsilon$.

Nice! Thank you very much! (Smile)
 

Related to Is the Limit of an Integrable Function Equal to Its Pointwise Limit?

What is the definition of the limit of an integrable function?

The limit of an integrable function is a mathematical concept that describes the behavior of a function as its input approaches a certain value. It is defined as the value that the function approaches as the input value gets closer and closer to the specified value.

How is the limit of an integrable function different from the limit of a regular function?

The limit of an integrable function is different from the limit of a regular function because it takes into account the area under the curve of the function, rather than just its value at a specific point. This means that the limit of an integrable function can be influenced by the behavior of the function over a certain interval, rather than just at a single point.

Can the limit of an integrable function exist even if the function is not continuous?

Yes, the limit of an integrable function can exist even if the function is not continuous. This is because the concept of integrability takes into account the behavior of the function over an interval, rather than just its value at a single point. As long as the function is integrable over the specified interval, its limit can still exist.

How is the limit of an integrable function used in real-world applications?

The limit of an integrable function is used in real-world applications to calculate areas and volumes, as well as to model and analyze physical phenomena such as motion, population growth, and economic trends. It is also used in numerical integration methods to approximate the value of a definite integral.

What are some common techniques for evaluating the limit of an integrable function?

Some common techniques for evaluating the limit of an integrable function include using the definition of the limit, applying limit laws, using L'Hôpital's rule, and using the Fundamental Theorem of Calculus. Other techniques, such as substitution and trigonometric identities, may also be useful depending on the specific function being evaluated.

Similar threads

I Regarding Wirtinger's inequality
Replies
4
Views
775
A Proving that this integral is divergent
Replies
2
Views
470
I ##(a_n) ## has +10,-10 as partial limits. Then 0 is also a partial limit
Replies
13
Views
2K
MHB "Approximation to the Identity" and "Convolution" Proof
Replies
2
Views
2K
MHB Real Analysis - Convergence to Essential Supremum
Replies
2
Views
2K
I On limit of convolution of function with a summability kernel
Replies
2
Views
805
I Courant and Fritz, Construction of the real numbers
Replies
2
Views
2K
MHB How Do Riemann Integrals Handle Function Splits and Summations?
Replies
2
Views
1K
MHB Proving Differentiability of f at $x_0$
Replies
11
Views
1K
I Question about convex property in Jensen's inequality
Replies
4
Views
321

Hot Threads

Recent Insights

Back
Top