Is the Integral Inequality Possible to Prove for Certain Parameters?

In summary: T}a(t)dt\Bigr)^{\frac{p+1}{2p}}\leq C\Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{p+1}{2p}}$$where ##C = 1##. This shows that the inequality holds for any ##C>0##, and thus, it is possible to show that it holds for some ##C>0##. In summary, it is possible to show that the inequality holds for ##p\in(\frac{1}{2},1)## and ##C## may depend on ##T##, but it can always be chosen to be ##1##.
  • #1
amirmath
8
0
I want to know that is it possible to show that
$$
\int_{0}^{T}\Bigr(a(t
)\Bigr)^{\frac{p+1}{2p}}dt\leq C\Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{p+1}{2p}}
$$
for some ##C>0## where ##a(t)>0## and integrable on ##(0,T)## and ##p\in(\frac{1}{2},1)##. It is worth noting that this range for ##p## yields ##\frac{p+1}{2p}>1##. In the case ##p>1## we have ##\frac{p+1}{2p}<1## and the Holder's inequality can be applied to obtain the result immediately. But in the first case I don't know the validity of the inequality.
 
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  • #2
The crucial point is whether ##C## may depend on ##T## or not. E.g. ##p=\frac{2}{3}\, , \,a(t)=t##.
 
  • #3


Yes, it is possible to show that the inequality holds for ##p\in(\frac{1}{2},1)##. First, we can rewrite the left-hand side of the inequality as
$$
\int_{0}^{T}\Bigr(a(t)\Bigr)^{\frac{p+1}{2p}}dt = \int_{0}^{T}\Bigr(a(t)\Bigr)^{\frac{1}{2}}a(t)^{\frac{p}{2p}}dt
$$
Then, we can use the Cauchy-Schwarz inequality to obtain
$$
\int_{0}^{T}\Bigr(a(t)\Bigr)^{\frac{1}{2}}a(t)^{\frac{p}{2p}}dt \leq \Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{1}{2}}\Bigr(\int_{0}^{T}a(t)^{\frac{p}{p-1}}dt\Bigr)^{\frac{p-1}{2p}}
$$
Since ##p\in(\frac{1}{2},1)##, we have ##\frac{p}{p-1} > 1##, which means that the second term on the right-hand side is integrable. Thus, we can apply Holder's inequality again to obtain
$$
\Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{1}{2}}\Bigr(\int_{0}^{T}a(t)^{\frac{p}{p-1}}dt\Bigr)^{\frac{p-1}{2p}} \leq \Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{1}{2}}\Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{p-1}{2p}} = \Bigr(\int_{0}^{T}a(t)dt\Bigr)^{\frac{p+1}{2p}}
$$
Therefore, combining all of these steps, we have
$$
\int_{0}^{T}\Bigr(a(t)\Bigr)^{\frac{p+1}{2p}}dt \leq \Bigr(\int_{0}^{
 

1. What is the Integral Inequality?

The Integral Inequality is a mathematical concept that states the integral of a function is always greater than or equal to the integral of its absolute value. In other words, the area under a curve is always equal to or greater than the area between the curve and the x-axis.

2. What are the parameters for which the Integral Inequality can be proven?

The Integral Inequality can be proven for any function that is integrable on a closed interval and has a continuous derivative on that interval. This means that the function must be defined and continuous at every point within the interval.

3. How is the Integral Inequality proven for certain parameters?

The Integral Inequality can be proven using various mathematical techniques such as the Fundamental Theorem of Calculus, the Mean Value Theorem, and the Intermediate Value Theorem. These techniques involve manipulating the function and its integral to show that the inequality holds true.

4. Why is the Integral Inequality important in mathematics?

The Integral Inequality is important because it allows us to calculate the area under a curve, which has many real-world applications such as calculating volumes, areas, and probabilities. It also provides a way to prove other important theorems in calculus and analysis.

5. Are there any exceptions to the Integral Inequality?

Yes, there are some cases where the Integral Inequality may not hold true. For example, if the function is not continuous or integrable on the given interval, the inequality may not be valid. Additionally, there are some special types of functions, such as oscillatory functions, for which the Integral Inequality may not hold true.

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