Evaluating Norms of an Integral: $\Vert x\Vert_{\infty}$

  • Thread starter ELESSAR TELKONT
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In summary: Similarly, using the reverse Holder's inequality, we can prove that:$$\lim_{p\rightarrow\infty}\Vert x\Vert_{p}\geq \Vert x\Vert_{\infty}$$Hence, we can conclude that:$$\lim_{p\rightarrow\infty}\Vert x\Vert_{p}=\Vert x\Vert_{\infty}$$In summary, for any function $x(t)$, the $L^p$ norms are a monotonic function that increases with $p$. Therefore, in the limit, the integral in the statement equals to one. This argument is valid and can be supported by the use of Holder's inequality.
  • #1
ELESSAR TELKONT
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Homework Statement



[itex]\lim_{p\rightarrow\infty}\left(\int_{a}^{b}\vert x(t)\vert^{p}\,dt\right)^{\frac{1}{p}}=\Vert x\Vert_{\infty}[/itex]

Homework Equations



[itex]
\begin{align*}
\Vert x\Vert_{s}\leq (b-a)^{\frac{r-s}{rs}}\Vert x\Vert_{r}\,\forall 1\leq s<r<\infty\\
\Vert x\Vert_{s}\leq (b-a)^{\frac{1}{s}}\Vert x\Vert_{\infty}\,\forall 1\leq s<\infty\\
0\leq\frac{\vert x(t)\vert}{\Vert x\Vert_{\infty}}\leq 1\,\forall t\in [a,b]
\end{align*}
[/itex]

The Attempt at a Solution



Using the last of the former inequalities I can say that

[itex]
\begin{align*}
\left(\int_{a}^{b}\frac{\vert x(t)\vert^{p}}{\Vert x\Vert_{\infty}^{p}}\,dt\right)^{\frac{1}{p}}\leq (b-a)^{\frac{1}{p}}
\end{align*}
[/itex]

and since in the limit
[itex]
\begin{align*}
\lim_{p\rightarrow\infty}a^{\frac{1}{p}}=1\, \forall a>0
\end{align*}
[/itex]

then

[itex]
\begin{align*}
\lim_{p\rightarrow\infty}\left(\int_{a}^{b}\frac{ \vert x(t)\vert^{p}}{\Vert x\Vert_{\infty}^{p}}\,dt\right)^{\frac{1}{p}}\leq 1
\end{align*}
[/itex]

But since I have the other relations between the norms and these relations say that, fixed a function [itex]x(t)[/itex], the norms as a function or [itex]p[/itex] are a monotonic function that grows, then the last inequality say that in the limit the integral equals to one. Therefore, since I can write

[itex]
\begin{align*}
\Vert x\Vert_{p}=\Vert x\Vert_{\infty}\left(\int_{a}^{b}\frac{\vert x(t)\vert^{p}}{\Vert x\Vert_{\infty}^{p}}\,dt\right)^{\frac{1}{p}}
\end{align*}
[/itex]

it follows the proposition taking the limit by the former reasoning.

MY PROBLEM IS IF IT IS TRUE MY ASSUMPTION OF THE MONOTONICITY OF THE NORMS WITH P. IS IT VALID THE ARGUMENT? IF NOT, WHAT OTHER ARGUMENT MUST BE USED? THANKS IN ADVANCE.
 
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  • #2


Firstly, it is important to note that the statement is not entirely clear as it does not specify the norms being used. Assuming that the norms being referred to are the $L^p$ norms, the statement can be restated as:

$$\lim_{p\rightarrow\infty}\left(\int_{a}^{b}\vert x(t)\vert^{p}\,dt\right)^{\frac{1}{p}}=\Vert x\Vert_{\infty}$$

Next, let's define the $L^p$ norm as:

$$\Vert x\Vert_{p}=\left(\int_{a}^{b}\vert x(t)\vert^{p}\,dt\right)^{\frac{1}{p}}$$

Now, let's consider the statement:

$$\lim_{p\rightarrow\infty}\Vert x\Vert_{p}=\Vert x\Vert_{\infty}$$

This statement is true and can be proven using the following argument:

Firstly, we can rewrite the $L^p$ norm as:

$$\Vert x\Vert_{p}=\left(\int_{a}^{b}\vert x(t)\vert^{p}\,dt\right)^{\frac{1}{p}}=\left(\int_{a}^{b}\vert x(t)\vert^{p-1}\vert x(t)\vert\,dt\right)^{\frac{1}{p}}$$

Now, using the Holder's inequality, we have:

$$\Vert x\Vert_{p}\leq(b-a)^{\frac{1}{p}}\left(\int_{a}^{b}\vert x(t)\vert^{p-1}\,dt\right)^{\frac{1}{p}}$$

Now, since $p-1< p$, we have:

$$\lim_{p\rightarrow\infty}\left(\int_{a}^{b}\vert x(t)\vert^{p-1}\,dt\right)^{\frac{1}{p}}=\Vert x\Vert_{\infty}$$

Therefore, we can write:

$$\lim_{p\rightarrow\infty}\Vert x\Vert_{p}\leq \lim_{p\rightarrow\infty}(b-a)^{\frac{1}{p}}\Vert x\Vert_{\infty}=\Vert x\Vert_{\infty
 

What is the definition of the $\Vert x\Vert_{\infty}$ norm?

The $\Vert x\Vert_{\infty}$ norm, also known as the maximum norm or supremum norm, is a way of measuring the size of a vector or function in a multi-dimensional space. It is defined as the maximum absolute value of all the elements in the vector or function.

How is the $\Vert x\Vert_{\infty}$ norm calculated?

To calculate the $\Vert x\Vert_{\infty}$ norm, you take the absolute value of each element in the vector or function and then choose the largest value as the norm. For example, if we have a vector x = [2, -5, 3], the $\Vert x\Vert_{\infty}$ norm would be 5, since the absolute value of -5 is larger than the absolute value of 2 and 3.

What is the purpose of evaluating norms of an integral using the $\Vert x\Vert_{\infty}$ norm?

Evaluating norms of an integral using the $\Vert x\Vert_{\infty}$ norm allows us to measure the size of a function over a certain interval. This can be useful in various applications, such as finding the maximum error in an approximation or determining the stability of a system.

Can the $\Vert x\Vert_{\infty}$ norm be used for any type of function?

Yes, the $\Vert x\Vert_{\infty}$ norm can be used for any type of function, as long as the function is defined over a finite interval. It is commonly used for continuous functions, but it can also be applied to discrete functions or even matrices.

How does the $\Vert x\Vert_{\infty}$ norm compare to other norms, such as the $\Vert x\Vert_{1}$ or $\Vert x\Vert_{2}$ norm?

The $\Vert x\Vert_{\infty}$ norm is just one of many possible ways to measure the size of a vector or function. It is generally larger than the $\Vert x\Vert_{1}$ and $\Vert x\Vert_{2}$ norms, since it takes the maximum absolute value instead of the sum or square root of the values. However, it can be useful in certain situations where we are primarily interested in the largest value of a function.

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