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Let \(f \in L_{p_{0}}[0,1]\) for some \(p_{0}>1 .\) Show that \(\lim _{p \rightarrow 1^{+}}\|f\|_{L_{p}}=\|f\|_{L_{1}} .\) If \(f \in L_{\infty}[0,1]\), then \(\lim _{p \rightarrow \infty}\|f\|_{L_{p}}=\|f\|_{L_{\infty}}\) Let \(x \in \ell_{q}\) for some \(q \geq 1\). Show that \(\lim _{p \rightarrow \infty}\|x\|_{\ell_{p}}=\|x\|_{\ell_{\infty}}\).

Short Answer

Expert verified
As \ p approaches 1, \|f\|_{L_p} approaches \|f\|_{L_1}. As \ p approaches \infty, \|f\|_{L_p} approaches \|f\|_{L_\infty} and \|x\|_{\ell_{p}} approaches \|x\|_{\ell_{\infty}}.

Step by step solution

01

Understand the Norm Definitions

The \L_p[-0,1] norm for a function f is given by \[ \|f\|_{L_p} = \left( \int_0^1 |f(t)|^p dt \right)^{1/p}. \] For \L_\infty norm, \L_{\infty}[-0,1], it is given by: \[ \|f\|_{L_\infty} = \text{ess sup}_{t \in [0,1]} |f(t)|. \] For sequences, the \ell_p norm is defined as \[ \|x\|_{\ell_p} = \left( \sum_{i=1}^{\infty} |x_i|^p \right)^{1/p}. \] The \ell_\infty norm is \[ \|x\|_{\ell_{\infty}} = \sup_{i} |x_i|. \]
02

Prove \lim _{p \rightarrow 1^{+}}\|f\|_{L_{p}}=\|f\|_{L_{1}}

Note that \|f\|_{L_p} is a continuous function of p for \ f \in L_{p_{0}}[0,1]. Use the Dominated Convergence Theorem which states that if \{f_n\} is a sequence of measurable functions such that \ f_n \rightarrow f \ a.e., and there is an integrable function \ g \ 1, we have \[ \lim _{p \rightarrow 1^{+}}\|f\|_{L_{p}}=\|f\|_{L_{1}}. \]
03

Prove \lim _{p \rightarrow\infty}\|f\|_{L_{p}}=\|f\|_{L_{\infty}}

For \ f \in L_{\infty}[0,1], note that as \ p \rightarrow \infty, the largest values of \ f will dominate the norm. Formally, this can be shown by using Chebyshev's inequality, and taking the limit as \ p \rightarrow \infty \ gives \[ \lim _{p \rightarrow \infty}\|f\|_{L_{p}}=\|f\|_{L_{\infty}}. \]
04

Prove \lim _{p \rightarrow \infty}\|x\|_{\ell_{p}}=\|x\|_{\ell_{\infty}}

For \ x \in \ell_{q}, \ q \geq 1, noting that the maximum component of the sequence will majorly influence the norm as \ p \rightarrow \infty, take the limit of \ \left(\sum_{i=1}^{\finfty} |x_i|^p \right)^{1/p} \ and show it equals to \ \sup_{i} |x_i|. This completes the proof \[ \lim _{p \rightarrow \infty}\|x\|_{\ell_{p}}=\|x\|_{\ell_{\infty}}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

L_p norm
The \(L_p\) norm is a fundamental concept in functional analysis and helps measure the 'size' of a function on a specific domain. For example, the \(L_p\) norm on the interval \([0,1]\) for a function \(f\) is defined as: \[ \|f\|_{L_p} = \left( \int_0^1 |f(t)|^p \ dt \right)^{1/p}. \] This means you take the absolute value of \(f(t)\) raised to the power \(p\), integrate it over the range, and then take the \(p\)-th root of the result. This norm becomes more sensitive to large values of \(f(t)\) as \(p\) increases.
The \(L_p\) space refers to functions where this norm is finite. Now, when \(p = \infty\), we get a different kind of measurement:
dominated convergence theorem
The Dominated Convergence Theorem (DCT) is a vital tool in real analysis. It deals with taking the limit under the integral sign. The theorem states that if you have a sequence of functions \({f_n}\) converging pointwise to a function \(f\) on a measurable space, and there is a dominating function \(g\) (i.e., \(|f_n(x)| \leq g(x)\) for all \(n\)), which is integrable, then the limit of the integrals of \({f_n}\) equals the integral of \(f\).
In mathematical form, if \(f_n \rightarrow f\) almost everywhere and \(|f_n| \leq g\) where \(g \in L^1\), it follows that: \[ \lim_{n \to \infty} \int f_n = \int \lim_{n \to \infty} f_n. \] This theorem helps in proving norm convergence in \(L_p\) spaces, especially when taking limits of \(L_p\) norms as \(p\) approaches different values.
Chebyshev's inequality
Chebyshev's inequality is an important inequality in probability and statistics that applies to measurable functions in real analysis. It helps in bounding probabilities and can show how the \(L_p\) norms relate as \(p \to \infty\).
Chebyshev’s inequality states that, for any random variable X and any \(k > 0\): \[ P(|X| \geq k) \leq \frac{E(|X|^p)}{k^p}, \] where \(E(|X|^p)\) denotes the expected value (or integral) of \(|X|^p\).
This inequality can be useful in the proof involving \(\backslash lim_{p \to \backslash infty} \backslash |f| \backslash _{L_{p}} = \backslash |f|_{L_{\backslash infty}}\). It shows how higher moments (expectations) help bound the probability of extreme values.
function spaces
Function spaces are a fundamental concept in mathematical analysis and provide the context for various norms, such as \(L_p\) and \(L_{\backslash infty}\). These are sets of functions that are equipped with a norm or a topology.
Common function spaces include:
  • \(L_p[0,1]\) – functions whose \(p\)-th power is integrable over \([0,1]\).
  • \(L_{\backslash infty}[0,1]\) – essentially bounded functions over \([0,1]\). The norm here measures the 'essential supremum'—the smallest bound that holds almost everywhere.
This distinction between different spaces allows mathematicians to handle functions with various properties and behaviors.
sequence norms
Just like functions, sequences can be assigned norms to measure their sizes or lengths in some sense. The \(\backslash ell_p\) spaces are sequence spaces defined analogously to the \(L_p\) spaces.
The norm for a sequence \(x = (x_1, x_2,\backslashldots)\) in \(\backslash ell_p\) is given by: \[ \backslash |x|_{\backslash ell_p} = \backslash left( \backslash sum_{i=1}^{\backslash infty} |x_i|^p \backslash right)^{1/p}. \] For \(p = \backslash infty\), the \(\backslash ell_{\backslash infty}\) norm is defined as the supremum of the absolute values of the elements in the sequence: \[ \backslash |x|_{\backslash ell_{\backslash infty}} = \backslash sup_{i} |x_i|. \] These concepts are essential when working with sequences, especially when investigating their convergence properties as p approaches certain limits.

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Most popular questions from this chapter

Let \(X\) be the normed space obtained by taking \(c_{0}\) with the norm \(\|x\|_{0}=\sum 2^{-i}\left|x_{i}\right| .\) Show that \(X\) is not a Banach space. Note that this shows that \(\|\cdot\|_{0}\) is not an equivalent norm on \(c_{0}\). Hint: The sequence \(\\{(1,1, \ldots, 1,0, \ldots)\\}_{n=1}^{\infty}\) is Cauchy and not convergent since the only candidate for the limit would be \((1,1, \ldots) \notin c_{0}\).

Let \(\sum x_{i}\) be an unconditionally convergent series in a Banach space \(X\). Show that for every \(\varepsilon>0\) there is \(n_{0}\) such that \(\left\|\sum_{i=n}^{m} \varepsilon_{i} x_{i}\right\|<\varepsilon\) for every \(\varepsilon_{i}=\pm 1\) and \(m \geq n \geq n_{0} ;\) in particular, \(\left\|\sum_{i=n_{0}}^{\infty} \varepsilon_{i} x_{i}\right\|<\varepsilon\) Hint: See the previous exercises.

Let \(X\) be an infinite-dimensional Banach space. Show that \(X\) admits no countable Hamel (algebraic) basis. Therefore, \(c_{00}\) cannot be normed to become a Banach space. Hint: If \(\left\\{e_{i}\right\\}\) is a countable infinite Hamel basis of a Banach space \(X\), put \(F_{n}=\operatorname{span}\left\\{e_{1}, \ldots, e_{n}\right\\} . F_{n}\) are closed and thus, by the Baire category theorem, at least one \(F_{n_{0}}\) has a nonempty interior; that is, there is \(x \in X\) and a ball \(B=\delta B_{X}\) such that \(x+B \subset F_{n_{0}} .\) Using linearity of \(F_{n_{0}}\), we have that \(-x+B \subset F_{n_{0}}\), so \(B \subset(x+B)+(-x+B) \subset F_{n_{0}} .\) Thus 0 is an interior point of \(F_{n_{0}} .\) This would mean that \(F_{n_{0}}=X\), a contradiction.

Let \(X\) be a normed linear space. Assume that for \(x, y \in X\) we have \(\|x+y\|=\|x\|+\|y\| .\) Show that then \(\|\alpha x+\beta y\|=\alpha\|x\|+\beta\|y\|\) for every \(\alpha, \beta \geq 0\) Hint: Assume \(\alpha \geq \beta .\) Write $$ \begin{aligned} \|\alpha x+\beta y\| &=\|\alpha(x+y)-(\alpha-\beta) y\| \geq \alpha\|x+y\|-(\alpha-\beta)\|y\| \\ &=\alpha(\|x\|+\|y\|)-(\alpha-\beta)\|y\|=\alpha\|x\|+\beta\|y\| \end{aligned} $$

Let \(\mathcal{L}\) be a normed space of all Lipschitz functions on a Banach space \(X\) that are equal to 0 at the origin, under the norm $$ \|f\|=\sup \left\\{\frac{|f(x)-f(y)|}{\|x-y\|} ; x, y \in X\right\\} $$ Show that \(\mathcal{L}\) is a Banach space.

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