Hot answers tagged

6

For $1\leq p<\infty$, $$\Vert Tf\Vert_p^p =\int |Tf|^p d\mu =\int |fg|^pd\mu\leq \Vert g\Vert_{\infty}^p \Vert f\Vert_p^p,$$ where we have used the fact that $|g| \leq \Vert g\Vert_{\infty}$ $\mu$-a.e. Take $p$th roots to get $$\Vert Tf\Vert_p \leq \Vert g\Vert_{\infty}\Vert f\Vert_p.$$ This is exactly what you wanted for $p$ finite. For $p=\infty$, ...


5

Big Hint : By Cauchy-Schwarz, $$\int|f(x)e^{-x^2}| \, dx\leq \sqrt{\int|f(x)|^2 \, dx}\sqrt{\int e^{-2x^2} \, dx}.$$


5

Let us assume that we are working with sequence spaces $\ell^p$ and $\ell^q$. Since the$$p=q=2$$case is taken care of already, assume$$p>2>q.$$Since $\ell^p\subset\ell^q$ for $q\le p$ as here, both sides of the desired inequality make sense. Start the indices of the sequences at zero, because we feel like it. So, an example to stretch things: ...


4

Here is a hint: by the Lebesgue differentiation theorem, the limit $$\lim_{r\to 0} \frac1{m(B_r(x))} \int_{B_r(x)} f(y) dy$$ where $B_r(x)=\{y\,:\,|x-y|<r\}$ exists and equals $f(x)$ for almost every $x$. Can you take it from there?


3

$\newcommand\ip[2]{\left\langle #1,#2\right\rangle}$ Leaving out many technical details, just illustrating what that "duality argument" is: Let's define $$\ip fg=\int fg.$$ Suppose that $K$ is a kernel and define $\tilde K(t)=K(-t)$. Then, assuming you can justify the Fubini (typically by restricting to some nice subspace of $L^p$), you see that ...


3

The $\ell^p$ spaces are a special case of the $L^p$ spaces obtained by using the counting measure on the set of natural numbers. If you squint closely at the integral it looks like a sum or indeed as Forever Mozart points out: summation is just integration with the trivial measure on $\mathbb{N}$.


3

Obviously, $||f||_{1,\infty} = \sup_{\alpha>0} \alpha \lambda_f(\alpha)$. For $n=1$ and a monotonic $f$ this is the area of the largest rectangle under the graph of $f$. The idea is thus the following. Take $f$ such that this rectangle has a large horizontal side. We add a small $g$, which bumps $f$ at the point where the rectangle touches the graph, ...


3

Let $\rho_f(s) := \mu\big(\{x : |f(x)| > s\}\big)$ and let $\|f\|_{1,\infty} := \sup_{s > 0}s\rho_f(s)$. Now let $f,g \in L^{1,\infty}$ and notice that $$\{x : |f(x) + g(x)| > s\} \subset \{x : |f(x)| > s(1 + \epsilon)^{-1}\} \cup \{x : |g(x)| > sC(\epsilon)^{-1}\},$$ where $(1 + \epsilon)^{-1} + C(\epsilon)^{-1} = 1.$ This shows that ...


2

Here is an outline of how to do this. Find $a$ such that $f(b)−f(a) = \int_a^bg(x) dx$ for almost all $b$. (You need to prove that you can do so.) Define $F(b) = f(a) + \int_a^bg(x) dx$. By its definition $F$ matches $f$ except on a set of measure 0. Now you have to prove that $F$ is continuous.


2

Alternatively, use $\text{H}\ddot{o}\text{lder}$'s Inequality: $f(x) = 1 f(x)$. Assume (WLOG) $y > x$. $$\int_x^y |f(t)|\,\text{d}t \leq \left(\int_x^y |f(t)|^2\,\text{d}t \right)^{1/2} \left( \int_x^y |1|\,\text{d}t \right)^{1/2}$$


2

1) If $|f| \ge |g|$ then $$(|f| + |g|)^p \le (2|f|)^p = 2^p |f|^p \le 2^p (|f|^p + |g|^p).$$ You can derive the same inequality if $|g| \le |f|$. 2) Once the inequality in 1) is established you have $$\int |f+g|^p \le 2^p \int |f|^p + 2^p \int |g|^p < \infty.$$ Thus by definition $f+g \in L^p$. 3) Since $p' = \frac{p}{p-1}$ it follows that $$ ...


2

Here's an example for $\mathbb{R}$, though it can likely be generalized to higher dimensions. Fix an integer $n\geqq 2$, and define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=\frac{n}{i}$ if $x\in [\frac{i-1}{n},\frac{i}{n})$, $1\leq i\leq n$, and $f(x)=0$ otherwise. Then $||f||_{1,\infty}=1$. Define a map $T:\mathbb{R}\to\mathbb{R}$ by $T(x)=x+\frac{1}{n}$ (mod ...


2

Let $e_n$ the sequence which is only $0$ except in the $n^{th}$ term where it is $1$. Let $f \in l^p$ and let $u_n = \sum_{k=1}^n f(k)e_k$. We will show that $u_n \in c_c$ converges to $f$ in $l^p$, so it shows that $c_c$ is dense in $l^p$. $$||f-u_n||_p=||f-\sum_{k=1}^n f(k)e_k||_p=\sum_{k=n+1}^{\infty}|f(k)|^p$$ Which tends to $0$ when $n$ tends to ...


2

Assuming you're using the essential supremum norm, here's an answer to part b) and part c). Here is a link to wikipedia's page on the essential supremum and below the proof of part c) is a link to a proof that $(L^\infty, \|\cdot\|_\infty)$ is a Banach space. Proof of b): Relying on the fact that a linear combination of essentially bounded and integrable ...


2

It is true, it is a consequence of the uniform boundness principle. https://en.wikipedia.org/wiki/Uniform_boundedness_principle Apply the contraposition. Consider the family $T^n(x) =(Tx)_n$ and suppose that $\sup_{T^n,\|x\|=1}\|T^n(x)\|=\infty$. Then there exists $x$ such that $\sup_n\|(Tx)_n\|=\infty$. This is impossible since $T$ must be defined at $x$. ...


2

The multiplication by 1 trick in action : $$\int_{-\pi}^\pi |f(x)| dx = \left|\int_{-\pi}^\pi 1|f(x)| dx \right| $$ Then you apply Cauchy-Schwartz $$ \leq \sqrt{\int_{-\pi}^\pi 1^2 dx } \sqrt{\int_{-\pi}^\pi |f(x)|^2 dx } $$ The first integral is easy to calculate, and the second is the $L^2$ norm $$\leq \sqrt{2\pi} \|f\|_{L^2}$$ And this inequality ...


2

Define $g_M := \max(\min(g,M),-M) ∈ L^1 ∩ L^\infty$ and apply your argument modified so that $s_k → \operatorname{sgn} g_M$ in $L^1$ to find that $‖g_M‖_1 = 0$. On the other hand, $|g_M| \uparrow |g| ∈ L^1$, so MCT says $∫|g_M| → ∫ |g|$, which means $∫|g| = 0$. (Actually since $g_M^2 \uparrow g^2$ as well you don't need to modify your $s_k$ but I'll just ...


2

The essential fact is not something about $e^{-x^2}$ decaying fast, but is not unrelated to that either. The essential fact is this: $$ \int_{-\infty}^\infty e^{-x^2}\,dx < \infty. $$ The point is that if $(X,\mathcal F,\mu)$ is a measure space and $\mu(X)<\infty$ then $L^2(X)\subseteq L^1(X)$. So let $X=\mathbb R$ and let $\mu$ be this measure: $$ ...


2

If $B$ is an orthonormal basis for the Hilbert space $A$ then $A$ is isomorphic to $L^2(\#)$, where $\#$ is counting measure on $B$. (Commonly known as $\ell^2(B)$.) So yes, every Hilbert space is an $L^2$ space. Can we make it $L^2(\nu)$ where $\nu$ is some measure on $X$? Yes, but not in any interesting way - $X$ has a subset with the same cardinality as ...


2

Note that for $n>1$ we have for $f(x)\ne 0$ $$ |f(x)|^{1/n}\le \begin{cases} |f(x)|&,|f(x)|\ge 1\\\\ 1&,|f(x)|<1 \end{cases}$$ Since $|f(x)|^{1/n}\le \max(1,|f(x)|)$, and since $f\in L^1[0,1]$ the Dominated Convergence Theorem guarantees that $$\lim_{n\to \infty}\int_0^1 |f(x)|^{1/n}\,dx=\int_0^1 \lim_{n\to \infty}|f(x)|^{1/n}\,dx=m\{x: ...


2

Claim: If $0 < p < 1$ then $(x + y)^p \le x^p + y^p$ for $x,y \ge 0$. The proof is the standard one for this type of inequalities: following PhoemueX's hint, define $$f(y) = (1 + y)^p - 1 - y^p.$$ Notice that $f(0) = 0$, $f'(y) \le 0$ for $y > 0$ and conclude that $f(y) \le 0$. This implies the claim and hence that $\rho(f,g)$ is a metric (the ...


1

Assuming continuity and differentiability of $f$: This comes from the Mean Value Theorem: If $f$ is continuous on $[a,b] \in \mathbb{R}$: $$\exists c \in [a,b]: f'(c)=\frac{f(b)-f(a)}{b-a}$$ Now: $||f'||_{L^{\infty}}$ is the infinity norm, which is the supremum value of $|f'|$ on some interval. So, your formula is saying that the absolute value of the ...


1

No, it is not true. A counter-example is given by $$ f(x) = \left\{\begin{array}{cc} 0, & 0 \le x < 1, \\ \frac{1}{\sqrt{x}(1+\ln x)}, & 1 \le x < \infty. \end{array}\right. $$ $f \in L^2$ with $\|f\|=1$ because \begin{align} ...


1

It is true almost in the way you stated it with $a_\epsilon = \epsilon$. You only have to restrict the norm on the left hand side to $\Omega_\epsilon =\{x \in \Omega \colon dist(x, \Omega^c)>\epsilon \}$, since this is where $u_\epsilon$ is defined. On the plus side, boundary conditions and boundary regularity are irrelevant. The proof is rather ...


1

Yes, this is easy to see. Let $N>>0$ be such that $(x_n)\in \ell^p$ has that $|x_n|<1$ for all $n>N$. Then we see that iff $\infty>q\ge p\ge 1$ $$\sum_{i=N+1}^\infty |x_i|^q\le \sum_{i=N+1}^\infty |x_i|^p<\infty$$ So $\ell^p\subseteq\ell^q$ when $q\ge p$. The case $q=\infty$ is already handled, as you noted.


1

Use the useful inequality $$\left\|f\cdot g\right\|_p\leq\left\|f\right\|_p\cdot\left\|g\right\|_{\infty}$$


1

For the form of the sphere for $l^p$ norms, well, it's not really conceptual, you just look at the definition. For the $l^1$ in $\mathbb{R}^2$ for instance, $(x,y)$ is a unit vector iff $|x| + |y| = 1$. If $x$ and $y$ are positive, this means $y=1-x$, so the intersection of the unit sphere with the first quadrant is the line $y= 1-x$. You can check the ...


1

By triangle inequality, $$|F-\sum_1^n f_k|\le\left|\sum_1^\infty|f_k|+\sum_1^n|f_k|\right| \le 2G.$$ Your estimation $|\sum_{n+1}^\infty f_k|^p\le \sum_{n+1}^\infty |f_k|^p$ is not true. For example, $(1+2)^2=9>1^2+2^2=5$.


1

In general, suppose that $X$ is finite measure space, and $p<q$. If $f\in L^q(X,\mu)$ then $f\in L^p(X,\mu)$, in fact $$|f|_p\leqslant \mu(X)^{1/p-1/q} |f|_q$$ Note that the first statement follows from the inequality stated, in fact this is how we prove such claim. This can be proven directly with Hölder's inequality, much like Tryss's answer uses the ...



Only top voted, non community-wiki answers of a minimum length are eligible