New answers tagged

0

The statement is false, at least if $c$ is to be independent of $f$. Fix $p\ge1$ and $\epsilon>0$ and suppose $$c^{1/p}(\|f\|_\infty-\epsilon)\le\|f\|_p$$ holds for some constant $c$. Let $f_n$ be a triangle function with base on $[-1/n^{p+1},1/n^{p+1}]$ and height $n$, and zero otherwise. Then $\|f_n\|_\infty=n$ and $$\|f_n\|_p^p=2\int_0^{1/n^{p+1}}(n-n^{...


1

You cannot find a constant $c$ independent of $f$. If you allow it to depend on both $f$ and $\epsilon$ but not on $p$ then you can use the following argument: There exists an interval $(c,d) \subset [a,b]$ such that $|f(x)| >M-\epsilon$ for $c < x<d$. We then get $\int_a^{b} |f(x)|^{p}dx \geq \int_c^{d} |f(x)|^{p}dx \geq (M-\epsilon)^{p} (d-c)$. So ...


2

Ok here's what I got. Let's clear up the relationship between $n \text{ and } d$. . If $n < d$ then by Holder's inequality $$ ||u||_{W^{1,n}(\mathbb{R}^d)} \leq C(d,p)||u||_{W^{1,d}(\mathbb{R}^d)} $$ which reduces to the case $n = d$. If $ n > d$ then we may apply Morrey's Inequality to show that $u \notin W^{1,n}$ here. Let's look - $$ ||u||_{C^{0,\...


1

For each $A \in \mathscr{A}$, since $\int_A f_i dP$ converge, let $\mu(A)= \lim_{i \to \infty} \int_A f_i dP$. We can prove that $\mu$ is a measure (and since $\int_A f_i dP$ is bounded by $1$ for every A $\in$ $\mathscr{A}$, $\mu$ is in fact a probability). One way to prove that $\mu$ is countably additive is to apply Vitali–Hahn–Saks theorem. Now, it is ...


2

The unit ball of $L^{2}[0,1]$ is not compact. Let $(g_n)$ be sequence in it with no convergent subsequence. Let $f_n(x)=g_n(x)$ for $ x \in [0,1]$ and $0$ for all other $x$. Then $(f_n)$ is a bounded sequence in $L^{2}(\mathbb R)$ and $(Tf_n)$ has no convergencet subsequence. Hence $T$ is not compact.


1

Usually convolution product is defined in this way: $$(f \star g)(s):=\int_{-\infty}^{\infty} f\left(s-t\right)g\left(t\right)dt \tag{1}$$ In the case of "causal" functions (i.e., functions which are zero or are made zero for negative values of $x$), we can write, instead of $f(t)$ and $g(t)$: $$U(t)f(t) \ \text{and} \ \ U(t)g(t)$$ where $U$ is the ...


0

Let $u \in W^{1,1}(0,T;X)$ for some Banach space $X$, i.e., $u' \in L^1(0,T;X)$. By the fundamental theorem of calculus we have in $X$ $$u(t)=u(s)+\int_s^t u'(\tau) d\tau,$$ for almost all $s,t \in (0,T)$. Taking the norm of $X$ $$\|u(t)\|_X = \left\|u(s)+\int_s^t u'(\tau) d\tau\right\|_X \leq \|u(s)\|_X + \int_s^t \|u'(\tau)\|_Xd\tau.$$ Taking the essential ...


1

The volume of an $\ell_2^n$ ball of radius $r$ is $r^n\cdot\frac{\pi^{n/2}}{\Gamma(n/2+1)}$. The volume of the $\ell_{\infty}^n$ unit-ball, i.e., the cube $[-1,1]^n$, is $2^n$. A straight-forward application of Stirling's formula shows that $$\lim_{n\to\infty}\left(\frac{\pi^{n/2}}{\Gamma(n/2+1)}\right)^{1/n}\sqrt{n}=\sqrt{2e\pi}$$ Therefore, what you wrote ...


1

So, here I consider $1<p<\infty$ and write $\frac{1}{p}+\frac{1}{q}=1$ for some $1<q<\infty$. Next let $F\in \ell_p'$ i.e. $F$ is a bounded or continuous real linear functional on $\ell_p$. Set, $$x_n:=\sum_{j=1}^n\big|F(e_j)\big|^{q-1} \operatorname{sgn}\big(F(e_j)\big)e_j.$$ Here, as usal $e_j$ is the sequence having $j$-th term as $1$ and all ...


2

This follows immediately from Holder's inequality. Take $q$ such that $\frac{1}{p}+\frac{1}{q}=1$. Then: $||X||_1=\int_{\Omega} |X|\leq (\int_{\Omega} |X|^p)^{\frac{1}{p}}(\int_{\Omega} 1^q)^{\frac{1}{q}}=||X||_p$ Edit: the general case follows from the special case. Assume $1\leq r<p<\infty$. We can define $Y=X^r$. By the special case we proved we ...


2

You basically showed already that there does not exist a $q>1$ such that $X\subset L^q$ holds. Using the statement from the linked question, this implies that the condition $$ \tag{*} X\subset \bigcup_{1<p\leq\infty} L^p(0,1) $$ is false (if it were true then $X\subset L^q$ would follow for some $q>1$). You might be falsely thinking that (*) should ...


1

Here is a different solution based on the following fact: If $\frac1p+\frac1q=$, $f\in L_p(\lambda)$ and $g\in L_q(\lambda)$, then $f*g$ is uniformly continuous: Here is a short proof of this: $$ |(f*g)(x+h)-(f*g)(x+k)|\leq \int|f(x+h-y)-f(x+k-y)||g(y)|\,dy\leq\|\tau_{-(k-h)}f-f\|_p\|g\|_q $$ The conclusion then follows by $L_p$ continuity of the translation ...


1

Begin a closed linear subspace of $L_1$, $X$ is itself a Banach space. Each $X_n$ is a closed subset of $X$: Suppose $f\in \overline{X}_n^X$ (the closure of $X_n$ relative to $X$). Then there is a sequence $\{f_k:k\in\mathbb{N}\}\subset X_n$ such that $\|f-f_k\|_{L_1}\xrightarrow{k\rightarrow0}$. By standard results (application of Chebyshev Markov, and ...


3

As far as I understand, you are able to prove that the operator $T_t: L^{\infty} \to L^{\infty}$ has norm $\|f\|_{L^1}$ if $f \in L^1$ is uniformly continuous. You can extend the assertion using a density argument. Let $f \in L^1$ be arbitrary, then there exists a sequence of uniformly continuous functions $(f_n)_{n \in \mathbb{N}} \subseteq L^1$ such that $...


3

Closedness of $X_n$'s follows immediately from Fatou's Lemma: If $\|f_j\|_{1+\frac 1 n} \leq n$ for all $j$ and $f_j \to f$ in $X$ then there is a subsequence $f_{j_i}$ which converges a.e. so $\|f\|_{1+\frac 1 n} \leq \lim \inf_k \|f_{j_k}\|_{1+\frac 1 n} \leq n$. Suppose there is an open ball $B(f_0,r)$ in $X$ which is $ \subset X_{n_0}$. Let $f \in X$. ...


1

By the open mapping theorem (or its corollary), a bijective linear bounded map between Banach spaces has a bounded inverse. In this case, the identity mapping $I:(S,\|\cdot\|_\infty)\to (S,\|\cdot\|_2)$ is clearly bijective, linear, and bounded by $\|f\|_2\le\|f\|_\infty$. The spaces are Banach as the asker rightly observes. Hence the inverse mapping (which ...


0

First of all note that the graph of $T$ is precisely $$\mathscr G(T) : = \{(f,f)\ :\ f \in S \} \subseteq \left (S,\|\cdot\|_2 \right) \times \left (S, \|\cdot\|_{\infty} \right ).$$ To prove that $\mathscr G (T)$ is closed we first take a convergent sequence $\{(f_n,f_n) \}_{n \geq 1}$ in $\mathscr G (T)$ converging to some $(f,g)$ with respect to the ...


3

Pick $\alpha(x)=x$. Then we have $\Vert \alpha \Vert_\infty = 1$. Assume that there exists $f\in L^2$ with $\Vert f \Vert_2=1$ such that $1=\Vert M_\alpha \Vert = \Vert M_\alpha (f) \Vert_2$. For $0<\varepsilon <1$ we have $$ 1^2=\Vert M_\alpha(f) \Vert_2^2 = \int_0^1 x^2 f(x)^2 dx = \int_0^{1-\varepsilon} x^2 f(x)^2 dx + \int_{1-\varepsilon}^1 x^2 ...


2

Yes you can. That can be achieve by direct application of Holder's inequality. For the specific case at hand, if $f\in L_2(\mu)$ $$\|f\|_1=\int_X|f|\,d\mu=\int_X|f|\mathbb{1}\,d\mu\leq \|f\|_2\|\mathbb{1}\|_2=\|f\|_2(\mu(X))^{1/2}\leq(\mu(X))^{1/2}\varepsilon$$ More generally, if $0<s<r$ and $f\in L_r(\mu)$, then $|f|^s\in L_{r/s}(\mu)$ and so $$ \...


6

$$\int|f|=\int|f|\cdot1\leq\bigg(\int|f|^2\bigg)^{1/2}\bigg(\int 1^2\bigg)^{1/2}=||f||_2\cdot \mu(X)^{1/2}.$$ More generally, for any $0<p<q<\infty$ we have, $$f\in L^q\implies\int|f|^p=\int|f|^p\cdot1\leq \big|\big||f|^p\big|\big|_{q/p}\cdot ||1||_{q/(q-p)}=||f||^p_q\cdot \mu(X)^{(q-p)/q} $$$$\text{ That is }||f||_p\leq ||f||_q\cdot \mu(X)^{(1/p)-(...


0

Write $C_0[-L/2,L/2]$ for the subspace of $C[-L/2,L/2]$ consisting of the $f$ with $f(-L/2)=f(L/2)$. Then by Stone-Weierstrass, the $f_n$ generate an $L^2$-dense subspace of $C_0[-L/2,L/2]$ and we know $C[-L/2,L/2]$ is $L^2$-dense in $L^2[-L/2,L/2]$. To complete the proof we need that $C_0[-L/2,L/2]$ is $L^2$ dense in $C[-L/2,L/2]$. But $C_0[-L/2,L/2]$ has ...


1

You salvage the approach by showing that the continuous functions such that $f(-L/2)=f(L/2)$ are dense in $L^2$. This is easy to show as you can always consider a small interval around a point where you don't need to approximate your function. That is, given $f\in L^2$ and $\varepsilon>0$, find $g$ continuous with $\|f-g\|<\varepsilon/2$. Then define $...


2

The proof is essentially correct. The fix is to prove the density instead on the torus, ie. $[-L/2,L/2]$ with the endpoints quotiented together. Then they are the same point, so there’s nothing to separate! (Note that S-W still works on a compact Hausdorff space.) Then since null sets don’t matter, $L^2$ on the torus is naturally identified with $L^2$ on ...


1

Answer of $(2)$ In this answer, I will use $L^2,L^\infty$ for simplicity instead of $L^2\big([0,1],\Bbb K\big)$ and $L^\infty\big([0,1],\Bbb K\big)$. Also, these are sets of equivalence classes, but I will not distinguish in between equivalence classes and their representatives, as any two representatives are almost everywhere equals. From the above, we have ...


2

For $p\in (1,\infty].$ Let $C$ be the set of all $x=(x_n)_{n\in \Bbb N}$ in $l^p$ such that $f(x)=\sum_{n\in \Bbb N}x_n$ converges. Then $$ (\bullet)\quad \sup_{0\ne x\in C}|f(x)|/\|x\|=\infty.$$ Let $C_0=\{x\in C: f(x)=0\}.$ Then $C_0$ is dense in $C.$ Proof: If $p\in C$ and $f(p)\ne 0$ then for any $\epsilon \in \Bbb R^+$ there exists $q\in C$ with $\|q\|&...


3

The result is true only for $p\in(1,\infty]$. Upper index $n$ will denote the $n$th sequence; lower index $k$ will denote the $k$th term of the sequence. Consider first an auxillary sequence $$x^n = \left(1,-1,\frac12,\frac{-1}2,\dots,\frac1n,\frac{-1}n,0,0,\dots\right)$$ The termwise and $\ell^p$ limit $$x= \left(1,-1,\frac12,\frac{-1}2,\dots,\frac1n,\frac{-...


0

Actually, the property you are pointing out is fairly more general. In fact, for any $1\leq p < q \leq +\infty$, if you have a function $$ f\in L^p(\mathbb{R})\cap L^q(\mathbb{R}), $$ then, $f$ belongs to all $L^r(\mathbb{R})$ for all $r\in(p,q)$. In fact, let assume that $q<+\infty$ (since the proof for $q=\infty$ is already in the other answers). Let'...


2

For $f\in L^1$ we have $\int|f|<\infty$. $$f\in L^\infty\cap L^1\implies \int|f|^2=\int|f|\cdot |f|\leq ||f||_\infty\int|f|<\infty\implies f\in L^2.$$ Note that for any measurable $f$ we define $||f||_\infty=\inf\big\{M:\lambda(x:|f(x)|>M)=0\big\}$. Now, from definition of infimum we have $\lambda\big(\{x:|f(x)|>||f||_\infty+1/n\}\big)=0$ for ...


1

It is not true. Short version: you can join a uniformly convergent and a pointwise but not uniformly convergent sequence with the same limit in one sequence and generate a counterexample. Now let us explicitly write down a counterexample. We choose $$ f_n(x)= \sqrt{x^2+\frac 1n} \quad\mbox{for even }n\qquad\mbox{and}\quad f_n(x)=|x|+\frac xn \quad\mbox{for ...


3

If $x \in \ell^p$, then $$ \sum_{n=1}^{\infty}|x_n|^p<+\infty.$$ In particular, $$\lim_{n}|x_n|=0.$$ The last equality implies that there is a $n_0$ such that $|x_n| \le 1$ for every $n\ge n_0$. Therefore, for every $n \ge n_0$ and $p'>p$, $|x_n|^{p'} < |x_n|^p.$ Then, by the comparison test: $$ \sum_{n=1}^{\infty}|x_n|^{p'}<+\infty$$ and $x \in \...


2

Since $L^p(\mu)$ is reflexive, its unit ball $B$ is weakly compact. Since $T$ is continuous for the norm topologies, it is also continuous for the weak topologies and hence $T(B)$ is a weakly compact set in $C(X)$. In particular, since the map $f \mapsto f(x)$ is a continuous linear functional on $C(X)$ for each $x \in X$, for any sequence $f_n$ in $T(B)$ ...


2

$\|f_n-f\|_1 \to 0$ implies there is a subsequnce $f_{n_k}$ converging a.e. to $f$. So $\int |f|^{p} \leq \lim \inf \int |f_{n_k}|^{p} \leq 1$.


0

If $f_n \to f$ in $L^1$ then $f_n \to f$ in measure. Then use this Exercise on convergence in measure (Folland, Real Analysis)


0

$V^\perp$ consists of all functions written in terms of the inner product $\langle f,g\rangle=\int_0^{\pi}f(t)g(t)dt$: $$ f-\frac{\langle f,\sin(x)\rangle}{\langle \sin(x),\sin(x)\rangle}\sin(x)-\frac{\langle f,\cos(x)\rangle}{\langle\cos(x),\cos(x)\rangle}\cos(x),\;\;\; f\in L^2[0,\pi]. $$ This is because $\langle \sin(x),\cos(x)\rangle =0$.


1

Hint: Use Young's inequality for each summand. This will look similar to your assumed inequality (maybe you meant to add the exponents so that it becomes Young's inequality). edit: To address the edit of the question: Yes, there does seem to be a mistake. In the second part the exponents $p$ and $q$ are missing over the parts in the parentheses.


1

Here is what is wrong: the correct statement is that the separability of $L^p$, $1\le p <\infty$, is equivalent to the existence of a countable collection $C$ of measurable sets of finite measure such that for each measurable set $Q$ of finite measure and for each positive $r$ there exists a set $A\in C$ such that the measure of $Q\Delta A$ is less that $...


0

An elegant way to prove the inequality $\|\int f d\mu\|\le \int \|f\| d\mu$ for functions with values in a Banach space $(X,\|\cdot\|)$ is the norm formula $\|x\|=\sup\{ |\phi(x)|: \phi\in X^*, \|\phi\|^*\le 1\}$ (which is a consequence of Hahn-Banach). For $\phi\in X^*$ with $\|\phi\|^*\le 1$ you get $$ \left|\phi(\int fd\mu)\right|=\left|\int \phi(f)d\mu\...


1

$\Bbb T$ is the circle parametrized by $e^{2\pi ix}$ with $x\in[0,2\pi]$. So you can view any function defined on $\Bbb T$ as a function on the interval $I=[0,2\pi]$ and $$\int_{\Bbb T}f=\int_If(x)dx\;.$$ Now take $f\in L^q$, that is, the integral $\int_I|f|^q$ is finite. Partition $I$ into subset $A$ where $|f(x)|\le 1$ and subset $B$ where $|f(x)|>1$. ...


1

The space $L_p (X)$ is dense subspace of $L_1 (X)$ and $S$ is defined on $L_p (X).$ So by Hahn - Banach Theorem you can extend $S$ to an operator $S_1 $ on whole $L_1 (X)$ but since $L_p (X)$ is dense in $L_1 (X)$ the extension must be unique, since every $f\in L_1 (X) $ can be approximate by functions $f_n\in L_p (X) $ such $||f_n -f||_1 \to 0.$


-1

The correct statement is that the space of summable functions (modulo a.e. equivalence) is separable if and only if there is a countable collection C of measurable sets of finite measure such that for each measurable set Q of FINITE measure and each integer N there is a set A in the collection C such that the measure of the symmetric difference between Q and ...


1

For $L^p$-spaces, you're reasoning is correct: We may extend the functions by $0$ and get a continuous embedding. In Sobolev Spaces, the same technique works if the Sobolev functions vanish on the boundary of the smaller domain $A$: $$W^{k,p}_{0}(A) \hookrightarrow W^{k,p}_{0}(B)$$ You can easily check that the weak differentiability is not affected. If we ...


2

If one consider real valued functions, i.e. $L_2(X;\mathbb{R})$ with $\langle f,g\rangle =\int fg\,d\mu$ Then $$F(f+h)=\langle f+h,f+h\rangle = \langle f,f\rangle + 2\langle f,h\rangle +\langle h,h\rangle=F(f)+2\langle f,h\rangle + F(h) $$ Cearly, $h\mapsto \langle f,h\rangle$ is linear, and $\frac{F(h)}{\|h\|_2}\xrightarrow{\|h\|_2\rightarrow0}0$. From this,...


1

Let $u_{n}(x)$ be a function of piecewise linear bumps of height $\frac{1}{n}$ and width $\frac{1}{n}$ on $\Omega=(0,1)$. The weak derivative is piecewise constant and satisfies almost everywhere $|u_{n}'(x)|=1$. Thus $||u_{n}'||_{L^q} = 1$. Then again, $||u||_{L^p}$ vanishes as $n$ goes to infinity. So you're inequality can't exist.


3

You have $g := |\varphi|^2 \in L^1(\mathbb{T})$ and $$\widehat{g}(n) = \int_{\mathbb{T}} g(x) \cdot \overline{x}^n \, dx = \int_{\mathbb{T}} e_{-n}(x) \cdot |\varphi(x)|^2 \, dx = 0$$ for any $n \in \mathbb{Z}$ with $n \neq 0$. Furthermore $$\widehat{g}(0) = \int_{\mathbb{T}} |\varphi(x)|^2 \, dx = 1.$$ Now check that the constant function $1 \in L^1(\mathbb{...


0

Write $h(p)=\int f^pd\mu$. $h'(p)=\int f^p\log(f)d\mu.$ Then we have $$\frac{d}{dp}(h(p)^{1/p})=\frac{d}{dp}(e^{\frac{\log(h(p))}{p}})=e^{\frac{\log(h(p))}{p}} \frac{\frac{p}{h(p)}h'(p)-\log(h(p))}{p^2}$$ When p=1, the above evaluates to $$h'(1)=\int f^p \log(f)d\mu.$$


5

The map $$F: (L^\infty(\mathbb{T}, |f \, d\lambda|), \|\cdot\|_1) \to \mathbb{C}, \, g \mapsto \int_{\mathbb{T}} fg \, d\lambda$$ is a bounded linear functional and hence continuous since $$|F(g)| = \left|\int_{\mathbb{T}} fg \, d\lambda \right| = \left|\int_{\mathbb{T}} g \, d(f\, d\lambda)\right| \leq \int_{\mathbb{T}} |g| \, d |f \, d\lambda| = \|g\|_1$$ ...


3

Just to write it out: So you know that $X^{(m)}$ converges to $X$ in $L^2([0,T]\times \Omega)$ for every $T>0$. Hence, let $X^{(m,1)}$ be a subsequence of $X^{(m)}$ converging a.e. on $[0,1]\times \Omega$ and recursively, let $X^{(m,n+1)}$ be a subsequence of $X^{(m,n)}$ which converges to $X$ a.e. on $[0,n+1]\times \Omega$, which, of course, we can do ...


0

Let $\mu_Y:=\mathsf{E}Y$, $\mu_X:=\mathsf{E}X$, $\Sigma_X:=\operatorname{Var}(X)$ $\Sigma_Y:=\operatorname{Var}(Y)$, and $\Sigma_{X,Y}=\mathsf{E}X(Y-\mu_Y)$. Consider the following transformation (assuming that $\Sigma_X$ is invertible): $$ Z:=Y-\Sigma_{X,Y}^{\top}\Sigma_X^{-1}X. $$ Since \begin{align} \mathsf{E}X(Z-\mathsf{E}Z)^{\top}&=\mathsf{E}X(Y-\...


0

Suppose $f_{X,Y}$ is the joint density of the variables with $Y$ included. This is a Gaussian density function by assumption. The conditional mean is almost surely equal to the mean of the conditional density $$ f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{\int_{\mathbb{R}}f_{X,Y}(x,y)dy}. $$ Namely, you can show that $$ E(Y|X) = \int_{\mathbb{R}} yf_{Y|X}(y|X)dy \;\;...


6

$\frac{1}{\log (2+n)}\in c_0$. But since for $x\gg 1$, $\log x\le C_q x^q$ for any $q> 0$, in particular $(\log x)^p \le C_p x$ for all $p\in[1,\infty)$, so $$\frac1{(\log (2+n))^p} \ge \frac{C_p}{2+n}$$ which is not summable.


Top 50 recent answers are included