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6

You have $$ \phi(p) = \int_0^1 \lvert f(x) \rvert^p \, dx $$ $p \mapsto x^p = e^{p\log{x}}$ is smooth for $x \geqslant 0$, so we can differentiate twice with respect to $p$, $$ \phi''(p) = \int_0^1 \lvert f(x) \rvert^p (\log{\lvert f(x) \rvert})^2 \, dx, $$ which is obviously nonnegative. (This is consistent even for $f$ possessing zeros if we make the ...


5

Hint: $$ \max(a,b)=\frac{a+b+|a-b|}{2},\quad a,b\in\mathbb{R}. $$


4

Yes your definition is correct. Usually people do simply say convergence, but if you want to really emphasise that you do not mean another form of convergence e.g. weak convergence or almost everywhere convergence, you can use the word strong.


4

Suppose $$ \frac\alpha p+\frac{1-\alpha}r=\frac1q\implies\frac{\alpha q}p+\frac{(1-\alpha)q}r=1 $$ Then Hölder's Inequality says $$ \begin{align} \int_{\mathbb{R}^d}|f|^q\,\mathrm{d}x &=\int_{\mathbb{R}^d}|f|^{\alpha q}|f|^{(1-\alpha)q}\,\mathrm{d}x\\ &\le\left(\int_{\mathbb{R}^d}|f|^{\alpha q\large\frac p{\alpha ...


4

Since $\hat{f}\in\ell^1\implies f\in L^\infty$ and $\hat{f}\in\ell^2\implies f\in L^2$, Riesz-Thorin interpolation guarantees that If $\hat{f}\in\ell^q$ where $\frac1p+\frac1q=1$ and $1\le q\le2$, then $f\in L^p$ Therefore, since the $\hat{f}$ you give above is in $\ell^q$ for all $q\gt1$, we have that $f\in L^p(\mathbb{T})$ for all $p\lt\infty$. ...


3

Hint: given $f$, define $A=\{ x : |f(x)| \geq 1 \}$. Then on $A$, $|f|^q \leq |f|^r$, and on $A^c$, $|f|^q \leq |f|^p$. Use this to get a bound on $|f|^q$ by a function which you know is in $L^1$. Conclude that $f \in L^q$ from there. You can think about this proof as breaking up the obstruction to $f$ being in $L^q$ between its "singularities" (which lie ...


3

The function $f$ is measurable if and only if there exists a sequence of step functions that converge to $f$ almost everywhere. Your $u$ is integrable, hence measurable. Therefore you have that sequence of step functions with support in $(0,1)$ which converges to $u$ a.e. Note that the same sequence converges a.e. to $\tilde u$ on $(0,\infty)$, hence ...


3

In one dimension, consider in order $$\chi_{[0,1]}, \chi_{[0,1/2]}, \chi_{[1/2,1]},\chi_{[0,1/3]},\chi_{[1/3,2/3]},\chi_{[2/3,1]}, \dots $$ This sequence $\to 0$ in $L^1,$ and pointwise nowhere.


2

Normally, Hölder's inequality is written as $$\int_E|fg|\le \|f\|_p\|g\|_q \tag{1}$$ that is, with absolute value inside the integral. For this version, you don't need the additional constraint on absolute value in the equality case (analyzed in On the equality case of the Hölder and Minkowski inequalites). The inequality for $\left| ...


2

It is very difficult to do analysis on a space if it is not complete. Consider for instance $\mathbb{Q}$. The function $f\colon[0,2]\cap\mathbb{Q}\to\mathbb{Q}$ defined as $$ f(x)=\frac{1}{x^2-2} $$ is continuous, but is not bounded. An important result in Banach spaces (and in particular in $L^p$) that depends on completeness is: Theorem. If $X$ is a ...


2

Why is it important that $L^p$ spaces be complete? Among other things, because we can use it to prove existence of solutions for some problems. For instance, we use (among other things) the completeness of $L^p$ to define a certain Banach space $X$ and to prove that the operator \begin{align*} A&=\begin{pmatrix} 0 & 1 & 0 & 0 & 0 ...


2

Look at the inverse image of a generating measurable set. $$\bar{u}^{-1}(a,b)$$ If $a<0<b$ then because inverse images and unions commute, i.e. $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$ we get: $$\bar{u}^{-1}(a,b)=\bar{u}^{-1}(a,0)\cup \bar{u}^{-1}(\{0\})\cup\bar{u}^{-1}(0,b)$$ $$=u^{-1}(a,0)\cup (1,\infty)\cup u^{-1}(\{0\})\cup u^{-1}(0,b)$$ ...


2

This is not true: Consider the function $f(x)=\chi_{\mathbb{R}\setminus(-1,1)}(x)|x|^{-1/2}$. Then for $p\leq 2$ we have $\| f\|_{p}=\infty$ and for $p>2$ we get $$ \| f\|_p = \left(\frac{4}{p-2}\right)^{1/p}. $$


2

Let $f(x):= x^{-1/p} \mathbf{1}_{(1,\infty)}$. Then $f^q$ is integrable because $\frac{q}{p} > 1$. But $f^p = \frac{1}{x}$ (on $(1,\infty)$) is not. That is, $f \in L^q(\mathbb{R})$ but $f \notin L^p(\mathbb{R}) $.


2

In what follows, we assume that $ G $ is a locally compact Hausdorff group that does not have to be abelian. You can certainly define an involution $ ^{*} $ on $ {L^{1}}(G) $ by $$ \forall f \in {L^{1}}(G), ~ \forall x \in G: \quad {f^{*}}(x) \stackrel{\text{df}}{=} \overline{f(x^{-1})} \cdot \Delta(x^{-1}), $$ where $ \Delta $ denotes the modular ...


2

Let $\langle\, f,g\rangle=\int_E fg\,d\mu$. We have that $$ \|\,f_n-f\|^2=\langle\, f_n-f,f_n-f\rangle=\langle\, f_n,f_n\rangle-2\langle\, f_n,f\rangle+\langle\,f,f\rangle. $$ Hence, as $\langle\, f_n,f\rangle\to 0$, we have that if $\langle\, f_n,f_n\rangle\to\langle\, f,f\rangle$, then $\|\,f_n-f\|\to 0$.


2

Suppose that $(f_n)_{n\geqslant 1}$ is a sequence of elements of $S$ converging in $\mathbb L^p(I^2)$ to some function $f$. Then we extract a subsequence $(f_{n_k})_{n\geqslant 1}$ which converges almost everywhere to $f$. There exists a function $g_{n_k}$ such that for almost every $(x,y)\in I$, $f_{n_k}(x,y)=g(x)$. Thus, for almost every $(x,y)\in I$, we ...


2

Remarks: The argument below proves the following. Let $1<r<\infty$ and $f\in L^r(\Omega, \Sigma, \mu)$ for some probability space $(\Omega, \Sigma, \mu)$. Then, the function $\phi:[1,r]\to$$\mathbb R$ defined by $\phi(p)=\|f\|^{p}_{L^p}$ is convex. Choose $\theta\in[0,1]$ and $p,q\in[1,r]$. Then, $\theta p+(1-\theta)q\in[1,r]$. Note that, by the ...


2

This answers the original question, wich asked for an example of strong ($L^1$) convergence instead of pointwise a.e. convergence. There is no such sequence because strong convergence implies weak convergence in normed spaces. For the other way around, look at $f_n(x) = \sin(nx) \to 0$ weakly, but not strongly in $L^2 [0,2\pi]$. $$\|f_n\|_{L^2(\mathbb T)}^2 ...


2

Consider the function $f : \mathbb{R} \to \mathbb{R}$, $$f(x) = \begin{cases} x^{-1/3} & 0 < |x| < 1\\ 0 & \text{otherwise}. \end{cases}$$ Note that $$\int_{\mathbb{R}}|f|^2dm = 2\int_0^1x^{-2/3} = 2\left[3x^{1/3}\right]_0^1 = 6 < \infty,$$ so $f \in L^2(\mathbb{R})$. Now let $M > 0$ and let $K = \sqrt[3]{M}$. Note that $\{x \in ...


1

Riesz-Thorin interpolation (RTI) bounds the norms of linear maps acting between $L^p$ spaces. Unlike Marcinkiewicz interpolation, RTI only works for strong type operators. Overview of interpolation: to understand how to obtain control on the expression $||Tf||_{L^q}$ for operator $T$ & function $f,$ one would divide $f$ into two (or more) components ...


1

Write $h(x) = f(x)g(x)$. Without loss of generality (by multiplying $g$ and $h$ by constants) we can assume $$ \int |g|^{p'} dx = 1 = \int |h| dx $$ So what we want to prove reduces to $$ \int |h / g|^p \leq 1 $$ To prove this, apply the regular Holder inequality: put $h^p$ in $L^{1/p}$ and $1/g^p$ in $L^{1/1-p}$. Observe that $$ \frac{1}{|g|^{p/(1-p)}} ...


1

For $L^p(\mathbb{R})$ and $L^q(\mathbb{R})$, suppose $p < q$. Let $f(x) = x^{-2/(p+q)}$ for $x > 0$ and $0$ otherwise. Let $f_p(x) = f(x)$ if $x < 1$ and $0$ otherwise, and $f_q(x) = f(x)$ if $x > 1$ and $0$ otherwise. Then $f_p\in L^p \setminus L^q$ and $f_q \in L^q \setminus L^p$. For $L^q[0,1]$ as a intersection of $L^p$ with $p < q$. The ...


1

In terms of getting a proof done, it really is just Cauchy-Schwarz: $$\left | \int_0^t f(s) ds \right | = (f,1)_{L^2} \leq \| f \|_{L^2} \| 1 \|_{L^2} = t^{1/2} \left ( \int_0^t f(s)^2 ds \right )^{1/2}.$$ Cauchy-Schwarz can be proven in the abstract setting of a general inner product space, and then it follows in your case from the fact that the $L^2$ ...


1

Use $f$ to construct a densely defined functional $\delta:g\mapsto\int fg$ on $L^p$, $0<p<1$, that is bounded. Extend by continuity. Show that the space $L^p$, for $0<p<1$ doesn't have non-zero bounded linear functionals: A non-zero bounded linear functional $\delta$ produces non trivial open convex sets like $\{x:\delta(x)<1\}$. Show that ...


1

For $0<s<1$ this is true in all dimensions $n$, and follows by writing the norm as in integral of divided differences [Leoni, A first course in Sobolev spaces, 14.8] $$ \|f\|_{H^s}^2 \approx \|f\|_{L^2}^2+\iint\frac{|f(x)-f(y)|^2}{|x-y|^{n+2s}}\,dx\,dy \tag{1} $$ Indeed, (1) shows at once that for any Lipschitz function $\varphi$ fixing $0$ we have ...


1

It's not true as written. Counterexample: Let $X=[0,1]$ with Lebesgue measure. Let $f_1(x)=f_2(x)=x$. Let $g$ be a convex function whose graph goes through $(0,0)$, $(\frac32,\varepsilon)$, $(2,1)$, and is linear in between those points. (That's not strictly convex, but we can perturb the example a bit to get strict convexity if we insist.) Then ...


1

A $C^*$-algebra $A$ is isometric to an $L_1$-space (even as Banach space!) iff it is one dimensional. Assume $A$ is isometric to $L_1$ space and $\operatorname{dim}(A)>1$, then $A$ is weakly sequentially complete. By result of Sakai (proposition 2), this is possible only if $A$ is finite dimensional. By classification theorem for $C^*$ algebras we know ...


1

Let $f \in \mathfrak{L}^2$$(0, 1)$, and let $g(x) = \int_{[0,x]} \frac{f(t)}{\sqrt{1-t^2}}dt$. Prove $g \in \mathfrak{L}^2$$(0, 1)$. $$\int_0^1 g(x)^2 dx = \int_0^1 \left( \int_0^x \frac{f(t)}{\sqrt{1-t^2}}dt \right)^2 dx$$ then, you use Jensen inequality (with a rescaling): $$\leq \int_0^1 \frac{1}{x}\int_0^x \left( x\frac{f(t)}{\sqrt{1-t^2}} \right)^2 ...


1

The function names in $\|f+g\|_p \leq \|f\|_p + \|g\|_p$ are placeholders. Functions don't have to be named $f$ and $g$ in order for the inequality to work; they don't have to be named at all — you can plug in some algebraic combinations of functions in place of $f$ and $g$. For example, $$\|24f^2+3f\cos(g)\|_p \leq \|24f^2\|_p + \|3f\cos(g)\|_p$$ ...



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