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1

Since $f\in L^\infty(X,\mu)$ and $\mu(X) < \infty$, then $f\in L^n(X,\mu)$ for all $n \ge 1$. Since $$\alpha_{n+1} \le \|f\|_\infty \int_X |f|^n\, d\mu = \|f\|_\infty \alpha_n,$$ we have $$\varlimsup_{n\to \infty} \frac{\alpha_{n+1}}{\alpha_n} \le \|f\|_\infty.$$ On the other hand, since $\alpha_n^{1/n} = \|f\|_n \to \|f\|_\infty$ as $n\to \infty$, ...


0

As Frank provided a counterexample the inclusion in general doesn't hold. In fact if your measure is finite the inclusion is vice versa because via Hölder you get \[ \|f\|_q \leq \mu(X)^{\frac{1}{p} -\frac{1}{q}} \cdot \|f\|_p.\] Maybe you are talking about the sequence spaces $\ell^p$ because there the inclusion holds as every function which is in ...


0

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 ...


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}) $.


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

It is enough to prove that if $f\in L^q(\mu)$ is such that $\int_{\cal X}fg=0$ for all $g\in L^p(\mu)$, then $f=0$ a.e. Let $g_1:{\cal X}\to \mathbb{R}$ be defined by $$ g_1(x)=\begin{cases} f(x)|f(x)|^{q-2} &\mbox{ if } f(x)\ne 0\\ 0 &\mbox{ otherwise } \end{cases} $$ We have $$ \int_{\cal X}|g_1|^p\,d\mu=\int_{\{f\ne ...


1

Let $f= f_1 - f_2$. Then, $fg \in L^1$ by Holder. You are told that $$ \int f\, g \, d\mu = 0 $$ for all $g \in L^p$. Let $X = \cup X_i$ where each $X_i$ has finite measure. (Here, we use the sigma-finite.) Fix $i$ and let $A \subseteq X_i$ be the points in $X_i$ where $f > 0$. Then, $1_{A} \in L^p$ since $A$ has finite measure. Moreover, $$ \int f\, ...


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 ...


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, $$\|24f^2+3f\cos(g)\|_p \leq \|24f^2\|_p + \|3f\cos(g)\|_p$$ ...


0

For a quick proof, observe that the map $T: L^p(I) \to L^p(I^2)$ that sends $g$ to such an $f$ is a linear map that preserves norms, so it can be thought of as an isometric embedding. An isometrically embedded image of a complete space is always closed in the ambient space.


0

Use Holder's Inequality, we have $$\int_{A}{}|f|d\mu\leq\left(\int_{A}{}|f|^pd\mu\right)^{1/p}\left(\int_{A}{}1^qd\mu\right)^{1/q}=\left(\int_{A}{}|f|^pd\mu\right)^{1/p}\left(\mu(A)\right)^{1/q}<∞ $$ $\text{ if }\left(\int _{A}{}|f|^pd\mu\right)^{1/p}<∞$. So if $f\in L^{p}(X)$, then $f\in L^{1}(X)$.


0

If $f\in L^p(\mathbb{R}^n;\mathbb{R}^m)$, then $f=(f_1,\ldots, f_m)$ with $f_j\in L^p(\mathbb{R}^n)$ for $j=1,\ldots, m$, therefore $$|f|=\sqrt{f_1^2+\ldots+ f_m^2}\leq \sqrt{n}\max_j |f_j|$$ is in $L^p(\mathbb{R}^n)$ (because the maximum of $L^p$ functions is in $L^p$). On the other hand, as $$|f_j|\leq\sqrt{|f_1|^2+\ldots+|f_m|^2}=|f|^2$$ if the latter is ...


1

No, your understanding is not correct. All simple functions are measurable, but in most $\sigma$-algebras of interest, far more functions are also measurable. For instance, all continuous functions are measurable for the Lebesgue $\sigma$-algebra on $\mathbb{R}$, but only constant continuous functions are simple. What is true is that all measurable ...


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 ...


0

It seems the following. Put $\mathcal L={\oplus}_{p=1}^\infty \ell_p$ and $\widehat{\mathcal L}=\Pi_{p=1}^\infty \ell_p$. There can be many norms on $\mathcal L$, but at least I expect that the restriction $\|\cdot\||_{\ell_p}$ of the norm $\|\cdot\|$ which you consider on $\mathcal L$, coincides with the standard norm $\|\cdot\|_p$ of $\ell_p$, which is ...


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 $p\lt\infty$. ...


0

This question Limit of convolution answers the question for general convolution operator on $L^p$.


1

You need to cut the study in two : On $A=[-1,1]$, you have $\|f\|_{1,A}^2 \leq 2\|f\|_{2,A}^2 \leq 2\|f\|_2^2$ (Jensen inequality) On $A^c=[-1,1]^c$, you have $$\|f\|_{1,A^c} = \left\| \frac{g}{x} \right\|_{1,A^c} \leq \|g\|_{2,A^c}\|\frac{1}{x}\|_{2,A^c} \leq \sqrt{2} \|g\|_2$$ Hence $$\|f\|_1 \leq \sqrt{2}(\|f\|_2+\|g\|_2)$$


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

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 ...


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 ...


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 ...


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)}} ...


0

The answer is yes. This follows from the Hahn-Banach theorem. Indeed, for each non-zero element $x$ in a Banach space $X$, there is a norm-one functional $f\in X^*$ such that $\langle f,x\rangle = \|x\|$. To see this, consider the one-dimensional subspace spanned by $x$. Let $\langle f,cx\rangle = c\|x\|$ ($c$ is a scalar). Then $f$ is a norm-one ...


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

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| ...


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.


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 $L^2$ ...


0

Let $e_i :\mathbb N\to\mathbb R$ defined as $$ e_i(j)=\left\{\begin{array}{lll} 1 &\text{if}&i=j,\\ 0 &\text{if}&i\ne j.\end{array}\right. $$ Then $e_i\in X$. If $f\in Y$, then setting $$ f_n=\sum_{i=1}^n f(i)\, e_i\in X, $$ we observe that $$ \|f-f_n\|_{L^1}=\sum_{i>n}\lvert\,f(i)\rvert. $$ Clearly, the right hand side of the above tends ...


1

Hint: Think about sequences that are $0$ except for finitely many terms.


1

No. Here is a hint on how to construct a counterexample, put $$g(x,y)= \sin(y\,h(x)),$$ then $g'_y(x,y)=h(x)\cos(y\,h(x))$.


0

Here is one of my favorites, $$f(x) = \begin{cases} n & n < x < n+\frac{1}{n^4},\, n\in\mathbb{Z}^+\\ 0 & \text{otherwise}. \end{cases}$$ Then $$\limsup_{x\to\infty}f(x)=+\infty$$ while $$\int_\mathbb{R}|f(x)|^2\,dx=\frac{\pi^2}{3}$$


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 ...


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

Let $(g_{n})$ be a sequence in $K$ satisfying $\left\|f-g_{n}\right\|_{p}\rightarrow\delta:=\inf_{g\in K}\left\|f-g\right\|_{p}$. It follows from the triangle inequality that the $g_{n}$ are bounded in norm, and therefore lie in a closed, convex subset $E\subset K$, which is weakly compact by Alaoglu's theorem. Whence $(g_{n})$ has a subset ...


4

It's equivalent to prove $T$ is continuous. One theorem to get $T$ is continuous is the closed graph theorem. So can you show the graph of ($x$,$Tx$) is closed? Suppose $x_n \rightarrow x$ and $Tx_n \rightarrow y$ and we want to show $Tx=y$. Indded, $\langle Tx, z \rangle = \langle x, T^*z \rangle =\lim_{n \to \infty} \langle x_n, T^*z \rangle = \lim_{n ...


1

Yes, because $g(x) \leq ||g||_{\infty}$ for a.e $x$, so $$||fg||_2 = \bigg( \int |f(x)|^2|g(x)|^2 d\mu(x) \bigg)^{\frac{1}{2}} \leq \bigg( \int |f(x)|^2||g||_{\infty}^2 d\mu(x) \bigg)^{\frac{1}{2}}=\bigg( ||g||_{\infty}^2 \int |f(x)|^2 d\mu(x) \bigg)^{\frac{1}{2}}=||g||_{\infty} \bigg( \int |f(x)|^2 d\mu(x) \bigg)^{\frac{1}{2}}=||g||_{\infty} ||f||_2$$


3

The inequality you are having trouble with is actually trivial. The "of course" one is actually false. $$\|fg\|_2^2 = \int_\Omega |fg|^2 \, dx \le \|g\|_\infty^2 \int_\Omega |f|^2 \, dx = \|f\|_2^2 \|g\|_\infty^2.$$ On the other hand, if $\Omega = (0,1)$ and $f(x) = g(x) = x^{-1/3}$, you have $\|fg\|_2 = \infty$ but $\|f\|_2 = \|g\|_2 < \infty$.


2

Yes, in fact if $p \ge 1$, $f \in L^p(\Omega)$ and $g \in L^\infty(\Omega)$, then $\|fg\|_p \le \|f\|_p \|g\|_\infty$. For $|f(x)g(x)|\le |f(x)|\|g\|_{\infty}$ for almost every $x\in \Omega$, so $$\|fg\|^p_p = \int_\Omega |fg|^p \le \int_\Omega |f|^p \|g\|_\infty^p = \|f\|_p^p\, \|g\|_\infty^p,$$ which implies $\|fg\|_p \le \|f\|_p\|g\|_\infty$.


3

Using the Holder inequality, we have $$ \left(\int_0^1 |x\,f(x)|\,dx\right)^3 \leq \left(\int_0^1 x^{3/2}\,dx\right)^2 \left(\int_0^1 |f(x)|^3\,dx\right) = \frac{4}{25} \int_0^1 |f(x)|^3\,dx. $$ But from our equation above we know this must be an equality. The Holder inequality is sharp exactly when one function is a constant multiple of the other, so we ...


2

For the last term, use $$ |\mathbb{E} X(V_n-V) 1_{|X|>N}| \leq C |\mathbb{E} X 1_{\{|X|>N\}}|, $$ where $C$ is our bound on $V_n$. So using your argument above you get a bound like $$ \lim_{n\to\infty} |\mathbb{E}X_nV_n - \mathbb{E}XV| \leq C |\mathbb{E} X1_{\{|X|>N\}}|, $$ which holds for all $N$. But $X\in L^1$, so as $N\to\infty$ the right ...


0

Let me give this a shot. So $f\in L^{p,\infty}$ means that $$ \mu(\{|f(x)|>t\}) \leq Ct^{-p} $$ for a constant $C$ independent of $t$. This is useful for large $t$. We also know that $\mu(\{|f|>0\}) < \infty$, so for another constant we have $$ \mu(\{|f(x)|>t\}) \leq D, $$ and this is useful for small $t$. Like you suggested, we'll use the ...


2

$\int_{m - \beta}^{m + \beta} |f(x)|^2 \,dx$ is the same as $\int_{-\infty}^{\infty} \chi_{[m - \beta, m + \beta]}(x)|f(x)|^2 \,dx$, where $\chi_A(x)$ means the characteristic function of $A$. So $$\sum_{m \in {\mathbb Z}}\int_{m - \beta}^{m + \beta} |f(x)|^2 \,dx = \sum_{m \in {\mathbb Z}}\int_{-\infty}^{\infty} \chi_{[m - \beta, m + \beta]}(x)|f(x)|^2 ...


2

If $\beta \leq 1/2$, then $$\sum_{m \in \mathbb{Z}} \int_{m - \beta}^{m + \beta} |f(x)|^2 dx \leq \sum_{m \in \mathbb{Z}} \int_{m - 1/2}^{m + 1/2} |f(x)|^2 dx = \int_{\mathbb{R}} |f(x)|^2 dx < \infty$$ If $\beta > 1/2$, then let $K = \lceil \beta \rceil$. Since the interval $[m - \beta, m + \beta]$ is contained in $\bigcup_{k = -K}^{K} [m + k - 1/2, ...



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