I found this as an exercise in Kolmogorov-Fomin, let me state the problem precisely:  Let $S_{\infty}$ be the space of functions on the real line for which there exists a colelction of constants $C_{pq}$ for all $p, q$ such that  $$ |x^pf^{(q)}(x)|\leq C_{pq}. $$ Is it true that if for all $p\geq 0$ $\int_{-\infty}^{\infty} x^pf(x)=0$, then $f=0$?  The exercise follows a section on the Fourier transform of functions of $S_{\infty}$. Now, I guess one should prove this exercise (I am fairly sure it is true) using some properties of the Fourier transform, but I cannot see how.   So my question is, how is this to be proved with the fundamental properties of the Fourier transform in $S_{\infty}$?
120818
Let $f\in S_\infty\subset L_1(\mathbb{R},\mu)$ with $\mu$ as the Lebesgue linear measure be a Lebesgue-summable function such that $$\forall (p,q)\in\mathbb{N}^2_{\ge 0}\quad\exists C_{pq}&gt;0: \Bigg|x^p\frac{d^q}{d x^q}f(x)\Bigg|&lt; C_{pq}$$ I was wondering whether, if $\forall p\in\mathbb{N_{\ge 0}}\quad\int_{\mathbb{R}}x^pf(x)d\mu=0$, then $f$ is constantly, or almost everywhere, null. I cannot find a counterexample and therefore I think that the implication might well hold, but I cannot prove it either.  Since $x^p f(x)$ belongs to $S_\infty\subset L_1$, thanks to the fact that $f$ belongs to it, and is continuous I think that the Lebesgue integral and the Riemann improper integral $\mathscr{R}\int_{-\infty}^\infty t^pf(t)dt$ are the same. Please correct if I am wrong.  Nevertheless I cannot use calculus facts to prove the desired implication... I have also tried using the fact that the Fourier transform induces a bijection $S_\infty\to S_\infty$, but with no result. What do you think about it? Has anybody got a counterexample or proof? Thank you very much!
25170
In a space with measure $1$, $||f||_p$ is an increasing function with respect of $p$. To show that $\lim_{p \rightarrow \infty} ||f||_p=||f||_{\infty}$ we have to show that  $||f||_{\infty}$ is the supremum, right??   To show that, we assume that $||f||_{\infty}-\epsilon$ is the supremum.  From the essential supremum we have that $m(\{|f|&gt;||f||_{\infty}-\epsilon\})=0$.  So, we have to show that $m(\{|f|&gt;||f||_{\infty}-\epsilon\})&gt;0$.   Let  $A=\{|f|&gt;||f||_{\infty}-\epsilon\}$.   We have that $\int_A |f|^p \leq \int |f|^p \leq ||f||_{\infty}^p$.   $\int_A |f|^p &gt;\int_A (||f||_{\infty}-\epsilon)^p=(||f||_{\infty}-\epsilon)^p m(A)$   So, $m(A)^{1/p} (||f||_{\infty}-\epsilon)&lt;||f||_p \leq ||f||_{\infty}$  How could we continue to show that $m(A)&gt;0$??   EDIT:   Is it as followed??  We have that $0&lt;||f||_{\infty}-\epsilon&lt;||f||_{\infty}$ for some $\epsilon&gt;0$.  $||f||_{\infty}$ is the essential supremum. So, from the definition we have that $m\left ( \{|f(x)&gt;||f||_{\infty}-\epsilon\} \right )&gt;0$.   Let  $A=\{|f|&gt;||f||_{\infty}-\epsilon\}$.   We have that $\int_A |f|^p \leq \int |f|^p \leq ||f||_{\infty}^p$.   $\int_A |f|^p &gt;\int_A (||f||_{\infty}-\epsilon)^p=(||f||_{\infty}-\epsilon)^p m(A)$   So, $m(A)^{1/p} (||f||_{\infty}-\epsilon)&lt;||f||_p \leq ||f||_{\infty}$  Taking the limit $p \rightarrow +\infty$ we have the following:  $$\lim_{p \rightarrow +\infty}m(A)^{1/p} (||f||_{\infty}-\epsilon)&lt;\lim_{p \rightarrow +\infty} ||f||_p \overset{ m(A)&gt;0 \Rightarrow \lim_{p \rightarrow +\infty}m(A)^{1/p}=1}{\Longrightarrow} \\ ||f||_{\infty}-\epsilon&lt;\lim_{p \rightarrow +\infty} ||f||_p$$   Is this correct?? How do we conclude that $\lim_{p \rightarrow +\infty} ||f||_p=||f||_{\infty}$ ??
63070
$\mathbf{Question}:$ Let $f$ be a continuous function on $[0,1]$. Then prove that the limit $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx$ is equal to $f(1)$.   $\mathbf{Attempt}$: First, we try to show that the sequence of functions $\{nx^nf(x)\}_{x\in [0,1]}$ is uniformly convergent to $0$ on the restricted domain $[0,1-\epsilon]$, $0&lt;\epsilon&lt;1$.   Let $\sup_{x\in[0,1]}f(x)= \mathcal{M}$. Then $|{nx^nf(x)}|\leq n(1-\epsilon)^n|\mathcal{M}|$ for any $x$ and $n(1-\epsilon)^n \to 0$ as $n\to \infty$.  Thereby, $\lim_{n \to \infty}\int_0^{1-\epsilon}nx^nf(x)dx=\int_0^{1-\epsilon}{\lim_{n \to \infty} }nx^nf(x)dx=0$  Now, denote $\sup_{x\in [1-\epsilon,1]} f(x)=M(\epsilon)$ and $\inf_{x\in[1-\epsilon,1]}f(x)=m(\epsilon)$.  $\int_{1-\epsilon}^1nx^n\ m(\epsilon)dx\leq\int_{1-\epsilon}^{1}nx^nf(x)dx\leq \int_{1-\epsilon}^1 nx^n\ M(\epsilon)dx$.  Now,  $\lim_{n \to \infty}\int_{1-\epsilon}^1 nx^n\ M(\epsilon)dx =\displaystyle \lim_{n\to\infty}\bigg[ \frac{nx^{n+1}}{n+1}M(\epsilon)\bigg]_{1-\epsilon}^1=M(\epsilon)$ [Similarly the other one is $m(\epsilon)$]  Clearly, as $\epsilon \to 0 , \ \ M(\epsilon), m(\epsilon) \to f(1)$, so $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx=f(1)$.  Is the procedure correct? Kindly verify.
107662
Let $f:[0,\infty) \to \mathbb{C}$ be a continuous and bounded function such that the limit $$\lim_{T\to\infty} \frac{1}{T} \int_{0}^{T}f(t)dt = d \quad\text{exists}.$$  Let $\hat{f}$ be the Laplace transform of f, i.e., $$\hat{f} = \int_{0}^{\infty}e^{-st}f(t)dt.$$  Prove that $$\lim_{s\to0,\:s&gt;0} s\hat{f} = d.$$  I have tried different ways, but for now I still did not get the whole proof.   I started with the following idea (non rigorous). Define $T=\frac{1}{s}$, then $$\lim_{s\to0,\:s&gt;0} s\hat{f} = \lim_{T\to\infty} \frac{1}{T}\hat{f}=\lim_{T\to\infty} \frac{1}{T} \int_{0}^{T}e^{-t/T}f(t)dt$$ where the integrand converges to $f(t)$ and therefore the dominant convergence theorem I guess could be used. However, there are two problems: (a) $T \in \mathbb{C}$ and therefore I am not sure I can do that, (b) I don't want to use the dominant convergence theorem since it was not introduced in class. A different approach is to use the fundamental theorem of calculus since f continuous and define: $$F(T) = F(0) + \int_{0}^{T}f(t)dt \quad \forall \; T \in [0,\infty).$$ Then from here use the formula for the Laplace transform of the derivative, but I did not manage to move on.
95627
Does there exist a continuous function on $[0, \infty)$  such that it is in $L^p(0,\infty)$ for some $p\in [1,2]$ but is not in $L^q(0,\infty)$ for any $q\in (2, 2/(2-p))$?    Thanks!
3887
If I know $\| f_n - f \|_{L^p(\mathbb{R})} \to 0$ as $n \to \infty$, do I know that $\lim_{n \to \infty}f_n(x) = f(x)$ for almost every $x$?
80116
Suppose $\{e_i\}_{i\in I}$ is an orthonormal basis for some Hilbert space. $\{f_i\}_{i\in I}$ is an orthonormal family with the property $\sum^\infty_{i\in I} \|e_i-f_i\|^2&lt;\infty$, show that $\{f_i\}_{i\in I}$ is also orthonormal basis.   I can think of for a $x$ with $(x,f_i)=0$ for all $i$, prove $x=0$. However, I don't know how to continue to the next step.
43867
I ran into the following problem when I was doing my homework, and I have no thoughts on where I should start with:  (1) If $f\in L^{2}$, show that $\displaystyle \lim_{p \rightarrow 1^{+}}\int_{[0,1]}|f|^{p}=\int_{[0,1]}|f|$  (2) If $0&lt;p$, show that $\displaystyle \lim_{q\rightarrow p^{-}}||f||_{q}=||f||_{p}$  My first thought was Generalized LDCT, but it didn't seem to work. I also made some other attempts but none of them were successful... Can anybody give me some hints on how I should look at this question?  Also, I know if $p\rightarrow\infty$ then $||f||_{p}\rightarrow||f||_{\infty}$ on $[0,1]$, but does similar continuity in p holds for other $L^{p}[0,1]$ norms in general?  Thank you!  Edit:   Sorry if I did not make it clear enough in the question. All $L^{p}$ refers to $L^p[0,1]$.  The first question is found  (thanks to t.b.), but the second question remains, mainly because $f$ is not guaranteed to be in any $L^{p}$.
29279
Consider a finite Borel measure $\mu$ on $\mathbb R$. The Fourier transform $\hat{\mu}$ of $\mu$ is defined by $\hat{\mu} (\xi)= \int _{\mathbb R} e^{-ix\xi} d\mu(x)$. I would like to prove the following identity:  $$\lim_{R\to \infty}\frac{1}{2R}\int_{-R}^R {|\hat{\mu}(\xi)|}^2d\xi=\sum_{x\in \mathbb R} \mu({\{x\}}) ^2$$.  My attempt:  By Jensen's inequality,  $$\int_{-R}^R {|\hat{\mu}(\xi)|}^2d\xi=\int_{-R}^R|\int _{\mathbb R} e^{-ix\xi} d\mu(x)|^2d\xi\leq\int_{-R}^R\int _{\mathbb R} e^{-2ix\xi} d\mu(x)d\xi$$  By Fubini,  $$\int_{-R}^R\int _{\mathbb R} e^{-2ix\xi} d\mu(x)d\xi=\int _{\mathbb R}\int_{-R}^R\ e^{-2ix\xi} d\xi d\mu(x)$$ But I don't know how to proceed from here. Any advice will be appreciated.
26271
Given measure space $(X, \mu)$ and measurable function $f$. If $p \in (0, +\infty]$, below are the facts I know about $\Vert f \Vert_p$:  (log convex in $\frac 1 p$) $\forall p_1, p_2\in (0, +\infty] \text{ such that } p_1 &lt; p_2. \forall t \in(0, 1), \text{ let } \frac 1 p = \frac t {p_1} + \frac {1-t} {p_2}$, then $\Vert f \Vert_p \le \Vert f \Vert_{p_1}^t\Vert f \Vert_{p_2}^{1-t}$ ($0\cdot(+\infty)$ is defined to be $0$). (lower semi-continuity) $\forall p_0 \in (0, +\infty], \liminf_{p\to p_0}\Vert f\Vert_p \ge \Vert f \Vert_{p_0}$. If $0&lt;\mu(X) &lt; +\infty$, then $ \Vert f \Vert_p/\mu(X)^{\frac 1 p}$ is monotone increasing in $p$, combine it with (2) we get $\lim_{p\to p_0-}\Vert f\Vert_p = \Vert f \Vert_{p_0}$.  From (1) we know the set of $p$ such that $\Vert f \Vert_p &lt; +\infty$ is an interval $I$ (possibly empty or a singleton).  If $I \not\in a = \inf I$ and $a \neq 0$, from (2) we know $\lim_{p\to a+} \Vert f \Vert_p = +\infty$. If $I \not\in b = \sup I$, then $\lim_{p\to b-} \Vert f \Vert_p = +\infty$.  From (1) and the dominated convergence theorem we know that if $I \ni p_0 \neq \inf I$ and $p_0 &lt; +\infty$, then $\lim_{p\to p_0+} \Vert f \Vert_p = \Vert f \Vert_{p_0}$. If $I \ni p_0 \neq \sup I$, then $\lim_{p\to p_0-} \Vert f \Vert_p = \Vert f \Vert_{p_0}$.  If $I$ is nondegenerate and $+\infty \in I$, then $\lim_{p\to +\infty}\|f\|_p = \|f\|_{\infty}$.  If $\inf I = 0$, from (1) we know $\lim_{p\to 0+}\|f\|_p$ exists.   Question: How $\Vert f \Vert_p$ behaves when $p$ is the endpoint of $I$? To be more precise, assume $I$ is nonempty. Let $p_0 = \sup I$ (only right endpoint is considered for simplicity):  $p_0\in I$ or $p_0 \not\in I$? Can $I$ be a singleton? What if $p_0 = +\infty$? What if $0&lt;\mu(X) &lt; +\infty$?  Please also notify me if there is any mistake above.
43389
I found  the following exercise and I'd like to know if my answer is correct.      Let $(X, \mathscr A, \mu)$ a finite measure space. Let $\{f_n\}$ a sequence of measurable functions such that $\|f_n\|_p\le M$ for a real constant $M$ $(1&lt;p&lt;\infty)$ and suppose that $f_n \xrightarrow{\text{a.e.}} f$. Then $$\int_X f_n g \,d\mu \to \int_X fg \,d\mu$$ for all $g$ in $\mathscr{L}^q (X)$, where $q^{-1}=1- p^{-1}$.   Since $\int |f_n|^p \,d\mu\le M^p $ by Fatou's lemma it follows that $\int |f|^p \,d\mu\le M^p&lt;\infty$, so $f$ belongs to $\mathscr{L}^p (X)$. Let $g$ in $\mathscr{L}^q$ arbitrary but fixed.   Then by Hölder inequality it follows that $$\|f_n g\|_1 \le \|f_n\|_p\|g\|_q\le M \|g\|_q&lt;\infty$$ and $$\|f g\|_1 \le \|f\|_p\|g\|_q \le M \|g\|_q&lt;\infty$$ thus $f_n g$ and $fg$ belongs to $\mathscr L ^1(X)$. Now since $g$ is in $\mathscr L ^q (X)$, then $g$ is finite a.e.,without loss of generality we may assume that $g$ is real-valued so $f_n g \xrightarrow{ a.e} fg$.   Let $\nu (A) =  \int_A |g|^q \,d\mu $, so $\nu $ is absolutely continuous with respect to $\mu$, also $\nu$ is a finite measure on $(X, \mathscr A )$. We can use the $\epsilon-\delta$ definition of absolutely continuous, so given $\epsilon&gt;0$ there is a $\delta&gt;0$ such that for $\mu(A)&lt;\delta$ then $\nu (A) &lt;\epsilon^q$ for any $A$ in $\mathscr A$.   Now by Egoroff's thm exists a $B$ in $\mathscr A$ such that $(f_ng) (x)\xrightarrow{\text{uniformly}} (fg) (x)$ for $x$ in $B$ and $\mu(X\setminus B) &lt;\delta$. Let $N$ such that for all $n\ge N$, $|(f_ng)(x)-(fg)(x)|&lt;\epsilon$ for all $x\in B$. Thus   \begin{align*}\left |\int_X (f_n-f) g \, d\mu \right|&amp;\le \int_B |f_ng-fg| \, d\mu +\int_{X-B}|f_n g|\, d\mu +\int_{X-B}|fg|\,d\mu \\[6pt] &amp;\le \int_B |f_ng-fg| d\mu + 2M \left( \int_{X-B}|g|^qd\mu \right)^{1/q}\\[6pt] &amp;\le\epsilon \mu(X)+2M  (\nu(X\setminus B))^{1/q}\\[6pt] &amp;\le \epsilon (\mu(X)+2M)\end{align*}  Since $\mu(x)&lt;\infty$ and $M&lt;\infty$ the result follows.
96450
Just wanted to confirm that this is a correct solution:    Proof: Suppose $f(x_0) &gt; 0$ for some $x_0 \in [a,b]$. Then, by continuity of $f$, for $\epsilon &lt; f(x_0)$, there exists $\delta &gt; 0$ so that, if $|x-x_0| &lt; \delta$, we have $|f(x) - f(x_0)| &lt; \epsilon$. That is to say, $f(x)$ is non-zero in a neighborhood $B_\delta (x_0)$. Now,   \begin{eqnarray*} U(f,P) &amp;=&amp; \sum_{P([a,b])} M_i \Delta_i \\ &amp;=&amp; \sum_{P(B_\delta(x_0))} M_i \Delta_i \\ &amp;=&amp; \sup_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \sup_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \sup_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &amp;&gt;&amp; 0 \end{eqnarray*}  In the same manner,   \begin{eqnarray*} L(f,P) &amp;=&amp; \sum_{P([a,b])} m_i \Delta_i \\ &amp;=&amp; \sum_{P(B_\delta(x_0))} m_i \Delta_i \\ &amp;=&amp; \inf_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \inf_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \inf_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &amp;\geq&amp; 0 \end{eqnarray*}    We can conclude from here that $\int_{a}^{b} f(x) dx &gt; 0$.  Is this a valid proof?
77706
Theorem 2.1 : Suppose that $f$ is an integrable function on the circle with $\hat f(n)=0$ for all $n \in \Bbb Z$. Then $f(\theta_0)=0$ whenever $f$ is continuous at the point $\theta_0$.      Proof : We suppose first that $f$ is real-valued, and argue by contradiction. Assume, without loss of generality, that $f$ is defined on $[-\pi,\pi]$, that $\theta_0=0$, and $f(0) \gt 0$.      Since $f$ is continuous at $0$, we can choose $ 0\lt \delta \le \frac \pi2$, so that $f(\theta) \gt \frac {f(0)}2$ whenever $|\theta| \lt \delta$. Let $$p(\theta)=\epsilon + \cos\theta,$$   Where $\epsilon \gt 0$ is chosen so small that $|p(\theta)| \lt 1 - \frac \epsilon2$, whenever $\delta \le |\theta| \le \pi$. Then, choose a positive $\eta$ with $\eta \lt \delta$, so that $p(\theta) \ge 1 + \frac \epsilon2$, for $|\theta| \lt \eta$. Finally, let $p_k(\theta)=|p(\theta)|^k$, and select $B$ so that $|f(\theta)| \le B$ for all $\theta$. This is possible since $f$ is integrable, hence bounded.      By construction, each $p_k$ is a trigonometric polynomial, and since $\hat f(n)=$ for all $n$, we must have $\int_{-\pi}^{\pi} f(\theta)p_k(\theta)\,d\theta=0$ for all $k$.   I understood the first paragraph clearly. But the rest is not making it's way into my head.   In the beginning of second paragraph, how does the given range works for choosing $\delta$? If the continuity is used to get the range, then how? How can we choose $\epsilon$ so small such that, $|p(\theta)| \lt 1 - \frac \epsilon2$, whenever $\delta \le |\theta| \le \pi$? How can we choose positive $\eta$ with $\eta \lt \delta$, so that $p(\theta) \ge 1+ \frac \epsilon2$, for $|\theta| \lt \eta$. Why do we must have $\int_{-\pi}^{\pi}f(\theta)p_k(\theta)\,d\theta=0$ for all $k$?
64748
Recall that $\|f\|_\infty = \inf\{M\ |\ |f(x)|\leq M \text{ for almost every } x \in X\}$, for every function $f$ defined on $X$. Show that, if $\|f\|_\infty$ is finite, then $\|f\|_\infty$ is also the smallest of all numbers of the form $\sup\{|g(x)| : x \in X \}$, where $f = g$ almost everywhere with respect to the Lesbegue measure.   I am trying to understand how $L^{\infty}$ is defined. Could someone help me see how to prove the above fact?
73379
I'm studying Quantum Mechanics and I came across this which I don't quite understand: For a vector space of functions $f(x)$ to be square integrable (i.e $\int{|f(x)|^2dx &lt; \infty)}$, the necessary condition is $\lim_{|x| \to \infty}f(x) \to 0$. Can someone help me understand why that must be the case? I sort of understand that it must not blow up at the infinities but can a (continuous or piece-wise continuous) function have a non-zero finite value at the infinities and still be square integrable?
74985
For a compactly supported distribution $f$, I want to show the non-surjectivity of the map $H:f \mapsto f*f \in \mathcal{E}'$.     My attempt: I have chosen to show that this is not the case by showing that $\text{image } (H)$ does not contain $ \delta_0'$. For instance I restrict myself to $d=1$.   Suppose that $H$ is surjective. Applying the Fourier transform on the convolution, I get $(\hat{f}(z)) ^2 = i z$. The theorem of Paley-Wiener-Schwartz guarantees that $(\hat{f}(z)) $ should be an entire function. However, as it is well-known, no such entire function exists:   If this were true, then $e^{i \pi /4}\hat{f}(z)$ would be the square root function, which cannot be well defined on $\mathbb C$, and even less so entire.    Q1: Is there another way of achieving the result (i.e. with a choice significantly different than $\delta_0'$, and with an approach different than the one using the Fourier transform)?  Q2: Can I generalize this without much effort in the case of $d \ge 1$ (probably using the tensor product)?    Edit: The source is .
83934
Rudin asked (Real Complex Analysis, First edition, Chapter 6, Problem 4):  Suppose $1\le p\le \infty$, and $q$ is the exponent conjugate to $p$. Suppose $u$ is a $\sigma$-finite measure and $g$ is a measurable function such that $fg\in L^{1}(\mu)$ for every $f\in L^{p}(\mu)$. Prove that then $g\in L^{q}(\mu)$.   I am being troubled with the fact that $|g|_{q}$ might be unbounded if we select wierd enough $f$. Holder inequality only gives us $$|fg|_{1}\le |f|_{p}|g|_{q}\leftrightarrow |g|_{q}\ge \frac{|fg|_{1}}{|f|_{p}}$$All  the constructions I know proving $g\in L^{q}$ starts by assuming $f\rightarrow fg$ is a bounded linear operator，so I cannot use circular reasoning at here.   It suffice to prove the statement for finite measure spaces and simple functions. So emulating Rudin we can assume $|g|=\alpha g$, where $|\alpha|=1$ and $\alpha$ is measurable. Let $E_{n}=x:|g(x)|\le n$ and let $f=\chi_{E_{n}}\alpha g^{q-1}$. Then we have $$\int_{E_{n}}|g|^{q}d\mu=\int_{X}|fg|d\mu\le K_{n}$$ for some $K_{n}&lt;\infty$. But this constant obviously shift with the $n$ I choose, hence probably does not have a finite upper bound (for example $K_{n}=n$). And I got stuck.   In problem 6, Rudin now ask:  Suppose $1&lt;p&lt;\infty$, and prove that $L^{q}(\mu)$ is the dual space of $L^{p}(\mu)$ even if $\mu$ is not $\sigma$-finite.   I keep thinking about it but do not know what is the best way to prove it.