let ${f_k}$ be a sequence of integrable functions on $[a,b]$ that converges in pointwise to $f$ and converges in mean to $g$. If $f$ and $g$ are continuous, then f = g. The 'convergence in mean' means, $\lim_{k\to \infty}\int_{a}^{b} [f_k(x) - g(x)]^2 dx = 0$ .  I tried to show that, if there exists some point $x_0$ s.t $f(x_0) \neq g(x_0)$, then on some neighborhood of $x_0$, $f_k-g$ is being uniformly nonzero, so that the value of integration can't be 0, so that $f_k$ can't converge in mean to $g$.  However, I couldn't prove this way, since  and $f_k$ is integrable, but not provided that $f_k$ is continuous.  Am I trying in a correct way? give me any idea please. (This question is from elementary analysis textbook, so please assume that properties about $L_2$ convergence or almost everywhere convergence is not known)
122166
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of Riemann-integrable functions on $[a, b]$.   Assume that $(f_n)_{n\in\mathbb{N}}$ converges pointwise to $f$ on $[a, b]$ and converges in the mean to $g$ on $[a, b]$, where $f$ and $g$ are both continuous. Show that $f = g$.   ($(f_n)_{n\in\mathbb{N}}$ converges in the mean to g on $[a , b]$ means that $\lim_{n\to\infty} d(f_n, g) = 0$, where $d$ is the $L_2$ metric.)  I am studying real analysis now and want to prove this statement. But I don't know how to use condition "$(f_n)_{n\in\mathbb{N}}$ converges in the mean to g on $[a , b]$." How can I prove this?
29866
Suppose $\{f_n\} \subset L^+$ , $f_n \rightarrow f$ pointwise, and $\int f = \lim \int f_n &lt; \infty$. Then $\int_{E} f = \lim \int_{E} f_n$ for all $E \in \mathcal{M}$. However, this needn't be true if $\int f = \lim \int f_n = \infty$.   This is the exercise 13 on page 52, Real Analysis, by Gerald B. Folland. $L^+$ is the space of all measurable functions from $(X,\mathcal{M}, \mu)$ to $[0, \infty]$.  I can find a counterexample satisfying $\int f = \lim \int f_n = \infty$. Let $X = \Bbb N$, $\mathcal{M} = \mathcal{P}(\Bbb N)$ and $\mu$ be the counting measure. $f_n = \chi_{\{2i+1:i \leq n\} \cup \{2j:n \leq j \leq 2n \}}$. Then $\int f = \lim \int f_n = \infty$. But when $E = \{2k: k \in \Bbb N\}$,  Then $\int_{E} f =  0$ and $\lim \int_{E} f_n = \infty$.  But how to show the part "$f_n \rightarrow f$ pointwise, and $\int f = \lim \int f_n &lt; \infty$. Then $\int_{E} f = \lim \int_{E} f_n$ for all $E \in \mathcal{M}$"?
90992
Theorem $:$ Let $f : [a,b] \longrightarrow \Bbb R$ be a Riemann integrable function. Then $f \in L_1[a,b]$ and $$\int f\ d\lambda = \int_{a}^{b} f(x)\ dx.$$  The proof given in my book is as follows $:$     At the fag end of the proof $f$ has been proved to be measurable which uses the completeness of the measure space $\left ([a,b], \mathcal L \cap [a,b], \lambda |_{[a,b]} \right ).$ Then it has been stated that $f \leq \Phi_n,$ $\forall$ $n$ and hence $f \in L_1[a,b]$ since $\Phi_n \in L_1[a,b]$ by the proposition $5.4.3 \ $(i) which states the following $:$  Let $f \in \Bbb L$ and $g \in L_1[a,b].$ If $|f(x)| \leq g(x)$ for a.e. $x(\mu).$ Then $f \in L_1[a,b].$  I don't know how to use this proposition to claim that $f \in L_1[a,b].$ Also I don't understand the way dominated convergence theorem used here. What I know about dominated convergence theorem is as follows $:$  Let $(X, \mathcal S, \mu)$ be a complete measure space. Let $g \in L_1(X)$ and $\{f_n \}_{n \geq 1}$ be a sequence in $\Bbb L$ such that $|f_n(x)| \leq g(x)$ a.e. $x(\mu).$ Let $f_n(x) \to f(x)$ as $n \to \infty$ a.e. $x (\mu).$ Then $f \in L_1(X)$ and $$\int f\ d\mu = \lim\limits_{n \to \infty} \int f_n\ d\mu.$$  From here how to conclude that $$\Psi_n\ \bigg\uparrow\ f \implies \int f\ d\lambda = \lim\limits_{n \to \infty} \int \Psi_n\ d\lambda.$$ Please help me in this regard. Thanks in advance.
32518
Let $g:[1, \infty)\rightarrow \mathbb R$ a continuous non-negative function, such that $\int_{1}^{\infty} g(x)\ dx$ converges. Is it true that $\lim\limits_{x\to \infty}g(x)=0$ ?   I tried to find a counter-example but I can't figure a trivial one. I also tried to prove it by the definition of the convergence of $g(x)$ but couldn't show that limit is really $0$.
116728
$\lim_{x \rightarrow \infty} \lim_{n \rightarrow \infty} f_{n}(x) = \lim_{n \rightarrow \infty}\lim_{x \rightarrow \infty}f_{n}(x) $ in case of uniform convergence of $f_{n}(x)$ to $f(x)$  I used this while I was given that $f_{n}(x)$ converges to $f(x)$ uniformly. Is this correct and I was also thinking to prove this!
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$f_n(x) = x - x^n$ for $x\in [0,1]$. Does the sequence converge pointwise or uniformly on $[0,1]$?   We also have to prove why it is or isn't pointwise or uniformly convergent.   I have worked out that on $x\in[0,1)$, the limit as $n$ approaches infinity is $x$. And when $x=1$ the limit is $0$.   So this means its pointwise convergent right? But how do I prove this?  Also how on earth do I prove it is not uniformly convergent? (I'm just assuming it's not).
21229
Please help me out with this problem:  Let $f_n(x) : [0,1] \to \mathbb R$ be a sequence of continuous functions convergent at every $x \in [0,1]$ to a continuous function $f: [0,1]  \to \mathbb R$ . Does $f_n$ converge uniformly to $f$?  My first solution: I read this example in  posted earlier and I think it works: Take a sequence $f_n:[0,1] \to \mathbb R$ such that $f_n$ increases linearly from $0$ to $1$ on the interval $\left[ 0,\frac{1}{n} \right]$, decreases linearly from $1$ to $0$ on the intreval $\left[ \frac{1}{n}, \frac{2}{n} \right]$, and is $0$ on $\left[ \frac{2}{n},1 \right]$. Then $f_{n} \to 0$ nonuniformly.  Is this answer true? Also, if anyone can give some examples of such function with detailed proof, I will very appreciative?
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I'm preparing for a test for real analysis and I came across this problem in Royden's book:  Let $\{f_n\}$ be a sequence of real valued functions on $[a,b]$ that converges pointwisely on $[a,b]$ to the real valued function $f$. Show that $TV(f) \leq \liminf  ~TV(f_n)?$  This looks quite similar in form to Fatou's Lemma to me, but can't find any way to establish TV with integration, can anybody please help?  (TV is short for total variation)
50361
Let $f:[0,\infty)\to [0,\infty)$ be a continuous function such that $\displaystyle \int_0^{\infty}f(t)\,dt&lt;\infty$. Which of the following statements are true ?   (A) The sequence $\{f(n)\}$ is bounded.  (B) $f(n)\to 0$ as $n \to \infty$.  (C) The series $\displaystyle \sum f(n)$ is convergent.  I am unable to prove directly but I am thinking about the function $f(x)=\frac{1}{1+x^2}$. For this function all options are correct. Is it correct ? I think not , as I have no proof in general.  Please help by giving a proof or disprove the statements.
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Let $(X, M, m)$ be an arbitrary measure space. Let $f_n, f \in L^1_m(X)$. Assume that $$\int_X |f_n - f| \, dm \leq \frac{1}{n^2} \text{  for all }n \geq 1. $$ Then I want to show that $f_n \rightarrow f$ a.e. on $X$.    I thought since we have norm convergence, we get a subsequence $f_{n_k}$ of $f_n$ converging to $f$ a.e. Does the condition $\int_X |f_n - f| \,dm \leq \frac{1}{n^2}$ for all $n \geq 1$ imply something stronger than norm convergence?
5851
Folland, "Real Analysis", Chapter 5.3, Exercise 36:     Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be   counting measure on $\mathbf{N}$.  Suppose that   $\left\{x_n\right\}_1^\infty$ is a countable dense subset of the unit   ball of $\mathcal{X}$, and define $T \colon L^1(\mu) \to \mathcal{X}$   by $Tf = \sum_1^\infty f(n) x_n$.  (a)  $T$ is bounded.  (b)  $T$ is   surjective.   I have proved (a).  I would like help on (b).  Here are my ideas so far.    Say we want to show $y \in \mathcal{X}$ is in the image of $T$.  Reduce to the case $||y|| = 1$ and let $x_{n_m} \to y$.  My thought is to produce for each $N$ a function $f_N \in L^1(\mu)$ such that $||y - \sum_n^\infty f(n)x_n|| &lt; 1/N$ and also such that the $f_N$ converge in $L^1(\mu)$ to some $f \in L^1(\mu)$.  Then use the continuity of $T$ to conclude $T(f) = y$.    The way I have been setting up my inequalities is as follows. Pick $M_N$ such that for all $m &gt; M_N$, $||y - x_{n_m}|| &lt; 1/N$.  Then I want to define $f_N$ so that $$\sum_n^\infty f_N(n) = \sum_{m &gt; M_N}^\infty f_N(m) = 1, $$ hence $$||y - T(f_N)|| = ||\sum_{m &gt; M_N}^\infty f_N(m) y + \sum_n^\infty f_N(n)x_n|| \leq \sum_{m &gt; M_N}^\infty f_N(m) || y - x_{n_m} || &lt; \frac{1}{N}. $$  Presumably, if the $f_N$ are appropriately chosen, then I can find a dominating function and apply the Dominated Convergence Theorem to finish up.    However, getting all these hypotheses to hold simultaneously has been difficult.  In essence, the difficulty seems to be in requiring that the $f_N$ be "normalized" as well as possess a dominating function.  If this method is feasible, at least it seems to require a bit of analysis to show the desired series all converge.  Perhaps I am missing something cleaner.  If so, a simpler solution would be much appreciated.    -Thanks.
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1) Construct a continuous function $f$ on $\mathbb{R}$ that is integrable on $\mathbb{R}$ but $\displaystyle\limsup_{x \to \infty} f(x)  = \infty$.  I took the function that is equal to $n$ on $[n, n + 1/n^{3})$ and made it continuous by saying that $f$ is the line segment joining $n$ and $n+1$ on $[n + 1/n^{3}, n+1)$. But I am failing to prove this integrable. For this, $\lim f(x) = \infty$ but how do you prove in general, if limit does not exist that $\limsup$ is infinity?  2) Prove that if $f$ is uniformly continuous and integrable on $\mathbb{R}$ we have $\displaystyle\lim_{|x| \to \infty} f(x) = 0$.  Any help is appreciated.  Thanks
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As I was going over the definitions of pointwise and uniform convergence I came to the following problem: since the canonical example for continuous functions on $[0,1)$ which are pointwise but bot uniform convergent(wrt the constant function $f=0$) is sequence of functions $f_n(x) = x^n$ I ask myself is there such sequence of functions for the interval $[0,1]$. So far I couldn't think find any example and I am stating to believe that the answer might be negative. So what do think, is there such sequence and if not can it be proven that such sequence does not exist?
74329
Stuck with this problem from Zgymund's book.   Suppose that $f_{n} \rightarrow f$ almost everywhere and that $f_{n}, f \in L^{p}$ where $1&lt;p&lt;\infty$. Assume that $\|f_{n}\|_{p} \leq M &lt; \infty$. Prove that:  $\int f_{n}g \rightarrow \int fg$ as $n \rightarrow \infty$ for all $g \in L^{q}$ such that $\dfrac{1}{p} + \dfrac{1}{q} = 1$.  Right, so I estimate the difference of the integrals and using Hölder end up with:  $$\left|\int f_{n} g - \int fg\right| \leq \|g\|_{q} \|f_{n} - f\|_{p}$$  From here I'm stuck because we are not assuming convergence in the seminorm but just pointwise convergence almost everywhere. How to proceed?
117279
Let $1&lt;p&lt;\infty$, $(u_n)_{n\in\mathbb{N}}\subseteq l^p(\mathbb{N})$ such that $u_n$ converges weakly to $u\in l^p(\mathbb{N})$. Prove that   (1) $(u_n)$ is bounded and (2) for all fixed component $j\in n\in\mathbb{N}$ it is $\lim_{n\to\infty} u_{n,j}=u_j$ (in lecture we called it "pointwise convergence of sequences").   My questions are: (1) is "$(u_n)$ bounded", i.e. $\|u_n\|&lt;\infty$ for all $n\in\mathbb{N}$ already satisfied because $(u_n)_{n\in\mathbb{N}}\subseteq l^p(\mathbb{N})$?   For (2) it is to show that $|u_{n,j}-u_j|\to 0,\; n\to\infty$ for fixed $j\in\mathbb{N}$. But I don't know how to do this.
63979
Let $A,B\subset\mathbb R$ and $f:A\to B$ be an invertible function (so 1-1 and onto). Prove that if $A$ is compact and $f$ is continuous, then the inverse $f^{-1}:B\to A$ is continuous. And give a counterexample when $A$ is not compact (no transcendentals).  If $f$ is continuous, then for every sequence $\{x_n\}$ in $A$ that converges to $L$, $\lim_{n\to\infty}f(x_n)=f(L)$. We need to prove that for every sequence $\{y_n\}$ in $B$ that converges to $K$, $\lim_{n\to\infty}f^{-1}(y_n)=f^{-1}(K)$ (Is this sufficient to prove that $f^{-1}$ is continuous?), I'm not sure how to proceed from there, and I'm having trouble coming up with a counterexample as well.
25113
Suppose $f$ is a continuous function at $x = c$ in $[a,b]$.  Prove that for any sequence ${x_n}$ in $[a,b]$ converting to $c$, the sequence $\{f(x_n)\}$ converges to $f(c)$.  That is, $$ \lim_{n\to\infty}f(x_n)= f\left(\lim_{n\to\infty}x_n\right)$$    This proof seems simple but there are a few things that I need to know first. If $\{x_n\}$ converges to $c$, is it sufficient to substitute $c$ in for $\lim_{n\to\infty}x_n$?  Also needing some guidance on the structure of this proof.  Thanks!