I'm going through my calculus lecture notes, and at some point the following claim is made: Let $f:\left[a,b\right]\to\mathbb{R}$ be a bound function. For any $\epsilon&gt;0$ there exists some $\delta&gt;0$ s.t. for any partition $P$ of $\left[a,b\right]$ for which $\Delta P&lt;\delta$, it holds that   $$\left|U\left(f,P\right)-\bar{\int}f\left(x\right)dx\right|&lt;\epsilon$$  Where $\Delta P$ is the size of the biggest segment in the partition (The text, which is not in English, refers to $\Delta P$ as the “partition parameter”, I don't think this is the correct translation into English since I did not find any reference to it). $U\left(f,P\right)$ is the upper Darboux sum for said partition and $\bar{\int}f\left(x\right)dx$ is the infimum for the upper Darboux sums of all possible partitions.   I have failed in proving this, and it seems important, could someone supply a source where this is proved?
121099
I was reading the theory of Riemann integration when I cam across the following ,   If $f$ is bounded on $[a,b]$, and $P = \{x_0,x_1,x_2.......x_n\}$ is a partition of $[a,b]$, let $$M_j = \sup_{x_{j-1}\leq x\leq x_j}f(x)$$ The upper sum of f over P is $$S(P) = \sum_{j=1}^{n} M_j(x_j-x_{j-1})$$ and the upper integral of $f$ over $[a,b]$, denoted by $$\int_{a}^{b^-} f(x)dx$$ is the infimum of all upper sums.  The theorem similarly goes on to state the result for lower sums.  My doubt is : I do not understand how is $$\int_{a}^{b^-} f(x)dx$$ the infimum of all upper sums. I understand that if we refine the partition P, then the upper sum would decrease, so it may be a lower limit for all the upper sums computed on the refinements of P (but still being the lower limit does not prove that it is the infimum) and what about those partitions for which P itself is the refinement of?  How do I know that it will be a lower limit for those, let alone a infimum?
38091
I've encountered the following problem in a set of course notes on complex analysis, but I can't seem to solve it:     Prove that if f is a continuous real-valued function with $|f(z)| \leq 1$, then $\left|\int_{|z| = 1} f(z)\,dz\right| \leq 4$.   This is my work so far:     \begin{align*} \left|\int_{|z| = 1} f(z) \,dz \right| &amp;= \left|\int_0^{2\pi}f\left(e^{i\theta}\right)ie^{i\theta}\,d\theta\right| \\ &amp;= \left|\int_0^{2\pi}f\left(e^{i\theta}\right)\cos\theta\,d\theta + i\int_0^{2\pi}f\left(e^{i\theta}\right)\sin\theta\,d\theta\right| \end{align*}   I initially tried to use the triangle inequality at this point and then bound the real integrals, but this gives me an upper bound of $4\pi$. Using the estimation lemma on the initial integral gives me an upper bound of $2\pi$. I can't manage to get it any tighter.  The emphasis on real-valued makes it seem that this is a key point. Any ideas?
93723
For every positive integer $d$, we let $\tau\left(d\right)$ be the number of positive divisors of $d$. Prove that \begin{align} \sum_{d|n} \tau^3(d) = \left(\sum_{d|n} \tau (d)\right)^2 \end{align} for each positive integer $n$, where the sums range over all positive divisors $d$ of $n$.  Now I only know that both sides are multiplicative arithmetic functions in $n$. Could you tell me what I need to do next?
69543
If $f$ is continuous, nonnegative on $[a, b]$, show that $\int_{a}^{b} f(x) dx = 0$ iff $f(x) = 0$  "$\Rightarrow$" Assume by contradiction that $f(x) \neq 0$ for some $x_0 \in [a, b]$. Without loss of generality, assume that $f(x_0) &gt; 0$. Since $f$ is continuous, $\exists \delta &gt; 0$ such that $f(x) &gt; 0,  \forall x \in (x_0 - \delta, x_0 + \delta)$. Since $f$ is continuous on $[a, b]$, then it is Riemann integrable. Therefore $\int_{a}^{b} f(x) dx = \sup L(f, p)$ for some partition $p$. Now what I want to do is to show that the lower sum, $L(f, p) &gt; 0$, in that case its supremum is $&gt;0$, and that contradicts the condition that $\int_{a}^{b} f(x) dx = 0$. However, I don't know how to articulate it.  "$\Leftarrow$" If $f(x) = 0, \forall x \in [a, b]$, then $\int_{a}^{b} f(x) dx = \int_{a}^{b} 0 dx = 0$
79633
I'm going over old exam problems and I got stuck on this one. Suppose that $\mathbb{f}\colon \mathbb{D} \to \mathbb{C}$ is analytic and bounded. Let $\{a_n\}_{n=1}^\infty$ be the non-zero zeros of $\mathbb{f}$ in $\mathbb{D}$ counted according to multiplicity. Show that  $$ \sum_{n=1}^\infty \left( 1 - \left|a_n \right|\right)\lt\infty $$  I can understand that $\left|a_n \right|$ goes to $1$ since zeros are isolated, but it doesn't help showing the series is convergent. Any help will be appreciated!
65966
Regarding: Let $f:(0, \infty) \to \mathbb{R}$ be a differentiable function for which $f'(x)f'(\frac{1}{x})&lt;0$ for every $0&lt;x \ne 1$. Then $f'(1)=0$. -- So I am trying to prove this, and I said: We know that $f$ is differentiable at $(0, \infty)$, so there exists a $0&lt; \delta &lt; 1$ such that $f$ is differentiable at $[1- \delta, 1+\delta]$ . So there exists a $x_{0} \in [1- \delta, 1+\delta]$ such that $f'(x_{0})f'(\frac{1}{x_{0}})&lt;0$ So by Darboux Theorem, there exists a $x_{0}&lt;c&lt; \frac{1}{x_{0}}$ such that $f'(c)=0$. But now, I do not know how to link $c$ with $1$... Any hints? Thanks! :)
98417
So we want to find a $\delta&gt;0$   such that for all $2-\delta&lt;x&lt;2+\delta$  , we will have $4-\epsilon&lt;x^{2}&lt;4+\epsilon$ for all $\epsilon&gt;0$  . If we can find a way to express $\delta$  as a function $\delta (\epsilon)$,  $\delta:\mathbb{R}_+ \to \mathbb{R}_+ $ then we will have solved the problem. But I can't see how to relate $\epsilon$ to $\delta$ in this case.  My initial reaction is write $(2-\delta)^2&lt;x^2&lt;(2+\delta)^2$, this makes the two inequalities look related, but I am not sure where to go from here.
52987
I'm currently a Combinatorics student trying to parse through this solution. I do not understand the proof currently. Any help understanding it is greatly appreciated.    Question Let the number of partitions be given by the function $p(n)$, i.e. the number of ways of writing the integer $n$ as a sum of positive integers. Prove that that if $g$ is any polynomial function in $n$, then there exists an integer $N$ such that $g(n)&lt;p(n)$ for all $n&gt;N$. Do not use the asymptotic Hardy-Ramanujan formula.    Provided Proof.    Enough to show $n^a&lt;p(n)$ for fixed positive integer $a$. (This is just saying that we want to show $p(n)$ grows faster than any degree polynomial of $n$, right?) We have that $(i_1,i_2,\ldots,i_{a+1},1^{n-i_1-\cdots -i_{a+1}})$ is a partition of $n$ as long as $i_{a+1}\leq i_a\leq \cdots i_1 \leq \lfloor\frac{n}{a+1}\rfloor$. (I really don't understand this step. What is this partition? Of the coefficients or terms of polynomial $p(n)$? And how do we know this is $\leq \lfloor\frac{n}{a+1}\rfloor$?) The number of partitions $n$ of this form is multiset binomial coefficient $$\left(\dbinom{\lfloor \frac{n}{a+1}\rfloor}{a+1}\right)=\frac{\lfloor\frac{n}{a+1}\rfloor(\lfloor\frac{n}{a+1}\rfloor+1)\cdots(\lfloor\frac{n}{a+1}\rfloor+a-1)}{(a+1)!}$$ (I'm assuming that if I can understand (2.) then I can understand this line) This means $p(n)&gt;\left(\binom{\lfloor \frac{n}{a+1}\rfloor}{a+1}\right)$ for all $n&gt;0$. Since $a$ is constant, the given fraction grows with order $n^{a+1}$, which means $p(n)&gt;n^a$ for $n$ sufficiently large. (Why does it grow with order $n^{a+1}$?)   Thanks all again in advance!
67741
Prove that if $f$ is a continuous real valued function on the interval $[a,b]$ such that $f(x)\ge 0$  for all $x\in [a,b]$ and $f(x)&gt;0$ for some $x\in[a,b]$ then $\int^b_a f(x) \, dx &gt;0$.  The definition I have for Riemann integral is: Let $a,b\in \mathbb{R}$ $a&lt;b$ and let $f$ be a real valued function on $[a,b]$. We say that $f$ is Riemann integrable on $[a,b]$ if there exists a number $A\in \mathbb{R}$ such that for any $\varepsilon&gt;0$ there exists $\delta&gt;0$ such that $|S-A|&lt;\varepsilon$ whenever $S$ is Riemann sum for $f$ corresponding to any partition of $[a,b]$ of width less that $\delta$.
76799
The following question is taken from Real Analysis by Royden, Chapter $4,$ question $5:$     Let $f$ be a bounded function on $[a,b].$       Suppose there is a sequence $\{P_n\}$ of partitions of $[a,b]$ for which $\lim_{n\to\infty}[U(f,P_n) - L(f,P_n)]=0.$       Show that $f$ is Riemann integrable over $[a,b].$   My attempt:  Let $\varepsilon&gt;0$ be given.     Then there exists a partition $P_N$ of $[a,b]$ such that      $$U(f,P_N)-L(f,P_N) &lt; \varepsilon.$$     Since $\varepsilon&gt;0$ is arbitrary, therefore      $$U(f,P_N)\leq L(f,P_N) \leq \underline{\int}_{a}^b f.$$     Similarly,      $$\overline{\int}_a^b f \leq U(f,P_N) \leq  L(f,P_N) \leq \underline{\int}_a^b f.$$     Therefore, $f$ is Riemann integrable on $[a,b].$  Is my proof correct?
123888
I need to prove that if $f$ is Riemann integrable $[0,1]$ then $$\lim\limits_{n\rightarrow\infty} \frac{1}{n}\sum f\left(\frac{k}{n}\right)= \int\limits_{0}^{1} f(x)dx$$  My idea is to recognize the right side defining the limit as a Riemann sum and use the uniform continuity.  I am not sure what to do next  Thanks
72404
I am working on some past prelim problems and one of them is the following: If $f \in \mathcal{L}^1(\mathbb{R})$ then $\lim_{h\rightarrow 0}\int_{-\infty}^\infty |f(x+h)-f(x)|dx\rightarrow 0$   Update: Following Jackie Chong's advice, I am attempting this problem with the dominated convergence theorem. Since $f\in L^1(\mathbb{R})$, $f$ can be approximated from below by a sequence of continuous functions $f_n$. Then, I believe I can say that, for all $n$ sufficiently large, $|f_n(x+\frac{1}{n})-f_n(x)|\le 2|f(x)|$. (I'm not sure of this part thought). Then $2|f(x)|$ is integrable and hence a dominating function for $|f_n(x+\frac{1}{n})-f_n(x)|$. So by the dominated convergence theorem we can take the limit inside $\lim_{n\rightarrow \infty}\int|f_n(x+\frac{1}{n})-f(x)|dx=\int \lim_{n\rightarrow \infty}|f_n(x+\frac{1}{n})-f(x)|dx=0$ Is this on the right track?  Another approach that I thought may work, but which I'm less certain of is the following: If $f$ can be approximated by $C^1$ functions (I have no real justification for this, so please correct me if its wrong), then for each $n$ and each $x$ there is some $y(x,n)\in B_{1/n}(x)$ such that $|f(x+\frac{1}{n}-f(x)|=\frac{1}{n}|f'(y(x,n)|$ but from here I don't see how to proceed, because just because $|f|$ is integrable, I can't justify $|f'|$ being integrable, but if there is some way to fix this approach, hints would be appreciated.   Another thing I was thinking was to first approximate $f$ from below by continuous functions $f_n$ then to define $g_{m,n}=|f_n(x+\frac{1}{m})-f_n(x)| \chi_{[-m,m]}$. Then, for each $n$, the function $g_{m,n}$ is continuous on its support, and since it support is compact, it is uniformly continuous there. Can this be used to justify bringing the limit $\lim_{n\rightarrow \infty}$ inside the integral? Then apply the monotone convergence theorem? I know this idea is probably even more sketchy than my previous one, but if it can be fixed please let me know :)  Thanks!
109196
Some days ago I answered a question that asked to find  $$\mathop {\lim }\limits_{x \to 0} {x^\alpha }\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^{\alpha  + 1}}}}dt} $$  given that $f$ is continuous in $[0,1]$  I proceeded as follows:  $$\eqalign{   &amp; t = x\cdot u  \cr    &amp; dt = x\cdot du \cr} $$  So this is produces:  $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du} $$  I then thought: "Well, if $f$ is continuous in the closed interval, then it is also uniformly continuous, so I can assume   $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  = f\left( 0 \right)\int\limits_1^\infty  {\frac{{du}}{{{u^{\alpha  + 1}}}}}  = \frac{1}{\alpha }f\left( 0 \right)$$  This turned out to be true. However, I wasn't very comfortable with such "move". So now I'm thinking, one can put  $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du} $$  And then   $$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du} } \right| &lt; \int\limits_1^{\frac{1}{x}} {\frac{{\left| {f\left( {xu} \right) - f\left( 0 \right)} \right|}}{{{u^{\alpha  + 1}}}}du}  &lt; \epsilon \frac{{1 - {x^\alpha }}}{\alpha }$$  However, this is still insufficient since I need to adress the behaviour of the upper limit too. Can someone show me how to adress both behaviours simultaneously?    Would this work?  Let $P$ be the statement that $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  = \frac{{f\left( 0 \right)}}{\alpha }$$  Then $P$ is true if and only if  $$ \forall \epsilon  &gt; 0\exists \delta  &gt; 0$$  Such that if $$\left| x \right| &lt; \delta $$ then $$ \left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  - \frac{{f\left( 0 \right)}}{\alpha }} \right| &lt; \epsilon $$  But then   $$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| &lt; $$  $$\left| {\int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| \leqslant $$  $$\varepsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } &lt; $$  And since  $$\frac{{1 - {\delta ^\alpha }}}{\alpha } &lt; \frac{1}{\alpha }$$ $$\epsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } &lt; \frac{\epsilon }{\alpha } &lt; \epsilon $$
70729
In , the following theorem is used implicitly (see Example (a) on page 488):     Let $0&lt;p&lt;\infty$. For every $\epsilon &gt; 0$, there exists some $C_{\epsilon} &gt; 0$ such that $\forall a,b \in \mathbb{C}$:   $$ \left||a+b|^p - |a|^p \right| \leq \epsilon |a|^p + C_{\epsilon} |b|^p $$   Is this a well-known result? I would like to know its proof, and its name if it has one. Can someone point me towards a reference to this result which includes a proof, or give me a proof outline?  I've tried using Jensen's Inequality for the case $p&gt;1, a,b&gt;0$, but it doesn't seem to work out.
78113
I was wondering whether anyone would be so kind as to briefly check my proof? I am supposed to prove the statement without using any theorems which would render the proof trivial.     If $\displaystyle\int_a^b f$ exists then $f$ has infinitely many points of continuity in $[a,b]$   For the sake of contradiction suppose $f$ has finitely many. It suffices to show that there is an interval $[u,v]\subset [a,b]$ over which $f$ is not Riemann integrable. Pick any $[u,b]\subset [a,b]$ such that $f$ is discontinuous everywhere in $[u,v]$. For any finite partition $D=\{x_1\cdots x_n\}$ of $[u,v]$ let:  $$s(f,D)=\sum_i (x_{i+1}-x_i)\inf_{x\in[x_{i},x_{i+1}]}f\quad\text{and}\quad S(f,D)=\sum_i (x_{i+1}-x_i)\sup_{x\in[x_{i},x_{i+1}]}f$$ We prove $\sup_D s(f,D) &lt;\inf_D S(f,D)\;(1)$. To achieve this consider an arbitrary chain: $$D_0\subset D_1\subset \cdots$$ Pick any $I_0\subset I_1\subset \cdots$ where $I_k=[u_k,v_k]$ and $I_k$ is an interval of the partition $D_k$. Without loss of generality $\bigcap_i I_i=z$. Given that $f$ is discontinuous everywhere, there exists $\epsilon&gt;0$ such that for any $\delta&gt;0$ there exists $y$ such that $|z-y|&lt;\delta$ yet $|f(z)-f(y)|\geq \epsilon$.   This immediately implies $\displaystyle\sup_{I_n} f&gt;\inf_{I_n} f$ and hence $\displaystyle\sup_{D_n} s(f,D)&lt;\inf_{D_n} S(f,D)$.   This is valid for any sequence $D_n$ so claim $(1)$ holds, contradicting the integrability of $f$.
107972
Theorem says that if $f:[a,b]\longrightarrow \mathbb{R}$ is a continuous function, then it's integrable in Riemann's sense. Here's my proof:  Proof  Let $\mathfrak{P}=\{x_0,x_1,...x_n\}$ be a subdivision of $[a,b]$. Let denote $m_k=\inf_{x\in[x_{k-1},x_k]}f(x)$, $M_k=\sup_{x\in[x_{k-1},x_k]}f(x)$ and a diameter as $\max_{k=0,1...,n}(x_{k+1}-x_k)$. Then, the lower Darboux sum is $s(f,\mathfrak{P})=\sum_{k=1}^n m_k(x_k-x_{k-1})$ and upper sum is $S(f,\mathfrak{P})=\sum_{k=1}^n M_k(x_k-x_{k-1})$.  We know that $f\in\mathfrak{R}([a,b])\iff \forall_{\epsilon &gt;0}\exists_{\mathfrak{P}}:S(f,\mathfrak{P})-s(f,\mathfrak{P})&lt;\epsilon$  Every funtion continuous on closed interval is also uniformly continuous, so $$\forall_{\epsilon &gt;0}\exists_{\delta &gt;0}\forall_{x_1,x_2\in [a,b]}:|x_1-x_2|&lt;\delta \Rightarrow |f(x_1)-f(x_2)|&lt;\frac{\epsilon}{b-a}$$  This means that for every subdivision $\mathfrak{P}$ of a diameter smaller than $\delta$ we get $$S(f,\mathfrak{P})-s(f,\mathfrak{P})=\sum_{k=1}^n(M_k-m_k)(x_k-x_{k-1})\leq \frac{\epsilon}{b-a}(b-a)=\epsilon$$ which ends the proof.  Is it understandable and, what's more important, correct?
123174
I know that big O is used to describe the upper bound of a function, so doesn't this mean that if the upper bounds f and g are the same, they are the same function? If this is the case, then we can say that  f ∈ Θ(g) because if that were not the case, then the two upper bounds would not be the same and thus one would out grow the other. Is this correct?