I have four functions $f_1,f_2,g_1,g_2:\mathbb{N}\rightarrow\mathbb{R}^+$. Suppose that $f_1\in\Theta(g_1)$ and $f_2\in\Theta(g_2)$. I need to prove that $f_1+f_2\in\Theta(\text{max}\{g_1,g_2\})$.   This is fairly easy to see with a specific example such as the following:  Let $f_1=x+5$. Then $f_1\in\Theta(x)$.  Let $f_2=2x^2$. Then $f_2\in\Theta(x^2)$.  $f_1+f_2 = (2x^2)+(x+5)$ which  is $\in\Theta(x^2)$ and $x^2$ is the max of $x$ and $x^2$.  I cannot figure out how to write this as a general proof.  I can say that $f_1+f_2\in O(\text{max}\{g_1,g_2\})$ since I want a bigger value of $g$.  I am having trouble saying that $f_1+f_2\in\Omega(\text{max}\{g_1,g_2\})$ since I don't necessarily know that with a bigger value of g the inequality of $\Omega$ will still apply.
79998
I need to prove that     $f_1 \in \Theta(g_1) \land f_2 \in \Theta(g_2) \implies f_1 + f_2 \in \Theta(\max\{g_1, g_2\})$    is relevant, but I have a slightly different case, so I don't know how to translate it into this one, because I need to turn the $f$'s into $g$'s.  Am I allowed to say $\max\{f_1, f_2\} \in \Theta(\max\{g_1, g_2\})$?  Or even $\Theta(\max\{f_1, f_2\}) = \Theta(\max\{g_1, g_2\})$?
9848
Suppose $D$ is a nonempty bounded subset of reals. Let $f:D \to \mathbb R$ and $g:D \to \mathbb R$. Define $(f+g)(x)=f(x)+g(x)$. Prove $\sup(f+g)(D) \le \sup f(D) + \sup g(D)$ (also prove that $\sup (f+g)$ exists).   I understand why this is the case, just not how to prove it. Left side is pretty much $\sup  (f(x)+g(x))$ and right side is $\sup (f(x) + g(y))$ for $x,\,y \in D$. Basically $f+g$ has to use the same variable and $f(D)+g(D)$ use different ones. But I don't know how to go about proving this.     The second part of the question is to find a specific example where strict inequality holds.    Let $D=\{a,b\}$ and $f: a \to 1,\, b\to 0, \,g: a \to 0,\, b\to 1$. $\sup f(D) = 1,\, \sup g(D) = 1,\, \sup f(D) + \sup g(D) = 2.$ $\sup (f+g)(D) = 1$ (if we choose a, $f+g = 1+0,\, b,\, f+g=0+1$).
40300
Let $T : [0, 1]^2 → [0, 1]$. Consider the following properties:  $T_1 : T(x, 1) = x$  $T_2 : T(x, y) = T (y, x) $  $T_3 : T(x, T (y, z)) = T(T (x, y), z) $  $T_4 $: If  $x ≤ u  , y ≤ v \Rightarrow T(x, y) ≤ T(u, v) $  Function $T: [0, 1]^2 → [0, 1]$ that satisfies $T_1 −T_4$ is T-norm . For $T(x) = \begin{cases}min(x,y),&amp;max(x,y)=1,\\0,&amp;o.w.\end{cases} $ conditions 1, 2 and 4 obviously.  Can you help me prove condition 3.
124068
Let $f,g$ be two continuous positive functions over $[a,b]$ Let $m_1$ and $M_1$ be the minimum and maximum of $f$ Let $m_2$ and $M_2$ be the minimum and maximum of $g$ Prove that $$\sqrt{\int_a^bf^2 \int_a^b g^2}\leq \frac{1}{2}\left(\sqrt{\frac{M_1M_2}{m_1m_2}}+\sqrt{\frac{m_1m_2}{M_1M_2}}\right)\int_a^bfg$$  I can't make any significant progress on this one... Thanks for any hint.
76460
I need to use this theorem as part of a proof. Basically it goes as follows:  If $f(x_0) &gt; g(x_0)$ and $\frac{d}{dx} f(x) &gt; \frac{d}{dx} g(x)$ for all $x \geq x_0$, then $f(x) &gt; g(x)$ for all $x\geq x_0$.  What's the name of this theorem, if it has one? Or if it doesn't, is there a simple way to prove it that I could include in my proof? Thanks.
75449
Just would like to see if I'm right.  Proof: Let $G$ be generated by $x$ (Let $G=\langle x \rangle$ for some $x$ in $G$.) We wish to show that there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$. First, we will show that for all $x^n$ in $G$, $f(x^n)=f^n(x)$. We proceed by induction.   Base case: $n=1$. $f(x*e_G)=f(x)f(e_G)=f(x)$. Now, assume that $f(x^k)=f^k(x)$ for some arbitrary positive integer k. We show that $f(x^{k+1})=f^{k+1}(x)$. Observe that $f(x^{k+1})=f(x^kx)=f(x^k)f(x)=f^k(x)f(x)=f^{k+1}(x).$   Next we will show that for all $x^{-n}$ in $G$, $f(x^{-n})=f^{-n}(x)$. We proceed by induction. Base case: $n=-1$. $f(x^{-1}*e_G)=f(x^{-1})f(e_G)=f(x)$. Now, assume that $f(x^{-k})=f^{-k}(x)$ for some arbitrary positive integer k. We show that $f(x^{-k-1})=f^{-k-1}(x)$. Observe that $f(x^{-k-1})=f(x^{-k}x)=f(x^{-k})f(x)=f^{-k}(x)f(x)=f^{-k-1}(x).$ Because $f$ is onto, $Im(f)=G'$ and therefore $Im(f)=(...,f^{-2}(x),f^{-1}(x),e_G,f(x),f^2(x),...)=G'$.  Thus, there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$ and the proof is complete by the principle of mathematical induction. Q.E.D.  I have a feeling the proof shouldn't be this long. Any suggestions for a shorter, cleaner proof?
103459
Suppose $f:\mathbb{R}\to\mathbb{R}$ has period 1, and for some $q\in(0,1)$:  $$|f(x)-f(y)|\leq q|x-y|\quad \forall x,y$$  Let $g(x)=x+f(x)$, for any $a\in\mathbb{R}$,define the following sequence:  $$a_1=a,\quad a_2=g(a_1),\quad a_3=g(a_2),\quad \dots,\quad a_{n+1}=g(a_n)$$  Show that  $$\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}$$  exists and does not depend on the choice of $a$.  I don't even know what is the possible approach, any hints? Thanks!
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?
72502
Show that for any real constants a and b, where b > 0,  $(n + a)^b = Θ(n^b)$.  So taking a stab at this.. I let $a = 0$, and $b = 1$  So I got $f(n) = (n + a)^b \implies f(n) = (n+0)^1$ which we know $\implies n^1$ or $O(n) \implies \Theta(n^1)$ ? I just took a guess. I'm not sure if that's even close to being right.
80732
Trying to find a proof for Clarkson's inequality, which states that if $2 \leq p &lt; \infty$, then for any $f, g \in L^p$, we have that $$\left|\left|f+g\right|\right|_p^p + \left|\left|f-g\right|\right|_p^p \leq 2^{p-1}\left(||f||_p^p + ||g||_p^p\right) $$ I have a couple of ideas, but I know they are wrong. I tried saying that $|f+g|^p \leq |f|^p + |g|^p$ and the same thing for $|f-g|$ but I know those inequalities aren't true in general.
42672
This is from Putnam:     If a function $f: \mathbb{R} \to [-1, 1]$ is such that $f''(x)$ exists for all $x \in \mathbb{R}$ and $\{f(0)\}^{2} + \{f'(0)\}^{2} = 4$ then prove that there is a point $c$ such that $f(c) + f''(c) = 0$.   Now the condition concerning values of $f(0), f'(0)$ (and the fact that range of $f$ is subset of $[-1, 1]$) implies that $f'(0) \neq 0$ so that function is not a constant. If we take $g(x) = \{f(x)\}^{2} + \{f'(x)\}^{2}$ then we can see that $g' = 2f'(f + f'')$. One hope of solving the problem is to show that $g$ attains local extrema at some point different from an extrema of $f$. This would ensure that $g'$ vanishes without making $f'$ vanish and that will lead to $f + f'' = 0$. However the constraint on $f(0), f'(0)$ does not help in analyzing the extrema of $f$ or $g$.  Another line of thought which could be helpful here is to consider $F(x) = f'(x)\sin x - f(x)\cos x$ so that $F'(x) = \sin x\{f(x) + f''(x)\}$ but in this case I am not able to see how to use the condition on values $f(0), f'(0)$ to impose some constraint on $F(x)$.  Perhaps both the approaches which came to my mind are not in right direction (or may be they are, but I am unable to see). Please provide any hints.  Update: Based on the answer by Sandeep Thilakan, I have dropped the requirement for continuity of $f''(x)$.
24300
Let $a, b, c, d, e, f$ be nonnegative real numbers such that   $a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6$. $ab + cd + ef = 3$. What is the maximum value of $a+b+c+d+e+f$&nbsp;?.   I was thinking about somehow manipulating the numbers to get $\left(a + b + c + d + e + f\right)^{2}$, but I don't really know what to do.
123818
Let $\kappa, \lambda,\theta$ be infinite cardinals of $M$ where $M$ is a transitive model of ZFC. Let $\mathcal{P}=Fn(\kappa\times \omega,2)$ (finite functions from $\kappa\times \omega$ to 2) and G is a (M,P)-generic set. I want to show $(\lambda^\theta)^{M[G]}=(max(\kappa,\lambda)^\theta)^M$. One direction is easy $(\lambda^\theta)^M \leq (\lambda^\theta)^{M[G]}$ and $(\kappa^\theta)^M\leq (\kappa^\theta)^{M[G]}\leq ((2^\omega)^\theta)^{M[G]}=(2^\theta)^{M[G]}\leq (\lambda^\theta)^{M[G]}$ where $(\kappa \leq 2^\omega)^{M[G]}$ is by the forcing construction. But now I am not quite sure how to get the other direction. Any thoughts?
68668
If $f:[0,1] \rightarrow \mathbb{R}$ is a continuous function such that $f(0)=f(1)$, then there exists $ x\in [0,1]$ such that  $f(x) = f(x+\frac{1}{n})$, where $n$ is any natural number.  Let $g(x)=f(x+\frac{1}{n})-f(x)$ on $[0,1-\frac{1}{n}]$. Now, $f$ is continuous function on closed interval.So, it must attain its bounds. Let $m$ be minima at $c_1 \in [0,1]$ and $M$ be maxima at $c_2 \in [0,1]$.  Now if $c_1,c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $g(x)$ on $[c_1,c_2]$.  If $c_1$ or $c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $[c_1,c_1-\frac{1}{n}]$ or  $[c_2,c_2-\frac{1}{n}]$ respectively. If  $c_1,c_2$ both do not belong to $[0,1-\frac{1}{n}]$, then apply IVT to $g$ on $[c_1-\frac{1}{n},c_2-\frac{1}{n}]$.  Is this proof correct? Thanks.
75435
I'm learning calculus, specifically derivatives and applications of MVT, and need help with the following exercice:      Show that, for all $n &gt; 1$ $$\frac{1}{n + 1} &lt; \log(1 + \frac1n) &lt; \frac1n.$$    I tried to follow the below steps in order to prove the RHS inequality:   Proving that $f &lt; g$ on $I$ from $a$ to $b$:  Step $1$. Prove that $f' &lt; g'$ on $\operatorname{Int}(I)$.  Step $2$. Show that $f(a) \leq g(a)$ or that $f(a^+) \leq g(a^+)$  Following the above steps, let $f(x) = \log(1 + \frac1x)$ and $g(x) = \frac1x$, for all $x &gt; 1$. One has  $$f'(x) =  -\frac{1}{x^2 + x} \quad \text{and} \quad g'(x) = -\frac{1}{x^2}.$$  We note that, for every $x &gt; 1$, $f'(x) &gt; g'(x)$. Moreover, $f(1^+) = \log(2) &lt; 1 = g(1^+).$   My problem is that I got the wrong inequality sign in Step $1$.    Looking at the solution in my textbook, the author suggests using the MVT but I don't know how to apply it in this case.
72740
This is my attempt I know this is incomplete or may even be wrong.   Let $θ_1 \in \arg(z)$ and $θ_2 \in \arg(w)$. Then, $θ_1+θ_2 \in \arg(z)+\arg(w)$. Also, $θ_1+θ_2 \in \arg(zw)$. Is this sufficient for the proof or correct at all? Hope someone could help me out. Thanks.   EDIT: Just to avoid any confusions I will add how $\arg z$ is defined.   $\arg z=\{\text{Arg}\, z+2\pi k\mid k\text{ is an integer}\}$
43556
To seek the maximum value of $S=x_1x_2+x_2x_3+\cdots+x_nx_1$ on this domain:  $x_1+x_2+\cdots+x_n=0$ and  $x_1^2+x_2^2+\cdots+x_n^2=1$.  I have made some trivial observations:  1) $S\in[-1,1)$ by the rearrangement inequality.  2) We can make $S$ arbitrarily close to $1$ by increasing $n$.  3) An equivalent problem is to minimise $(x_1-x_2)^2+\cdots+(x_n-x_1)^2$.  But does the maximum have a meaningful closed form  for each $n$?