The tittle of this question is the main question.  Why it is necessary to not consider these kind of functions in $\Lambda$. My teacher told me that if one consider these functions to be in the ring of symmetric functions, then the inner product would fail since this kind of functions would have infinitely non-zero homogeneous parts. But I do not see it clearly.  Can anyone explain me why it is necessary to exclude the functions which do not have bounded degree?  Thank you very much.   $$\prod_i (1+x_i) \notin \Lambda.$$
124063
I am doing my thesis related with symmetric functions and representations.   It is for this reason that I am reading MacDonald's book Symmetric Functions and Hall Polynomials.  When reading chapter 1.2. The ring of symmetric functions, he also talks about symmetric polynomials. And he compares both notions and sometimes treats them as equivalent but sometimes not. Does anyone have a clearer explanation about what is the difference (if there are differences)?  So, what is the difference between symmetric functions and symmetric polynomials?
14214
Good day, I want to prove the following theorem:  1) For any nonzero vector $ v \in V $, there exists a linear funtional $ f \in V^*$ for wich $f(v) \neq 0 $  I know that if $f$ is a lineal functional then we have 2 posibilities   1)$\dim \ker(f)=\dim V$   2)$\dim \ker(f)=\dim V-1$   I've tried to suppose that, for all $v \neq 0 $ and $f \in V^*$ We have $f(v)=0$,  Then  $Im(f)=0$ for all $f \in V^*$ and $ v \in V$ but then the only linear functional is $f=0$ ($V^*=\{0:V \to F\}$) and from this follows: $\dim \ker(f)=\dim V$.  But 2) suggest the existence of a linear functional which $\dim \ker(f)=\dim V-1$, so ¿is that the contradiction?
16668
I wonder  whether there is a function $f\colon\Bbb R\to\Bbb R$ with the folowing characteristic?   for every two real numbers $\alpha,\beta,\alpha\lt\beta$,  $$\{f(x):x\in(\alpha,\beta)\}=\Bbb R$$   I can't  say such a function does not exist, neither can I construct a example  Thanks a lot!
55873
Today I was thinking about composition of functions.  It has nice properties, its always associative, there is an identity, and if we restrict to bijective functions then we have an inverse.  But then I thought about commutativity.  My first intuition was that bijective self maps of a space should commute but then I saw some counter-examples. The symmetric group is only abelian if $n \le 2$ so clearly there need to be more restrictions on functions than bijectivity for them to commute.  The only examples I could think of were boring things like multiplying by a constant or maximal tori of groups like $O(n)$ (maybe less boring).  My question: In a euclidean space, what are (edit) some nice characterizations of sets of functions that commute? What about in a more general space?   Bonus: Is this notion of commutativity important anywhere in analysis?
87676
I am wondering if there are any functions $f(x)$ such that it cannot be expressed as a power series of $x$? This might turn out to be a silly question, but I can't think of one at the moment.   Thanks!
58764
Let $A$ be a nonsingular $n \times n$ matrix, $\|\cdot\|$ be any natural norm, and $K_{p}(A)=$ $\|A\|_{p}\left\|A^{-1}\right\|_{p} .$ Let $\lambda_{1}$ be the smallest and $\lambda_{n}$ be the largest eigenvalues of the matrix $A^{t} A$ (b) Show that $K_{2}(A)=\sqrt{\frac{\lambda_{n}}{\lambda_{1}}}$. (Hint: One should use the $\|A\|=\sqrt{\rho\left(A^{t} A\right)}=\sqrt{\rho\left(A A^{t}\right)}$ relation) For part $(b)$, it seems minimal eigenvalue of matrix $A$ will be the largest eigenvalue of the inverse of matrix $A$, so we have to use this idea but could not understand formally why square roots are needed for $K_{2}(A)$ and how spectral radius is necessary for this part?!
73092
These facts are taken for granted in a QM text I read. The purportedly guaranteed non-normalizability of eigenfunctions which correspond to a continuous eigenvalue spectrum is only partly justified by the author, who merely states that the non-normalizability is linked to the fact that such eigenfunctions do not tend to zero at infinity.  Not a very satisfying answer. What I'm really after is an explanation based in functional analysis. I believe there is a generalized result about inner products being finite for discrete spectra but infinite for continuous spectra.  Can anyone shed some light on this?
119882
I came across a problem that says:  Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function. If $|f'|$ is bounded, then which of the following option(s) is/are true?     (a) The function $f$ is bounded.   (b) The limit $\lim_{x\to\infty}f(x)$ exists.   (c) The function $f$ is uniformly continuous.   (d) The set $\{x \mid f(x)=0\}$ is compact.   I am stuck on this problem. Please help. Thanks in advance for your time.
59973
I have a question in my assignment :     If $f$ is a nonconstant entire function such that $|f(z)|\geq M|z|^n$   for $|z|\geq R$ for some $n\in\mathbb N$ and some $M$ and $R$ in   $(0,\infty)$ show that $f$ is a polynomial of degree atleast $n$.   Now , I defined a function    $\ g(z) = \frac {1}{f(z)}\ $ such that $\ |g(z)| \le \frac{1}{M{|z|}^n}.$  Now , by using the cauchy inequality  $$|g^{(n)}(z)| \le  \frac{n!}{R^n |z|^nM}.$$    Considering that $ g(z) $ is an analytic function , it has a radius of convergence $ \infty $  $ \implies\  g^{(n)}(z) = 0.$    But if we go by this approach , then $ g^{(n)}(z) = 0 \ $ for any n . Also how can we be so sure that $ f(z) \neq 0 $ for any z ?  Is my reasoning correct or is there some other way to solve it ?
48673
It is apparently that this holds in Hilbert space, but I can not prove this for general inner product space or find a counterexample. (The only not complete inner product space known to me is $L^2$ restricted in continuous functions.)   Here give a example, but it seems don't meet the condition closeness.  By the way, Is there any other less complicated spaces equipped with inner product but not complete?
66254
It's true that I'm not familiar with too many exotic functions, but I don't understand why there exist functions that cannot be described by a Taylor series?  What makes it okay to describe any particular functions with such a series? Is there any difference for different number sets? In the case of complex numbers maybe?  Could somebody provide an example?
26441
A question that may sound very trivial, apologies beforehand. I am wondering why $( \mathbb{Q}_{&gt;0} , \times )$ is not isomorphic to $( \mathbb{Q} , + )$. I can see for the case when $( \mathbb{Q} , \times )$, not required to be positive, one can argue the group contains elements with order 2 (namely all negatives). In the case of the requirement for all rationals to be positive this argument does not fly. What trivial fact am I missing here?
66521
I realize that every inner product defined on a vector space can give rise to a norm on that space.  The converse, apparently is not true.  I'd like an example of a norm which no inner product can generate.
116151
Define $f(z) = z \ ^ 2 + \lambda$ and $V = \left\{x \in \Bbb C : |x| \gt \frac{1}{2} + \sqrt{\frac{5}{4} + |\lambda|} \right\}$, for some $\lambda \in \Bbb C$.   Then, if $x \in V$ then $|f(x)| \gt |x| +1$.   I tried the following :   Let $x\in V$ then $|f(x)| = |x \  ^ 2 +\lambda | \ge |x |  \ ^ 2 - | \lambda| \gt (\frac{1}{2} + \sqrt{\frac{5}{4} + |\lambda|}) \  ^ 2 - |\lambda| = 1.5 + \sqrt{\frac{5}{4} + |\lambda|} $.  But this doesn't get me anywhere.
60967
I am looking for an Integrable function $f : [0,1] \to \mathbb{R}$ which is not essentially bounded on any subinterval of $[0,1]$.  Thank you in advance.
47051
Let $f: \mathbb C \to \mathbb C$ be an entire function. The order of $f$ is defined by $$\lambda=\limsup_{r \to \infty} \frac{\log \log M(r)}{\log r}, $$ where $$M(r)=\max_{|z|=r} |f(z)| .$$  Ahlfors in his Complex Analysis claims that     "According to this definition $\lambda$ is the smallest number such that       $$M(r)\leq e^{r^{\lambda+\varepsilon}} $$ for any given $\varepsilon &gt; 0$ as soon as $r$ is sufficiently large."   Why is this true?  My attempt:  We know that $$\lambda=\lim_{\rho \to \infty} \sup_{r \geq \rho} \frac{\log \log M(r)}{\log r}. $$  From the definition of the limit we have that for any $\varepsilon&gt;0$, there exists some $\rho_0&gt;0$, such that $$\left\lvert \sup_{r \geq \rho} \frac{\log \log M(r)}{\log r}-\lambda \right\rvert \leq \varepsilon ,$$ for every $\rho \geq \rho_0$. In other words $$\frac{\log \log M(r)}{\log r} \leq \lambda+\varepsilon $$ for every $r \geq \rho_0$. From here it is easy to see that $$M(r)\leq e^{r^{\lambda+\varepsilon}}, $$ for all $r \geq \rho_0$. I cannot see why $\lambda$ is the smallest number with this property.  Thanks in advance.
55720
Do there exist functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y),$ but which aren't linear? I bet you they exist, but I can't think of any examples.  Furthermore, what hypotheses do we need to put on $f$ before no such functions exist? I feel continuity should be enough.