Could you please give me hints may leads to prove the following:  Let $X$ be a real vector space, $\,p_1,p_2:X\to\mathbb R\,$ be two , and $\,f:X\to\mathbb R\,$ be a linear functional satisfying $$ f(x)\le p_1(x)+p_2(x), \quad\text{for all $\,x\in X$.} $$ Prove that there exist two linear functionals $\,f_1,f_2:X\to\mathbb R$, such that $\,f=f_1+f_2\,$ and $\,f_i(x)\le p_i(x),\,$ for $i\in\{1,2\}$ and $x\in X$.  I think we should use Hahn-Banach Theorem.  Thanks in advance.
123786
Let $X$ be a linear space, $p, q$ sublinear functionals on $X$, and $L$ a linear functional on $X$ such that $|L(x)| ≤ p(x) + q(x),$ for all $x ∈ X$. Prove that there exist linear functionals $L_1, L_2$ on $X$ such that $L(x) = L_1(x) + L_2(x),$ and $|L_1(x)| ≤ p(x), |L_2(x)| ≤ q(x),$ for all $x ∈ X.$   My Work:  First I thought to use Hahn Banach Theorem. But since there is no known subspace it was useless. Then I tried to make $L(x)$ as $L(x)=\frac{L(x+\lambda)+L(x-\lambda)}{2}$ for some scalar $\lambda$ but failed to find suitable $L_1$ and $L_2$. I think this problem is little bit tricky. I want to try it myself and I only need a hint to start. Can somebody please give me a hint?
40932
I'm working on an exercise from functional analysis.  Let $E$ be a vector space and $\|\cdot\|_1$ and $\|\cdot\|_2$ be two complete norms on $E$. Now suppose that $E$ satisfies the following property:  $\bullet$ if $(x_n)$ is a sequence in $E$ and $x,y\in E$ such that $\|x_n-x\|_1\to 0$ and $\|x_n-y\|_2\to 0$, then $x=y$.  Now we want to show that the norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent.  My idea is as follows: If for any $n&gt;0$, there is an element $x_n\in E$ such that $\|x_n\|_1&gt;n\|x_n\|_2$. Then consider $(\frac{x_n}{\|x_n\|_1})_{n\geq 1}$. Clearly, $(\frac{x_n}{\|x_n\|_1})_{n\geq 1}$ converges to $0$. However, I cann't get a contradicition from this. Maybe my idea is wrong.   In fact, I even don't konw how to show that a Cauchy sequence in norm $\|\cdot\|_1$ is also a Cauchy sequence in norm $\|\cdot\|_2$.   Anyone can give me some hints or a counter example? Thank you very much.
112741
If $V_1$ and $V_2$ are vector spaces over the field of rational numbers and $f \colon V_1 \to V_2$ is a map such that $f(x+y)=f(x)+f(y)$ for all $x, y \in V_1$, show that $f$ is a linear transformation.   It is easy to show that $f(0)=0$ and $f(-x)=-x$. Here scalars are from rationals. If scalar is an integer then clearly $f(\alpha x)=\alpha f(x)$. Hence $f$ is a linear transformation.  Now I have to show that when $\alpha$ is not an integer then $f(\alpha x)=\alpha f(x).$
35477
Let  $f_1,\ldots,f_n$ be linearly independent linear functionals on a vector space $X$. Show that there exist $x_1,\ldots,x_n\in X,$ such that the $n\times n$ matrix $\big(f_i(x_j) \big)_{i,j=1}^n$ is non-singular.
6030
Prove that a linear transformation $P \colon V \to V$ of a finite dimensional vector space satisfies $P^2 = P$ if and only if there exists a basis with respect to which $P$ can be written as a block matrix   $$P = \begin{bmatrix} I &amp; 0 \\ 0 &amp; 0 \end{bmatrix}.$$  Hence determine the minimal and characteristic polynomials of $P$.  I don't know what to do with this problem at all, so some help would be great, Thanks.
63224
In the middle of another proof (Theorem 3.4, p. 60) in his Functional Analysis book, Rudin says that  "every nonconstant linear functional on $X$ (topological vector space) is an open mapping."   Is this obvious? Could someone please show me how the proof goes?
41637
Can you show me a continuous function $f \colon \mathbb{R}^n\to\mathbb{R}^m$ that satisfies $f(a+b)=f(a)+f(b)$ but is not linear?   We have that  $$f(0)=f(0+0)=2f(0)\implies f(0)=0\\ f(x-x)=f(0)=f(x)+f(-x)=0\implies f(-x)=-f(x)\\ f(nx)=f(x+x+\dots+x)=f(x)+\dots+f(x)=nf(x)\quad \forall n \in \mathbb{N}$$ But $$ f(-nx)=-f(nx)=-nf(x) $$ So: $$ f(ax)=af(x) \quad \forall a \in \mathbb{Z} $$
78138
Let $a$, $b$ be two arbitrary but fixed real numbers such that $a &lt; b$, let $C[a,b]$ denote the normed space of all continuous real (or complex) valued functions defined on $[a,b]$ with the maximum norm, and let $f$ be the functional defined on $C[a,b]$ as $$f(x):= \int_{a}^{b} x(t) x_0(t) dt $$ for all $x$ in $C[a,b]$, where $x_0$ is a fixed element of $C[a,b]$.   Now $f$ is evidently linear and bounded. How to compute the norm of $f$? What is this norm?   And what is the norm of the bounded linear functional $g$ defined as follows: $$g(x):= \int_{a}^{(a+b)/2} x(t) \cdot dt - \int_{(a+b)/2}^{b} x(t) \cdot dt$$ for all $x$ in $C[a,b]$?
116634
Show that any bounded linear functional on a normed linear space is continuous.   Can we say that it is uniformly continuous ?   Also, is it true, if we reverse the statement    any continuous  linear functional on a normed linear space is bounded ??
15335
I'm reading Axler "Linear agebra done right" and in Chapter 1 he discusses subspaces and direct sum. My question is, are there subspaces of the infinite-dimensional vector spaces, e.g. a functional Banach space with sup norm $V$ that directly sum to the entire space $$V = U_1\oplus U_2$$ (except for trivial $U_1=\{0\}$, $U_2 = V$).
8151
Prove: If a linear map, $φ: U→V$  is bijective, then $φ(x)=0$ has only the solution $x=0$.  Either a proof or explanation will be appreciated.
27467
I am trying to prove the following. I have seen it alluded to in other places of the internet (this site included) but without proof.  Let $L,L_1\ldots L_n$ be linear functionals on a vector space $X$. If $\bigcap_{i=1}^n ker(L_i) \subset ker(L)$ then there exists $t_i$ for  $i=1\ldots n \in \mathbb{R}$ such that $L = \sum_{i=1}^n t_i L_i$.   In other words, if the intersection of kernels of linear functionals is contained by the kernel of another linear functional then they are linearly dependent.  Related:
14924
I would like only a hint to the following exercise:  Let $V$ be a vector space over the field $K$, and $T$, $S$ linear functionals on V such that $Tv=0\Rightarrow Sv=0$. Prove that there exists $r\in K$ such that $S=rT$.  I know how to prove this when $V$ is finite dimensional. I show that if there is no such constant $r$ then $n-2=\operatorname{dim}\textrm{ }(\ker\textrm{ }T\textrm{ }\cap \ker\textrm{ }S)=\operatorname{dim}\ker T=n-1$, a contradiction. But this approach doesn't seem to help at all for the stated problem.
64034
I recently started studying functional analysis. I have many ebooks loaded on my laptop, but can't figure out which one to start with. I've asked my instructor, and he says there aren't any specific books to start with. Here's a brief list of the topics on our syllabus:   Normed linear spaces Banach Spaces Hilbert spaces Compact Operators Knowledge of $C[0,1],L^p[0,1]$ Continuous linear operators Hahn-Banach theorem    Open mapping and Closed Graph Theorem   Uniform Boundedness Principle   I'm seeing these topics first in my lifetime. One thing I want to ask is whether functional analysis is very difficult. Please suggest some books to start this topic which covers the above and contains some exercises. Thanks in advance.
123150
The second part of an exercise asks that I show:     There does not exist a surjective bounded linear map from $\ell^2(\mathbf{N})$ to $\ell^1(\mathbf{N})$.     The first part was to show that these two spaces are separable in their respective $L^p$ norms.    I've really hit a wall on this one.  I know, in fact, that there does exist a surjective bounded linear map from $\ell^1(\mathbf{N})$ to $\ell^2(\mathbf{N})$.  So my intuition says that if there were also one going the other way, then the two spaces would have to be isomorphic, which is false .  But I don't know of any such theorem that I could apply to "the category of Banach spaces".    My only other idea is to argue by contradiction using the open mapping theorem, but nothing is coming to me...  My guess is that separability is also going to be important, given the first part of the exercise.  -Thanks.
28309
Let $(X,\langle\cdot,\cdot\rangle)$ be an inner product space. Prove if $X$ is a real vector space, then  $\lVert x+y\rVert= \lVert x-y \rVert$ implies $x\perp y$.
37019
How does one show that $(C[0,1], ||.||_{2})$ where $||.||_2=(\int_{0}^{1}|f(t)|^2dt)^{1/2}$ and $C[0,1]$ is the space of all continous function which are mapped from $[0,1]\rightarrow \mathbb{R}$, is not complete?   First to show that $(C[0,1], ||.||_{2})$ is a normed vector space:   $||.||_2$ fulfills $||f(t)||=0 \Rightarrow f(t)=0$   $||.||_2$ fulfills the  scalar multiplication property: $$||\lambda f(t)||_{2} = (\int_{0}^{1}|\lambda f(t)|^2dt)^{1/2}=|\lambda|(\int_{0}^{1}| f(t)|^2dt)^{1/2}=|\lambda|||f(t)||_2$$  Thirdly, it also fulfills the triangle inequality:$$||x(t)+y(t)||_2^2 = \int_{0}^{1}|x(t)+y(t)|^2dt \le \int_{0}^{1}|x(t)|^2dt + 2\int_{0}^{1}|x(t)y(t)|dt +\int_{0}^{1}|y(t)|^2dt \le ||x(t)||_2^2 + 2||x(t)||_2||y(t)||_2+||y(t)||_2^2 = (||x(t)||_2+||(y(t)||_2)^2$$  this also shows the closure under addition in $C[0,1]$  so , $(C[0,1],||.||_2)$ is a normed vector space , but how can we show that it is not complete?