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17

I thought I never did it directly, but now that I have found the solution below (quite quickly), I begin to suspect that I must have done something similar years ago. Anyway, as a first step, let's square everything. Then we need to prove this: $$ \left(\sum_i x_i y_i\right)^2 \leq \left(\sum_i x_i^2\right)\left(\sum_j y_j^2\right). $$ Let's subtract the ...


6

It's not hard. In fact the first proof that I encountered in high school was without using inner products. It goes as follows: Consider the quadratic $$\sum_{i=1}^{n} (a_ix+b_i)^2 = (\sum_{i=1}^{n}a_i^2)x^2 + 2 (\sum_{i=1}^{n}a_ib_i)x + (\sum_{i=1}^{n}b_i^2)$$ Since the quadratic expression is always non negative, its discriminant must be $\leq 0$ [1]. ...


4

Consider the following sequence of elements in the space $V$ of finite sequences: $$ u_1=(1,0,0,\ldots),\ \ u_2=(0,\frac12,0,\ldots),\ \ u_3=(0,0,\frac13,0,\ldots) $$ Then $$ \sum_{k=1}^nu_k=(1,\frac12,\frac13,\ldots,\frac1n,0,\ldots) $$ Now consider these two norms on $u=(a_1,a_2,\ldots)$: $$ \|u\|_1=\sum_{k=1}^\infty|a_k|,\ \ \ ...


3

It seems weird that it should be easier to define a lot of new terms, just to prove an inequality. In general you can find many examples of problems (even inequalities) that are solved easier with the use of some mathematical machinery. Dan Shved gave an excellent answer to your question. Let me just add that you could prove the inequality for n=2 ...


2

One can significantly simplify your proof. Assume $B(X,Y)$ is Banach. The normed space $Y$ can be isometrically be embedded into the $B(X,Y)$ via the map $$ I:Y\mapsto B(X,Y):y\mapsto f(\cdot) y $$ for some $f\in X^*$ of norm 1. Since embedding is isometric, then $\operatorname{Im} I$ is a closed subspace of Banach space $B(X,Y)$. Hence $\operatorname{Im} ...


2

Given a Cauchy sequence $x_n$ it is straightforward to see that $x_n(i) $ converges to some $x(i)$ for all $i$. Let $\epsilon>0$ and choose $N$ large enough so that if $n,m \ge N$, then $\|x_n-x_m\|_1 < {\epsilon \over 2}$. \begin{eqnarray} \sum_{i=1}^L |x_n(i)-x(i)| &\le & \sum_{i=1}^L |x_n(i)-x_m(i)| + \sum_{i=1}^L |x_m(i)-x(i)|\\ &\le ...


2

It is mostly correct now, but could be written better. Consider $l:X'/U^\perp\to U'$, which sends each coset of $U^\perp$ to its restriction to $U$. This map is well-defined because all elements of the coset agree on $U$. The map is surjective, because every functional $f$ on $U$ can be extended to a functional $g$ on $X$ (by Hahn-Banach), and ...


1

Here is a sketch of the first part. For each $f\in C[0,1]$, \begin{eqnarray*} \|Tf\| &=& \max_{t\in[0,1]}\left|\int_{0}^{1}K(t,s)f(s)ds\right|\\ &\leq& \max_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)f(s)\right|ds\\ &\leq& \|f\|_{\infty}\max_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)\right|ds\\ \end{eqnarray*} Therefore $\|T\|$ is ...


1

The fact that dual to dual norm is equal to the original norm in case of finite-dimensional spaces is equivalent to the fact that the corresponding Banach space is reflexive. By James' theorem, a Banach space $B$ is reflexive if and only if every continuous linear functional on $B$ attains its maximum on the closed unit ball in $B$. That is surely true for ...


1

My question: I can not show that when $\oplus_\infty X_i$ is separable then $I$ is finite. Literally, that need not be true. The set of $i$ such that $X_i \neq \{0\}$ must be finite. Now, if $J = \left\{i \in I : X_i \neq \{0\}\right\}$ is infinite, then take a sequence $(j_n)_{n\in\mathbb{N}}$ of distinct elements of $J$, for each $j_n$ choose an ...


1

$A$ is closed, but not open. The sequence $x=(1,1,1,...)$ is not an interior point. If $(x_n)$ is a convergent sequence in $A$ with the limit, then it also converges pointwise. But since $0\le x_{n,m}\le 1$, we also have $0\le \lim \limits_{n\to \infty} x_{n,m}\le1$, so $\lim \limits_{n\to \infty}x_n\in A.$ $B$ is open, but not closed. If $f\in B$, not that ...



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