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The inner product is already determined by the norm using the Polarization identity so no need to try and build it in a non-constructive way. If you define a function $\left< \cdot, \cdot \right>$ on $X \times X$ by $$\left< u, v \right> := \frac{||u + v||^2 + ||u - v||^2}{4}$$ then clearly $\left<u, u\right> = ||u||^2$ for all $u \in ... 6 $$||x||+||y||-||x+y|| \\ = ||x||-\frac{||x+y||}{2}+||y||-\frac{||x+y||}{2} \\ \le ||x -\frac{x+y}{2}|| + ||y -\frac{x+y}{2}|| \\ \le \frac{||x-y||}{2} + \frac{||y-x||}{2} \\ \le ||y-x||$$ 4 Just note that every norm is continuous (because 1-lipschitz from the second triangle inequality$|\,||x||-||y||\,|\leq||x-y||$) and then$A=||\cdot||^{-1}(\{1\})$is a closed set as a reciprocal image of the closed set$\{1\}$of$\mathbb{R}$by the continuous map$||\cdot||.$3 A linear functional on a$\mathbb{K}$-vector space$f:E\rightarrow\mathbb{K}$is continuous if and only if it is Lipschitz, and thus if and only if there exists$M\geq 0$such that$\|f(x)\|\leq M\|x\|$for all$x$. The norm on the real Hilbert space$\ell^2(\mathbb{N}^*)$is associated to a dot product which verifies some inequality. 3 By definition$\|A\|=\sup_{\|x\|\leq 1} \|Ax\|$, and by Hahn-Banach theorem we can show$\forall y\in Y, \|y\|=\sup_{\|f\|\leq 1,f\in Y^*} \|f(y)\|$. Combining them, we have$\|A\|=\sup_{\|x\|\leq 1} \sup_{\|f\|\leq 1,f\in Y^*} \|f(Ax)\|$, which is what you want. 3 The claim that$T$is surjective implies range of$T^*$being closed is not true. Take$X=c_{00}=Y$the space of sequences with finite length. The dual space can be identified with$l^1$. Define $$Tx = (x_1, x_2/2, \dots, x_n/n,\dots).$$ Clearly,$T:X\to Y$is injective and surjective, however$T^{-1}$is not bounded. Let$g\in l^1$be given. Then for ... 3 Hint: The sequence of numbers$\| x_n\|$is real and bounded, so it must contain a convergent subsequence by the Bolzano–Weierstrass theorem. 3 Another reason: a compact subset of the reals is closed and bounded (Heine-Borel theorem), and the rationals are distinctly not bounded. 3 Hint: try checking that$P([0,1]) \ni p \mapsto p' \in P([0,1])$is not bounded in the unit sphere. Here$P([0, 1])$is the space of polynomials in$[0,1]$, with the sup norm. 2 A compact space is complete. Another reason: a compact subspace is closed. And precisely, the closure of$\mathbf Q$is$\mathbf R$. 2 It is not necessarily true. Consider$x_n = 1/\sqrt n \not\in \ell_2$. Then for any$y_n \in \ell_2$$$\| x_ny_n \|_2 = \sqrt{\sum_{i=1}^\infty |x_ny_n|^2} \leq \sqrt{\sum_{i=1}^\infty |y_n|^2} = \| y_n \|_2$$ That is,$x_ny_n \in \ell_2$for all$y_n \in \ell_2$. 2 Let's take a Cauchy-sequence$(a_{n})_{n}\subset V$. By the completeness of$V$with respect to the first norm$\Vert\cdot\Vert_{1}$, we have $$\forall\,\epsilon>0,\,\exists\,N\in\mathbb{N}:\forall\,n\geq N:\Vert a_{n}-a\Vert_{1}<\epsilon$$ for some$a\in V$. As$\Vert\cdot\Vert_{1}$and$\Vert\cdot\Vert_{2}$are equivalent, there exists$C>0$such ... 2 Let$X$be a Banach space and let$\{x_d\colon d\in D\}$be a dense subset of the unit ball of$X$. Consider the space$\ell_1(D)$of all absolutely summable sequences on$D$. We define a linear map$T\colon \ell_1(D) \to X$by $$T\Big((\lambda_d)_{d\in D}\Big) = \sum_{d\in D}\lambda_d x_d\qquad ((\lambda_d)_{d\in D} \in \ell_1(D)).$$ This is a well-defined ... 2 WLOG,$x_0 = 0.$Let$\epsilon>0.$Choose$\delta > 0$such that$x\in B(0, \delta) \implies \|Df(x) - Df(0)\| < \epsilon.$Fix$s,t\in B(0,\delta).$Note that the map$l(u) = t + u(s-t)$is a differentiable map from$\mathbb R$to$E,$with$Dl(u)[v] = v(s-t).$So if we define$g = f\circ l,$we see$g:\mathbb [0,1] \to \mathbb R$is the ... 2 Define the norm$\lvert \cdot \rvert_1 \, \colon X \rightarrow [0, \infty)$by \begin{equation*} \lvert x \rvert_1 =\sum_{l=1}^{n} \lvert \lambda_l \rvert, \end{equation*} where$x \in X$is given by$x = \sum_{l=1}^{n} \lambda_l v_l$. Let us now choose a vector$x_i$from the sequence$\{x_i \} $. Then, by definition of$\lvert \cdot \rvert_1$, ... 2 Hint:$g$is linear. A linear operator is continuous if and only if it is continuous at$x=0$, if and only if it is bounded. Try to find a constant$C>0$such that for all$x$: $$|g(x)|\leq C \|x\|$$ 2 The answer to the first question is "yes", and I guess this is originally due to Whitney. For a proof see Infinitely differentiable function with given zero set? The answer to the second question is "no" in general for$n \ge 2$. E.g., if$n=2$and$A$is the unit circle in the plane, then you can find a regular value$y=f(x)$where$x$is contained in the ... 2 Edit: The following answers a former version of the question, referring to a subset of$S^{n-1}$. The statement in the title, appearing again in the body of the question, is wrong. A subset $$A\subset S^{n-1}\subset {\mathbb R}^n$$ is compact iff it is closed. Consider, e.g., the set $$A:=\bigl\{x\in{\mathbb R}^n\>\bigm|\>\|x\|=1, \ x_n>0\bigr\}\ ... 2 Not in general. Here is a counterexample. Let X = C([-1,1]). For t \in [-1,1], let \delta_t denote the point mass / evaluation functional \delta_t(x) = x(t). Let D = \{ \delta_q : q \in [-1,1] \cap \mathbb{Q}\}. Then D is countable and we have \|x\| = \sup_{f \in D} |f(x)| for every x \in X. Let$$y(t) = \begin{cases} 4t, & -1 \le t ... 2 If$||I-P||<1 \implies P$is invertible$\implies P=I$1 Is there a typo on your right hand side? I assure you mean$f(||y||-||x||)$. If so, every$f(\cdot)$that is monotonically decreasing should do the trick. What is left to check are the limits on$x$and$y$1 English-speaking mathematicians use the word "any" too much. Pick any projection on a normed linear space onto a subspace and then prove that it satisfies this inequality? I don't think that's what you meant, but it bears that interpretation in normal English usage. Just saying "every" instead of "any" costs nothing, except two keystrokes. I take ... 1 (i) (1) follows because the sum of two norms is a norm. (2) is not a norm;$\|f'\|_\infty = 0 $does not imply$ f \equiv 0.$(3) is a norm, even though it the sum of a norm and a non-norm. (4) I do not understand your question "What is$|f(0)|?$" Anyway, this is a norm: Show$|f(0)| +\|f'\|_1 = 0 \implies f' \equiv 0 \impliesf$is constant; the condition ... 1 The$l_p$norm for$(x,y)\in\mathbb{R}^2$is this: $$||(x,y)||_p=(|x|^p+|y|^p)^{1/p}$$ So basically your intuition of putting zeros after the second place in an infinite dimensional vector is correct, but in fact there is no need to carry them around. 1 Let$A:U\rightarrow V$be bounded. If$X \subset U$is bounded there exists a$K \in \mathbb{R}$such that$||x|| < K$for all$x \in X$. Let$y \in A(X)$i.e.$Ax = y$for an$x \in X$then$||y|| = ||Ax|| \leq ||A||\cdot ||x|| \leq ||A||\cdot K$therefore$A(X)$is bounded too. Now let$A$send bounded sets to bounded sets and let$K$be a bound for ... 1 Hint: use the Cauchy-Schwartz inquality of$L^2([-1,1])$to show that$L$is bounded$\mid L(f)\mid\leq \int_{-1}^1\mid f(t)\mid dt=<\mid f(t)\mid,1>_{L^2}$1 Often (not always) it's simpler to show the continuity of a linear functional on a normed space by explicitly giving a bound on its norm rather than showing that its kernel is closed. Here, we have $$\lvert L(\varphi)\rvert = \biggl\lvert \int_{-1}^0 \varphi(t)\,dt - \int_0^1 \varphi(t)\,dt\biggr\rvert \leqslant \int_{-1}^1 \lvert \varphi(t)\rvert\,dt ... 1 On one side, let x \in \bar E, choose a sequence (x_n) \in E^{\mathbf N} such that x_n \to x. If now x' \in X' is given such that \Re x'|_E \ge 1. Then, as \Re x' is continuous, we have$$ 1 \le \Re x'(x_n) \to \Re x'(x) $$That is \Re x'(x) \ge 1. Along the same line we see that \Re x'|_E \le 1 implies \Re x'(x) \le 1. For the other ... 1 Let \delta=\inf\{\|x-x_0\| :x \in K\}, this implies that there exists a sequence (x_n) in K such that ||x_0-x_n||\to \delta. Since K is a closed and bounded subset of \Bbb R^n, it is compact so there is a convergent subsequence (x_{n_k}) of (x_n) such that x_{n_k}\to x^*\in K. We then have$$ ||x^*-x_0||=\lim_{k\to \infty}||x_{n_k}-x_0||= ... 1 Indeed!$||x_n-x_m|| < \epsilon$for$n,m > N$is equivalent to saying that$x_m \in B_\epsilon(x_n)$for$n,m > N\$.