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It's not immediate or trivial, so I wouldn't feel too bad for having trouble. This is an exercise in Friedberg, Insel, and Spence's Linear Algebra, 4th Edition, which has an extensive 8 part "Hint." Here's an edited sequence of hints, following theirs: First, prove that the result holds for $\lambda = 2$, that is, $\langle 2u,v\rangle = 2\langle ... 71 Since this question is asked often enough, let me add a detailed solution. I'm not quite following Arturo's outline, though. The main difference is that I'm not re-proving the Cauchy-Schwarz inequality (Step 4 in Arturo's outline) but rather use the fact that multiplication by scalars and addition of vectors as well as the norm are continuous, which is a bit ... 56 I sent Professor Purdy an email. I asked him what he recalled about the incident. With his permission I've copied his correspondence below. Dear Jacob, Yes, I was there, and I'm the one who told the story to Paul Hoffman, who then included it in his book "The man who loved only numbers." The 30 page proof was written by Jack Bryan just ... 53 Fix$\delta>0$and let$S_\delta:=\{x,|f(x)|\geqslant \lVert f\rVert_\infty-\delta\}$for$\delta<\lVert f\rVert_\infty$. We have $$\lVert f\rVert_p\geqslant \left(\int_{S_\delta}(\lVert f\rVert_\infty-\delta)^pd\mu\right)^{1/p}=(\lVert f\rVert_\infty-\delta)\mu(S_\delta)^{1/p},$$ since$\mu(S_\delta)$is finite and positive. This gives ... 52 It seems to me that you've got the things quite right. I restrict attention to real or complex vector spaces. (Edit: But see the update.) Given a scalar product$\langle \cdot, \cdot\rangle$we get a norm by setting$\|x\| = \sqrt{\langle x,x \rangle}$by an application of the Cauchy-Schwarz inequality. Conversely, a normed vector space structure comes ... 50 Let me ask you a dual question: I am a mathematics student in set theory, why don't category theory students do set theory? I find it strange that most books on category theory have only a naive handling of set theory. Now let me answer your question. Category theory is an impressive tool for abstraction, but analysis is not always in need for abstraction - ... 48 [...] Grothendieck, with his background in functional analysis, must have been familiar with Stone's work in that field. cited by Qiaochu reveals a gap in the knowledge of Grothendieck's work (which is of course nothing to be blamed for, given its vastness and ramifications in so many fields) but it also misses a link to one of Grothendieck's beautiful ... 47 Robert's and joriki's examples are of course nice and explicit, but you can get examples on any subset of$\mathbb{R}^n$with infinite measure. Here's how: Take a function$f$that is in$L^p$but not in$L^q$for$q \gt p$(on the unit ball$B$around zero, say). Now take a sequence$x = (x_n)$that is in$\ell^p$but not in$\ell^q$for$q \lt p$(there ... 42 There are three main ways of interpreting the Fourier transform. Decomposition relative to eigenfunctions of the Laplacian On$\mathbb{R}^n$, the plane waves$E_\xi(x) = \exp( i \xi\cdot x)$can be interpreted as generalized eigenfunctions of the Laplacian. That is, let $$\triangle = \sum_{i=1}^n \left(\frac{\partial}{\partial x_i}\right)^2$$ then ... 42 Section 1 Let us begin with the following theorem. Theorem 1 Let$ G $be a topological group. If$ G $admits a Lie group structure, then this structure is unique up to diffeomorphism. Proof: Suppose that$ \mathcal{A}_{1} $and$ \mathcal{A}_{2} $are smooth structures (maximal smooth atlases) on$ G $that make it a Lie group. Observe that the ... 41 This is a theorem by Riesz. Observe that $$|f_k - f|^p \leq 2^p (|f_k|^p + |f|^p),$$ Now we can apply Fatou's lemma to $$2^p (|f_k|^p + |f|^p) - |f_k - f|^p \geq 0.$$ If you look well enough you will notice that this implies that $$\limsup_{k \to \infty} \int |f_k - f|^p \, d\mu = 0.$$ Hence you can conclude the same for the normal limit. 40 Well, it seems that you have just discovered a beautiful theory of (semi)group generators by yourself. To give some basics of it, let us consider a collection of "nice" functions on real values - e.g. bounded and having continuous derivatives. The action of operators$L^h$on this space has a semigroup structure: $$L^s(L^tf(x)) = L^sf(x+t) = f(x+s+t) = ... 38 The essential idea of many transforms is to change the basis in the space of functions with the hope that in the new basis the problem will simplify. Let me give a finite-dimensional example. Suppose we have a 2\times2 matrix A and we want to compute A^{1000}. Direct approach would not be very wise. However, if we first diagonalize A as PA_dP^{-1} ... 37 Yes, if (\Omega, \Sigma, \mu) is a (complete) \sigma-finite measure space then (L^{\infty}(\Omega,\Sigma,\mu))^{\ast} is the space \operatorname{ba}(\Omega, \Sigma,\mu) of all finitely additive finite signed measures defined on \Sigma, which are absolutely continuous with respect to \mu, equipped with the total variation norm. The proof is ... 37 There is a Theorem of Wielandt which asserts that if A is any normed algebra, complete or not, we can't express I = 1_{A} in the form xy - yx. The proof is given in Rudin's book, but it is so beautiful that I give it here. Suppose that xy -yx = I. I claim that xy^{n} - y^{n}x = ny^{n-1} for all n \in \mathbb{N}. We have the case n = 1. Suppose ... 37 In short terms, kets are vectors on your Hilbert space, while bras are linear functionals of the kets to the complex plane$$\left|\psi\right>\in \mathcal{H}$$\begin{split} \left<\phi\right|:\mathcal{H} &\to \mathbb{C}\\ \left|\psi\right> &\mapsto \left<\phi\middle|\psi\right> \end{split} Due to the Riesz-Frechet theorem, a ... 36 Let V be a vector space over the field \mathbb{F}. A norm$$\| \cdot \|: V \longrightarrow \mathbb{F}$$on V satisfies the homogeneity condition$$\|ax\| = |a| \cdot \|x\|$$for all a \in \mathbb{F} and x \in V. So the metric$$d: V \times V \longrightarrow \mathbb{F},d(x,y) = \|x - y\|$$defined by the norm is such that$$d(ax,ay) = \|ax - ... 34 Here's a sketch of a proof. Let$\sigma(x)$denote the spectrum of$x$. Then$\sigma(xy)\cup\{0\} = \sigma(yx)\cup\{0\}$. On the other hand,$\sigma(1+yx)=1+\sigma(yx)$. If$xy=1+yx$, then the previous two sentences, along with the fact that the spectrum of each element of a Banach algebra is nonempty, imply that$\sigma(xy)$is unbounded. But every ... 34 Since$1/x$is the border case in both directions, the most promising candidate would be a modified version of$1/x$that just converges but won't converge if you nudge it ever so slightly. We have $$\int_2^\infty \frac1{x\log^2x}\mathrm dx=\left[-\frac1{\log x}\right]_2^\infty=\frac1{\log2}\;,$$ whereas $$\int_2^\infty ... 33 Considering the fact that you have only had one undergraduate course in analysis and will be taking an actual functional analysis class, I don't think you actually want to self-study functional analysis. It would be much more useful for you to bulk up on your linear algebra review your real analysis Functional analysis is, for a large part, linear ... 33 Consider g_k = 2^p(|f_k|^p + |f|^p) - |f_k - f|^p. Since g_k \geq 0 (why?), and g_k \to 2^{p+1}|f|^p a.e., we can apply Fatou's Lemma:$$\int \liminf g_k \leq \liminf \int g_k$$so that$$\int 2^{p+1}|f|^p \leq \liminf \left(\int 2^p |f_k|^p + \int 2^p |f|^p - \int |f_k - f|^p \right),$$and I'll let you take it from here. 33 If P is an infinitely differentiable function such that for each x, there is an n with P^{(n)}(x)=0, then P is a polynomial. (Note n depends on x.) See the links in the comments. 33 There are two questions here, in reality, I think. First, in brief, I am told by many people that I "do functional analysis in the theory of automorphic forms", and I certainly do find a categorical viewpoint very useful. Second, in brief, it is my impression that the personality-types of many people who'd style themselves "(functional) analysts" might be ... 33 It seems that the proof using the Baire category theorem can be found in several places on this site, but none of those questions is an exact duplicate of this one. Therefore I'm posting a CW-answer, so that this question is not left unanswered. We assume that a Banach space X has a countable basis \{v_n; n\in\mathbb N\}. Let us denote ... 33 Yes, you are right. It was Jack Bryant, not Jack Brian. He might have retired by now. By the way, even before Erdos came to town, it was generally agreed that there must be a proof that was shorter than 30 pages, but not a two liner! Professor Don Allen talked to Erdos the next day to see if he could help with the research problem that had generated this ... 32 Check out "Introductory Functional Analysis with Applications" by Erwin Kreyszig. I have not read it myself, but I have heard great things. Also, in the preface, he writes that Calculus and a familiarity with Linear Algebra are all that's needed as prerequisites. 32 Suppose T: L^\infty \to \ell^\infty is your isomorphism. Looking at this coordinatewise, this corresponds to a bounded sequence \phi_n of bounded linear functionals on L^\infty such that 1) there is \epsilon > 0 such that \max_n |\phi_n(f)| \ge \epsilon \|f\|_\infty for all f \in L^\infty 2) For every bounded sequence t_n of reals there ... 32 A compact set must be bounded. Otherwise we can take \{ x_n \}_{n=1}^\infty such that \| x_n \| \geq n. This will have no convergent subsequence, which we can prove by showing that it has no Cauchy subsequence. A compact set must be closed. Otherwise we can pick a sequence which converges to a point in the closure which is not in the set. This will ... 31 Try$$f(x) = \frac{1}{x^{1/p} (\ln(x)^2+1)} \qquad \text{on} \qquad (0, \infty)$\$