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8

If such a function exists, then you may add any function whose integral between $0$ and $1$ is zero. Now you should find one particular solution. For this, a constant (with respect to $x$) function suffices.

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There is no such operator. Note that $(C ([0,1]),\|\cdot\|_\infty )$ is a Banach space, so that it suffices to show that $T$ has closed graph. Thus, assume $f_n \to f$ and $T f_n \to g$ (both with respect to $\|\cdot\|_\infty$). Then $f_n \to f$ and $T f_n \to g$ bith with respect to $\|\cdot\|_2$ (why?) and hence $T f_n \to Tf$ also with respect ...

4

You could not find a proof because it is not true. In fact $BL(X)$ is separable in the topology of uniform convergence if and only if $X$ is totally bounded. If $X$ is totally bounded, every Lipschitz function can be extended to its completion $\tilde{X}$, which is compact. Thus we obtain an isometric embedding $BL(X) \to C(\tilde{X})$. If $X$ is not ...

4

Triangle inequality $$|\, ||f_n ||_2 -|| f ||_2\, |\leq ||f_n-f||_2$$

4

since $\langle T(x+y),x+y\rangle =0$ that implies : \begin{eqnarray} \langle T(x+y),x+y\rangle &=&\langle Tx+Ty,x+y\rangle \\ &=&\langle Tx,x+y\rangle+\langle Ty,x+y\rangle\\ &=& \langle Tx,x\rangle+\langle Tx,y\rangle+\langle Ty,x\rangle+\langle Ty,y\rangle\\ \end{eqnarray} Then $$\langle T x,y\rangle +\langle Ty,x\rangle=0 \qquad ... 3 Since \|f - f_n \|_p \to 0, we can extract a subsequence f_{n_j} so that \| f - f_{n_j} \|_p \le \frac{1}{2^j}. Put$$g = \lvert f \rvert + \sum^\infty_{j=1} \lvert f - f_{n_j} \rvert.$$Then g \in L^p and by the triangle inequality we have$$\lvert f_{n_j} \rvert \le g \,\,\, (\text{almost everywhere}).$$From this subsequence, you can extract a ... 3 If your example \mathcal{B} were a real Banach algebra instead of a complex Banach algebra, then you would be right that there are four connected components, since \mathcal{B} can be identified with \mathbb{R}^2 and the invertible elements split into four quadrants. But over \mathbb{C}, you have the complement in \mathbb{C}^2 of two (complex) one-... 3 A counterexample for d=2: let \Omega be the disk \{x:\|x\|<\exp(-\exp(\pi))\}, and$$ f(x) = \sin \log \log \frac{1}{\|x\|} $$This function is in W_0^{1,2}(\Omega)\cap L^\infty(\Omega) (relevant calculations here) but has a discontinuity at 0, and moreover cannot be made continuous by redefining it on a set of measure zero. On the other ... 2 Here is a simpler proof:$$ \sum_{n=1}^\infty| \langle x, e_n \rangle \langle y, e_n \rangle | \le \sum_{n=1}^\infty| \langle x, e_n \rangle|\cdot | \langle y, e_n \rangle | \\ \le \left(\sum_{n=1}^\infty| \langle x, e_n \rangle|^2\right)^{1/2}\left(\sum_{n=1}^\infty | \langle y, e_n \rangle |^2\right)^{1/2} \le \|x\| \cdot \|y\|. $$First inequality is ... 2 According to Daniel Fischer: The left hand side cries for an application of the Cauchy-Schwarz inequality. And according to siminore:$$ x=\sum_n x_n e_n = \sum_n \langle x,e_n \rangle e_n \quad ; \quad y=\sum_n y_n e_n = \sum_n \langle y,e_n \rangle e_n $$But we give it a twist:$$ x'=\sum_n |x_n| e_n = \sum_n \left|\langle x,e_n \rangle\right| e_n \quad ; ...

2

Hah! This is actually a specific example of something in my research! (My work attacks a more general set of integral equations, in some sense.) Let's go for something nontrivial (unlike previous answers/comments). If you consider what I like to call a diagonal kernel, i.e. $g(x,t) = f(xt)$ for some $f$ and assume $g$ is real analytic, then this is very ...

2

An easy solution can be obtained by making $g(x,t)$ degenerate: $$g(x,t)={1\over2}\exp(-bx^4)\exp(at^4-|t|)$$ which is susceptible to the generalization $$g(x,t)={1\over C}\exp(-bx^4)\exp(at^4)f(t)$$ where $f$ is integrable over $\mathbb R$ and $$\int_{-\infty}^{\infty}f(t)dt=C\neq0$$

2

As @AlexanderFrei pointed out, it should read $$\forall x: \lim_{n \to \infty} T_n(x) = T(x) \iff \forall K \subseteq X \, \text{compact}: \lim_{n \to \infty} \sup_{x \in K} \|T_n(x)-T(x)\| = 0.$$ The implication "$\Rightarrow$" is trivial, just choose $K= \{x\}$ for fixed $x \in X$. It remains to prove "$\Leftarrow$". Suppose that $T_n(x) \to T(x)$ for ...

2

Since $T - \lambda I$ is also normal, we have $$\| T - \lambda I \| = \text{spr} (T - \lambda I) = 0,$$ showing that $T = \lambda I$. (I recently asked basically the same question (Self-adjoint operator with single point spectrum), but your formulation is more general so I thought it might be worth sharing the answer here.)

2

Any $*$-homomorphism between C$^*$-algebras is contractive. This is standard (i.e., it appears in every book on the subject) and is due to three things: The C $^*$-identity $\|a\|^2=\|a^*a\|$, which reduces the problem to norms of positives; The equality $\|a\|=\text {spr}\, (a)$ for $a$ positive; The fact that a $*$-homomorphism reduces the spectral ...

2

(Huge) hint: Is it complete? Why or why not? (Even bigger hint below.) Follow-up:

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