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The problem is evidently invariant under scalar multiplication so we may suppose that both $\|v\|=1$ and $\|f\|=1$, in which case $f$ is necessarily given by $f(\lambda v)=\pm\lambda $. We may further replace $v$ by $-v$, if necessary, so as to be able to assume that $f(\lambda v)=\lambda $. Next notice that a linear functional $g$, defined on the whole ${\...


2

No, this is not the case, even assuming $X$ is complete. Take your favourite discontinuous linear functional $f$ on your favourite infinite-dimensional normed linear space $X$. The kernels of discontinuous linear functionals on $X$ are codimension-$1$ subspaces of $X$, and hence any Hamel basis of $\operatorname{ker} f$ can be extended to a Hamel basis of $X$...


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The problem is not too hard if you consider $$f(x)=\tan (A(x))\,\tan (B(x))$$ Using logarithmic differentiation $$\frac{f'(x)}{f(x)}=A'(x) \csc (A(x)) \sec (A(x))+B'(x) \csc (B(x)) \sec (B(x))$$ This gives $f'(0)=0$. Repeating the process knowing that $f(0)=\frac 13$ and $f'(0)=0$ (this simplifies a lot the calculations), we have $$f''(0)=\frac{16}{81} \left(...


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I'll give my suggestion: As stated by @Davide Giraudo in the post you linked to, you know that $f\star g= g\star f$, so we can assume without loss of generality that $f\in C^1$. Using this version of Leibniz intergral rule, we obtain that $$\frac{d }{d x}\big[ f\star g \big] = \frac{1}{2\pi} \int_{-\pi}^\pi \frac{\partial }{\partial x} \big[ g(y) \cdot f(x-...


1

Let $\mu$ be a measure on some measurable space $X$. You may then define: $$L^p(X,\mu):= \{f:X\to\Bbb C\,\mid f\text{ measurable and }\int_X|f|^pd\mu<\infty\}$$ For example if $X=\Bbb N$ and $\mu_c$ is the counting measure then a function in $L^p(\Bbb N, \mu_c)$ is just a sequence $f(n)$ satisfying $\sum_{n\in\Bbb N} |f(n)|^p<\infty$, ie this is $\ell^...


1

Let $H$ be any infinite-dimensional Hilbert space, and let $\psi:H\to\mathbb C$ be an unbounded linear functional (these exist in ZFC). Define $T:H\to H^a$ by $$ Tx=\psi(x)\,\psi. $$ Then $T$ is symmetric, defined everywhere, and $TH\subset H^a\setminus H^*$. If you don't want Choice, then you lose a lot of functional analysis; but in any case the above can ...


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Let us see that $\ker U\subset\mathcal M^\perp$. Let $x\in\ker U$, $x = u+v$ with $u\in\mathcal M$ and $v\in\mathcal M^\perp$. Then $0 = \|Ux\| = \|Uu\| = \|u\|$, so $u=0$ and thus $x = v\in\mathcal M^\perp$.


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For any $\varepsilon> 0$ there is $x\ne 0$ such that \begin{align*} \varepsilon\|x\|\ge\|Tx-\lambda x\|\ge\big|\|Tx\|-|\lambda|\|x\|\big|\ge |\lambda|\|x\| - \|Tx\|. \end{align*} Hence, $$ |\lambda|\|x\|\le\|Tx\| + \varepsilon\|x\|\le(\|T\| + \varepsilon)\|x\|. $$ Dividing by $\|x\|\neq0$ gives $|\lambda|\le\|T\|+\varepsilon$. Letting $\varepsilon\to 0$ ...


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