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The answer is "no". The unit sphere is norm closed in the unit ball under the (norm) subspace topology. But in an infinite dimensional normed space, the weak closure of the unit sphere is the unit ball. See this post for a proof of this. So, in the space $B(X, {\rm weak})$, the closure of the unit sphere is again all of $B_X$ (in general if $A$ is a ...


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$$\langle w,w\rangle = 1$$ leads to $$\langle -u-v,-u-v\rangle =1.$$ Expand and simplify and you're done.


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For completess, here is how the rest goes. The radial part of the Euclidean Laplacian (without minus sign) in $\mathbb{R}^{n+1}$ is $r^{-n}(r^n u_r)_r$ (derived here). If $u$ is a homogeneous polynomial of degree $k$, then $u(r\xi)=r^k u(\xi)$ (with $\xi\in S^n$), hence $$r^{-n}(r^n u_r)_r =- r^{-n}(k r^{n+k-1})' u(\xi) = k(n+k-1)r^{k-2}u(\xi)$$ Since $u$ ...


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As others have shown, the algebraic method to prove the assertion in question isn't too hard but it occurred to me to think about a geometrical proof. Since the 3 vectors add to the zero vector they form a triangle. Since they have the same magnitude, that triangle is equilateral. An equilateral triangle has its 3 internal angles all equal to 60 degrees. The ...


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I'll prove it directly for an arbitrary measure space $({\cal X}, A, \mu)$. It is not the most efficient way to do it - you can use Hölder's Inequality, but I think that this proof makes sure the idea hits home. Let $a,b \in \Bbb R^+$, e $p, q > 1$ such that $\frac{1}{p} + \frac{1}{q} = \frac{1}{r}$. So: $$(ab)^r \leq r \left(\frac{a^{p}}{p} + ...


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To “prove” this, you need convince yourself, that (i) every closed subset of a complete metric space is automatically complete with respect to the metric; (ii) the reals under the eukl. norm $(\mathbf{R},|\cdot|)$ is a complete metric space; and (iii) $[a,b]\subseteq\mathbf{R}$ is closed in this topology. Statement (i) is an easy exercise (take a ...


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Because $(\Delta f,f) \le -\lambda_1\|f\|^{2}$ for $f\in\mathcal{D}(-\Delta)$, then \begin{align} \frac{d}{dt}\|T(t)f\|^{2}& =(\Delta T(t)f,T(t)f)+(T(t)f,\Delta T(t)f) \\ & \le -2\lambda_1(T(t)f,T(t)f)= -2\lambda_1\|T(t)f\|^{2},\;\; f \in\mathcal{D}(-\Delta) \end{align} Hence, $$ \frac{d}{dt}\left( e^{2\lambda_1 ...



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