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The set $S$ is convex and symmetric, i.e., $-x \in S$ for all $x\in S$. If $x$ is an interior point of $S$ and $B(x,\varepsilon)\subseteq S$ (where $B(x,\varepsilon)$ is the ball around $x$ with radius $\varepsilon$) you get $B(0,\varepsilon/2)\subseteq S$ since for $\|y\| <\varepsilon/2$ you have $y= \frac 12 (-x) +\frac 12 (x+2y) \in \frac 12 S + \frac ... 2 Daniel Fischer's comment says yes, since$\ell^1=c_0^*$. One can give an explicit projection from$(\ell^\infty)^*$onto$\ell^1$by saying $$P\Lambda=(\Lambda e_1,\Lambda e_2,,\dots).$$ 2 The answer to both of your questions is based on the linearity of$f$. For the first question, notice that, if there is even a single$x$with$f(x)\neq0$, then by multiplying$x$by a large positive real number$r$, you get$|f(rx)|=r|f(x)|$, which gets arbitrarily large if you take$r$large enough. So the only way$f$could be bounded everywhere (rather ... 1 To answer your second question: Assume that for some$x_n$in your sequence it holds that$0 < \vert\vert x_n \vert\vert < 1$(without loss of generality$0 < \vert\vert x_n \vert\vert$, since otherwise$x_n=0$and$f(x_n)=0$, which cannot be the case for$n$large enough). Now define a new sequence$y_n = \frac{x_n}{\vert\vert x_n \vert\vert}$and ... 1 Look at this: http://planetmath.org/banachspacesofinfinitedimensiondonothaveacountablehamelbasis there is refference also. Example is any Banach space of infinite dimension . 1 Is$e_0$supposed to have norm 1? Notice that if we suppose that, we obtain$1=|g_i(e_0)|\leq \Vert g_i\Vert_E\Vert e_0\Vert=\Vert g_i\Vert$, hence$\Vert g_i\Vert_E =1$. 1 I think that a direct proof is possible: Let$(u_k)$be a Cauchy sequence in$\mathcal{B}$. Then,$(u_k)$is Cauchy in$L^2(0,T; H^1_0)$and$(\partial_tu_k)$is Cauchy in$L^2(0,T; H^{-1})$. Since these are Banach spaces, we conclude that there exists$u\in L^2(0,T; H^1_0)$and$w\in L^2(0,T; H^{-1})$such that$\$\left\{\begin{align}u_k\to ...