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It may fail to be closed. If T is injective and compact on an infinite dimensional Hilbert space, then $0$ is in the closure of $T(S)$, and not in it.

You want a version of the classical Müntz–Szász theorem for the space $L^2([0,1])$ (which is, incidentally, the case considered initially by Szász). Here is a nice paper on the situation for $L^p([0,1 …

The optimality of your configuration can be shown as a plain consequence of the Kirszbraun theorem. (I happened to ask myself this problem too, and eventually added this short section in a wiki artic …

If $T:H\to H$ is a normal operator, $\operatorname{ker}T=\operatorname{ker}T^*$ (recall that $\|Tx\|^2= (T^*Tx,x)=(T T^*x,x)=\|T^*x\|^2$). So there is a $T$-invariant orthogonal decomposition $H=\ope …

You may also send $X$ into a space $L_2(X,\mu)$ via the Fréchet-Kuratowski isometry $x\mapsto d(\cdot,x)\in C^0(X),\|\cdot\|_\infty$, followed by the bounded linear inclusion $C^0(X)\to L_2(X,\mu)$, w …

In general it is not true: $V_m$ and $U_m$ could even be transverse for all $m$, giving $$ \bigcap_{m=1}^\infty \left( V_m + U_m \right) = \ell_2. $$ Let $\kappa:\mathbb{N}\to\mathbb{N}$ be a map suc …

This was initially meant to be a comment, but became too long. So, firstly, I'd warmly suggest you to adopt and follow a good book on the subject, rather than using on-line material: which is good as …

I think the perfect reference for you is Lars Hörmander's The Analysis of Linear Partial Differential Operators, vol III (the chapter on elliptic operators). There you'll find in perfect Hörmander sty …

Here is a first step. Let $\tau=\sum_{i=1}^n h_i\otimes u_i$ with $h_1,\dots h_n\in H$ and $u_1,\dots u_n\in E$, where w.l.o.g. $h_i$ are orthonormal. Then, in your infimum, you may fix $\mu$ to be …

No infinite dimensional reflexive space can be embedded into $\ell_1$, because every infinite dimensional closed subspace of $\ell_1$ has a non separable dual.

Some simple arguments (which, to be honest, I'm also not completely convinced of) about the non-necessity of non-separable Hilbert spaces. If "in real life" means working into a given Hilbert space, a …

True: via a local chart we can assume $F^{-1}(0)$ is a closed linear subspace $N$ of $X$, and since $F_{|N}=0$, we also have $N\subset \text{ker} DF(x) $. (A formal explanation of the latter: if we de …

Short answer: because it's an instance of the spectral radius formula. Details. Let's start with the following inequality on operator norms. Given a finite family $A:=(A_j)_{j\in J}$ of operators $A …

I add this remark because it may be part of what the OP wants. Note that, as to the spectrum of a bounded linear self-adjoint operator on $\ell^2$, of course, it can be any compact set $K$ of $\mathbb …

If the bounded 2-form $E$ satisfies $E(x,x)\ge b |Q(x,x)|$ for any bounded 2-form $Q$ and for some $b>0$ depending on $Q$, then actually $E(x,x)\ge a\|x\|^2$ for some $a >0$. Indeed, suppose the lat …