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The answer is no, in general. Before we discuss a counterexample, let us note that whenever a set $\mathcal{O}_u(f,\epsilon)$ contains $0$, then there is a another number $\tilde \epsilon > 0$ such t …

Edit: I improved the constant to $c = \frac{2}{3}$. (Later edit: But the optimal constant turns out to be $c = \frac{1}{2}$, see Yuval Peres' answer.) Answer: Yes, we have $$ \inf_{(n_k)} \sup_{i,j …

There even exists a largest set $X$ to which $f$ can be continuously extended. The trick is the following result (which I state here in more generality, to point out which topological assumptions one …

Yes. This is a special case of the following result. Proposition. Let $X$ and $Y$ be normed spaces and let $i: X \to Y'$ be a continuous embedding. Then the mapping $j: Y \to X'$, given by $\langle j …

Counterexample. Let $f: \mathbb{R} \to \mathbb{R}$ denote your favourite test function with support in $(0,1)$ and with integral $1$. We define $f_n(x) := n f(nx)$ for all $n \in \mathbb{N}$ and all $ …

Setting. Throughout, let $E$ be a complex Banach space and denote the space of bounded linear operators on $E$ by $\mathcal{L}(E)$. Let $X \subseteq E$ be a closed subspace and let $\mathcal{T} = (T(t …

As $f$ is non-decreasing, the integral inequality implies that $f$ is $0$. Proof. The integral inequality implies, in particular, that the same inequality is satisfied for $C=0$, so $$ \int_{t_0}^t …

It seems that you are looking for so-called lower heat kernel bounds. For the heat equation the inequality you are looking for can, for instance, be found in Theorem 1.1 of "Zhang: The Boundary Beha …

The answer is no, in general. In order to construct a counterexample, let $X = Y = \mathbb{R}^n$ for any $n \ge 2$ and endow this space with the $p$-norm for your favourite $p \in [1,\infty] \setminu …

The answer is yes, as a close inspection of the standard proof of the uniform boundedness principle/Banach-Steinhaus theorem shows. The standard proof (or at least the proof which I would consider to …

In general, the answer is "no" even in the autonomous case, i.e. in the case where $A := A(t)$ does not depend on $t$. First note that if $A$ generates a $C_0$-semigroup on $H$, if $X$ is a dense sub …

The answer to the question is as follows: Theorem 1. For every $C > 0$ there is, up to scalar multiples, only one function $0 \le f \in L^1 := L^1((0,\infty))$ such that $Tf = f$ (where $T$ is the $C …

The operator $G$ is always invertible - and this is true for general strongly continuous mappings $T: [0,\tau] \to \mathcal{L}(X)$, no matter whether they fulfil the semigroup law. Proof. As indicate …

Example 1. The following kind of nilpotent construction satisfies the properties require in the question: Let $F$ be an infinite dimensional Banach space and let $E = F \times F$ (say, with the maxim …

An additional remark (too long for a comment) that might be of interest: Given the finite-dimensional counterexamples in the answers by Iosif Pinelis and Pietro Majer it seems worthwhile to note that …