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This answer should perhaps be posted as a comment, since the whole topic was already debated in this MathOverflow Q&A. However i briefly restate the main result here and leave the above link for further detals: a definitive answer is Cafiero's convergence theorem (see [1]) which, roughly states that$\DeclareMathOperator{\Dm}{\operatorname{d\!}}$ $$\... 4 \newcommand\ep\varepsilonThe conjunction of the following conditions is enough: The f_n's are uniformly bounded: |f_n|\le M for some real M>0 and all n; X is Polish; f_n\to f uniformly on every compact K\subseteq X; \mu_n\to\mu weakly for some \mu. Indeed, take any real \ep>0. By Prokhorov's theorem and in view of conditions 2 ... 1 Using the Radon-Nikodym Theorem this problem can be reduced to the case for a fixed measure. Suppose none of the \mu_n is the zero-measure, i.e. \mu_n(X)>0 for all n \in \mathbb{N}. For a series (\alpha_n)_{n \in \mathbb{N}} \subset (0,1) with \sum\limits_{n=1}^\infty \alpha_n = 1 define the probability measure$$ \mu = \sum\limits_{n=1}^\infty \...

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Since this got bumped to the front page... Let $\sigma_i$ be the singular values of $A$ and $\tilde{\sigma}_i$ be those of $A+E$. Then, $|\sigma_i - \tilde{\sigma_i}| \leq \|E\|$, by Weyl's inequalities. That gives you 1-Lipschitz continuity.

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The box dimension is not countably stable, so even one exceptional point could drive it up. For example, given $\epsilon>0$, the function $f(x)=x^{1/2} \sin(1/x)$ on $[0,1]$, with $f(0)=0$, is locally Lipschitz on $(0,1]$, yet the graph of $f$ has box dimension at least $1.25$. Indeed, for $j<k$, to cover the graph above $[1/(2\pi(k+j)), 1/(2\pi(k+j+1))... 1 The following is obviously not optimal, but at least it shows that the optimal$c$satisfies$c \leqslant 4$. On the other hand,$c \geqslant 2 / \log 2 \approx 2.885$, so if$c$is to be integer, it remains to show that in fact$c < 4$and hence necessarily$c = 3$. Edit: As remarked in the comments, the following argument works for continuous decreasing ... 3 The inequality$(2)$(even with factor$\frac12$in the r.h.s.) follows from the inequality quoted in this answer: $$M_a - M_g \leq \frac1{2n\min_k x_k} \sum_{i=1}^n (x_i - M_a)^2.$$ First, we notice that \begin{split} (x_i - M_a)^2 &\leq \max_k x_k\cdot |x_i-M_a|\\ &= \max_k x_k\cdot \left|\frac1n \sum_{j=1}^n (x_i - x_j)\right| \\ & \leq \frac{\... 2 Edit: In the first part,$C^{k,\alpha}$is understood locally. For a global result, see the final part. The equation reads $$f(t) = \int_{-\infty}^t e^{-(t - s)\theta} g(s) ds,$$ is equivalent to $$e^{t \theta} f(t) = \int_{-\infty}^t e^{s \theta} g(s) ds,$$ so if$g$is$C^{k-1,\alpha}$, then$s \mapsto e^{s \theta} g(s)$is$C^{k-1,\alpha}$, its integral$...

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By the Chernoff–Hoeffding theorem, the sum in question is $\le\exp(-nD(a||p))$ for $a\le p$, where $a:=\alpha$ and $$D(a||p):=a\ln\frac ap+(1-a)\ln\frac{1-a}{1-p},$$ the Kullback–Leibler divergence between the distributions of Bernoulli random variables with parameters $a$ and $p$. In particular, this gives your bound for $p=1/2$. On the other hand, if \$a>...

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