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Alternate proof of Dini's theorem over a closed real interval

This approach can be made to work. Indeed by continuity of the $f_n$, for each $n\ge 1$ there exists $x_n \in [a,b]$ such that $f_n(x_n) = ||f_n||_\infty$. It suffices to show that $||f_{n_k}||_\infty ...
raj's user avatar
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On properties of solution by Euler's method; uniform convergence, etc.

Question 1: A function is uniformly continuous if and only if it admits a modulus of continuity. $\varepsilon(\delta)$ is a modulus of continuity for $\pmb f$, which by definition satisfies $\lim_{\...
sunny's user avatar
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2 votes

Test the sequence of functions $x_n={e^{-nt}}$ for convergence in $C[0,1]$,$L_{1}(0,1)$ and $L_{2}(0,1)$.

(1) Yes, your argument is justified. (2) The limit is the zero function. $L^{1}$ consists of equivalence classes of functions and values at a finite number of points do not matter. (3) is similar to (...
geetha290krm's user avatar
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On uniform convergence of partial derivatives on a compact set

This is the relevant proof in the book by Rudin (Example 1.46): Where $\Omega$ is a non-empty subset of $\mathbb{R}^k$ for some $k \in \mathbb{N}$. To elaborate on ...each $D^\alpha f_i$ converges (...
jd27's user avatar
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1 vote

If $f:[0,1]\to\mathbb R$ is continuous, then $f_n(x) = f(x^n)$ converges uniformly on $[0,a],$ $a < 1$ and $∫_0^1 f_n(x)\,dx \to f(0).$

Let $\varepsilon>0$. Then there exists a $\delta>0$, such that $$ 0\le x <\delta\quad\Longrightarrow\quad |f(x)-f(0)|<\varepsilon. $$ If $a\in (0,1),\,$ then $a^n\to 0$, and hence there is ...
Yiorgos S. Smyrlis's user avatar
3 votes

The limit of the sequence of functions $\sqrt{x^2+\frac{1}{n^2}}$

$$ \left\lvert \sqrt{x^2 + n^{-2}} - \sqrt{x^2} \right\rvert = \frac{n^{-2}}{\sqrt{x^2 + n^{-2}} + \sqrt{x^2}} \leq \frac{n^{-2}}{n^{-1}} = \frac{1}{n} $$ for any $ x \in \mathbf{R}$; it follows that $...
terran's user avatar
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