I was reading about convergence of Fourier series and a "demonstration technique" appears to me to be wrong. The idea is: Take $F_n \colon (0,1) \to \mathbb{R}$ as the n-th summand from the Fourier series of $f \colon (0,1) \to \mathbb{R}$, and supose that: $$\forall \delta>0, \epsilon>0, \exists N\in \mathbb{N} \colon m,n > N \implies|F_n(x)-F_m(x)|<\epsilon, \forall x \in (\delta,1).$$ ie, $(F_n)_{n \in \mathbb{N}}$ is Uniformly Cauchy in, so it must be Uniformly Convergent (on $(\delta,1)$), and taking the limit of $\delta \to 0$, one can conclude that $F_n \to f$ uniformly with domain $(0,1)$.
Here's why i don't think this is right: Let $f_n \colon (a,b) \to \mathbb{R}$ be a function sequence with $|a|<\infty$ and supose that $(\delta_j)_{j \in \mathbb{N}}$ is a numerical sequence such that $a<\delta_j<b$, $\forall j \in \mathbb{N}$, and $\delta_j \to a$. Let's call $f_{n}^{j} = f_{n}|_{(\delta_j,b)}$ the restrinction of $f_n$ to $(\delta_j,b)$, then i propose the:
Proposition: $(f_n)_{n \in \mathbb{N}}$ is Uniformly Convergent if, and only if, for every $j \in \mathbb{N}$, the sequence $(f_{n}^{j})_{n \in \mathbb{N}}$ is Uniformly Cauchy and, $\forall \epsilon>0$ the sequence $(N_j(\epsilon))_{j \in \mathbb{N}}$ is bounded, where $N_j(\epsilon)$ is such that $|f_{n}^{j}(x)-f_{m}^{j}(x)|<\epsilon, \forall m,n > N_j(\epsilon)$ and $x \in (\delta_j,b)$.
If i didn't make any mistake, the proof is:
$(\Rightarrow)$ If $(f_n)_{n \in \mathbb{N}}$ is Uniformly Convergent, $\exists f \colon (a,b) \to \mathbb{R}$ such that, $$ \forall \epsilon>0, \exists N(\epsilon) \in \mathbb{N} \colon n>N(\epsilon) \implies |f_n(x)-f(x)|<\epsilon, \forall x \in (a,b). $$ Using the triangular inequality $|x+y|<|x|+|y|$, we have, $\forall m,n > N(\epsilon/2)$: $$ |f_n(x)-f_m(x)| \leq |f_n(x)-f(x)|+|f_m(x)-f(x)| < \epsilon, \forall x \in (a,b) $$ ie, $(f_n)_{n \in \mathbb{N}}$ is Uniformly Cauchy. The inequality above follows for $x \in (\delta_j,b) \subseteq (a,b)$, so we can take, $\forall j \in \mathbb{N}$, $N_j(\epsilon) = N(\epsilon)$, thus $(N_j(\epsilon))_{j \in \mathbb{N}}$ is a bounded (constant) sequence.
$(\Leftarrow)$ If, for every $j \in \mathbb{N}$, the sequence $(f_{n}^{j})_{n \in \mathbb{N}}$ is Uniformly Cauchy, we can take, $\forall \epsilon>0$, $N_j(\epsilon)$ such that $$n,m>N_j(\epsilon) \implies |f_{n}^{j}(x)-f_{m}^{j}(x)|<\epsilon, \forall x \in (\delta_j,b)$$ By hypotesis, $(N_j(\epsilon))_{j \in \mathbb{N}}$ is bounded, so there's $N(\epsilon) \in \mathbb{N}$ such that $N(\epsilon) \geq N_j(\epsilon)$, $\forall j \in \mathbb{N}$. Now, note that: $$\delta_j \to a \implies \forall x \in (a,b), \exists j \in \mathbb{N} \colon a<\delta_j<x<b \implies f_n(x) = f_{n}^{j}(x) $$ Then, $\forall x \in (a,b)$, $\exists j \in \mathbb{N}$, wich satifies: $$ n,m > N(\epsilon) \geq N_j(\epsilon) \implies |f_n(x)-f_m(x)| = |f_{n}^{j}(x)-f_{m}^{j}(x)| < \epsilon $$ ie: $n,m>N(\epsilon) \implies |f_n(x)-f_m(x)|<\epsilon, \forall x \in (a,b)$, and $(f_n)_{n \in \mathbb{N}}$ is Uniformly Cauchy.
The rest of the proof is, by no mean, original. Take, for every $x \in (a,b)$, the numerical sequence $(f_n(x))_{n \in \mathbb{N}}$, wich is Cauchy. By completness of $\mathbb{R}$, $\exists f(x) \in \mathbb{R}$ such that $f_n(x) \to f(x)$, this define a function $f \colon (a,b) \to \mathbb{R}$. Thus: $$ \forall \epsilon>0, \exists N(\epsilon) \in \mathbb{N} \colon n,m>N(\epsilon) \implies |f_n(x)-f_m(x)|<\epsilon$$ Taking the limit $m \to +\infty$ and by the punctual convergence $f_m(x) \to f(x)$, follows: $$|f_n(x)-f(x)|<\epsilon, \forall x \in (a,b)$$ ie, $(f_n)_{n \in \mathbb{N}}$ is Uniformly Convergent on $(a,b)$. $\square$
$$\dots$$
If this proposition is true, some examples (i mean, theorems) do not provide proofs of Uniform Convergence. For example:
(Lema 3.1 - from: análise de fourier e equações diferenciais parciais / Djairo Guedes de Figueiredo. 4a. ed. 2000) Let $\psi$ be a periodical function with period $2L$ defined by:
$$\psi(x) = \left\{\begin{matrix} -\frac{1}{2}(1+\frac{x}{L}) & -L \leq x < 0 \\ 0 & x=0 \\ \frac{1}{2}(1-\frac{x}{L}) & 0 < x \leq L \end{matrix}\right.$$ Then the Fourier series of $\psi$ converges uniformly to $\psi$ in any interval without points of the form $x=2Ln$, $n \in \mathbb{Z}$.
Here, the author used the Abel's lemma (for summation by parts) to find, to $\Psi|(\delta,L)$, the bound: $$ |\Psi_n(x)-\Psi_m(x)| < \frac{2}{m \cdot \sin\left(\frac{\delta L}{2\pi}\right)} $$ This is, taking $N_\delta(\epsilon) \geq 2/\left(\epsilon \cdot \sin\left(\frac{\delta L}{2\pi}\right)\right)$, one can proof the uniform convergence of $\Psi_n \to \psi$ on $(\delta,L)$. But: $$\delta \to 0 \implies \sin\left(\frac{\delta L}{2\pi}\right) \to 0 \implies N_\delta(\epsilon) \to +\infty,$$ so this argument cannot proof the uniform convergence on $(0,L)$.
(Lema 1.2 - from: EDP: um curso de graduação / Valéria de Magalhães Iório. 4a ed. 2018) The function defined by: $$ K(x,y,t) = \frac{2}{b}\sum_{k=1}^{+\infty} K_k(x,y,t) = \frac{2}{b}\sum_{k=1}^{+\infty} \frac{\sinh(k\pi x/b)}{k\pi a/b}\sin\left(\frac{k\pi t}{b}\right)\sin\left(\frac{k\pi y}{b}\right) $$ is such that $K(x,y,t) \in C^{\infty}([0,a) \times \mathbb{R}^2)$ and satify $K_{xx}+K_{yy}=0$.
Inside this lema, the author claims that $K_n \to K$ uniformly, arguing that, for $0 < \epsilon < a$ and $(x,y,t) \in [0,a-\epsilon) \times \mathbb{R}^2$: $$ \frac{b}{2}|K_k(x,t,y)| \leq \exp\left( -k\frac{\pi(a-x)}{b} \right) \leq \exp\left( -k\frac{\pi\epsilon}{b} \right) $$ Therefore you can apply Weierstrass M-test to get that, if $N(\epsilon)$ satifies, $$ n > N(\epsilon) \implies \sum_{k=n}^{+\infty} \frac{1}{e^k} < \frac{\epsilon}{\exp\left( \frac{\pi\epsilon}{b} \right)} $$ then: $$ m,n > N(\epsilon) \implies |K_n(x,y,t)-K_m(x,y,t)| < \epsilon, \forall (x,y,t) \in [0,a-\epsilon) \times \mathbb{R}^2. $$ But again, $$\epsilon \to 0 \implies \frac{\epsilon}{\exp\left(\frac{\pi\epsilon}{b}\right)} \to \frac{0}{1} \implies N(\epsilon) \to \infty$$
$$\dots$$
I'm not into analysis, so it can be that i'm wrong with this proposition, but if you can take a time to judge it carefully, i'd appreciate the feedback of yours. Also, if you can leave references of alternative proofs for Lema 3.1 and Lema 1.2 above, it'll be usefull to me.