How does a function 'uniformly' go to zero? I have read that Jordan's Lemma can be written a saying that:
$$I=\int_{\Gamma_R} f(z)e^{iaz}\, dz$$
where $a>0$ and $\Gamma_R$ is a  semi-circle in the upper-half plane will go to zero if  $|f(z)|\rightarrow0$ uniformly as $R\rightarrow \infty$. What in this does 'uniformly' mean (i.e. when does something go to $0$ uniformly and when doesn't it)? (Note $R$ is the radius of the semi-circle.)
 A: $\newcommand{\eps}{\varepsilon}$For technical simplicity, this answer focuses on sequences (of numbers, or of number-valued functions). The ideas extend easily to functional limits as well.
Let $X$ be a non-empty set containing infinitely many elements, and let $(f_{n})$ be a sequence of real- or complex-valued functions on $X$.
Definition 1: The sequence $(f_{n})$ converges pointwise to $0$ on $X$ if, for each $x$ in $X$, the numerical sequence $\bigl(f_{n}(x)\bigr)$ converges to $0$, i.e., if $\lim\limits_{n \to \infty} f_{n}(x) = 0$ for each $x$ in $X$.
Formally:

For every $x$ in $X$ and for every $\eps > 0$, there exists a natural number $N$ such that if $n \geq N$, then $|f_{n}(x)| < \eps$.

The crucial "weakness" of this definition is that $N$ generally depends on $x$.
Example 1: Let $\phi(x) = \max(1 - |x|, 0)$. (Replace as desired with your favorite function that is not identically $0$, but approaches $0$ as $x \to -\infty$.) The sequence $f_{n}(x) = \phi(x - n)$ converges to $0$ pointwise on the set of real numbers. Clearly, however, "the graphs of $f_{n}$ do not approach the $x$-axis" in the naive sense of "generally shrinking in the vertical direction toward the axis"; instead, the graph of $\phi$ is translated off to the right, "eventually disappearing from any viewing window of finite width".

In many circumstances (see Note 3 below), we need more: Not merely that $\bigl(f_{n}(x)\bigr) \to 0$, but that the "rate of convergence is the same" independently of $x$. The uniformity in your question refers to uniformly in $x$.
Definition 2: The sequence $(f_{n})$ converges uniformly to $0$ on $X$ if:

For every $\eps > 0$, there exists a natural number $N$ such that if $n \geq N$, then $|f_{n}(x)| < \eps$ for every $x$ in $X$.

Remark: Uniform convergence on $X$ implies pointwise convergence on $X$ a fortiori.
Lemma: Let $(f_{n})$ be a sequence of real- or complex-valued functions on a set $X$, and define the sequence $(a_{n})$ of extended real numbers by
$$
a_{n} = \sup_{x\in X} |f_{n}(x)|.
$$
The sequence $(f_{n})$ converges uniformly to $0$ on $X$ if and only if $(a_{n}) \to 0$.
Example 1 revisited: For each $n$, $\sup |f_{n}| = \sup |\phi| > 0$. By the preceding lemma, our pointwise-convergent sequence $(f_{n})$ does not converge uniformly to $0$ on the reals.
Example 2: Let $\phi$ be a bounded function on $X$, and define $f_{n}(x) = \frac{1}{n}\phi(x)$. The sequence $(f_{n})$ converges uniformly to $0$ on $X$. (Why?)

Notes:


*

*There is no loss of generality in taking the limit to be $0$: If, for each $x$, the sequence $\bigl(f_{n}(x)\bigr)$ converges to a limit $f(x)$ (this is the definition of $f$), then the sequence $\bigl(f_{x}(x) - f(x)\bigr)$ converges to $0$.

*If instead you have a family of functions, say $(f_{R})$ parametrized by positive real numbers $R$:


*

*Pointwise convergence to $0$ is defined in the obvious way: For each $x$ in $X$, we have $\lim\limits_{R \to \infty} f_{R}(x) = 0$. That is, for each $x$ in $X$ and for every $\eps > 0$, there exists an $R_{0}$ (depending on $\eps$ and on $x$) such that if $R > R_{0}$, then $|f_{R}(x)| < \eps$.

*Uniform convergence to $0$ is defined in the obvious way: For every $\eps > 0$, there exists an $R_{0}$ (depending only on $\eps$) such that if $R > R_{0}$, then $|f_{R}(x)| < \eps$ for all $x$ in $X$.


*In elementary analysis, one finds that if $(f_{n}) \to f$ pointwise, the limit function "inherits" few properties of the approximating sequence. By contrast, uniform limits behave more nearly as expected. Particularly:


*

*If $(f_{n}) \to f$ uniformly on some interval $[a, b]$, and if each $f_{n}$ is continuous, then $f$ is continuous, and for each $c$ in $[a, b]$,
$$
f(c) = \lim_{x \to c} f(x)
  = \lim_{x \to c} \lim_{n \to \infty} f_{n}(x)
  = \lim_{n \to \infty} \lim_{x \to c} f_{n}(x)
  = \lim_{n \to \infty} f_{n}(c).
$$

*If $(f_{n}) \to f$ uniformly on $[a, b]$, and if each $f_{n}$ is Riemann integrable, then $f$ is Riemann integrable, and
$$
\int_{a}^{b} f
  = \int_{a}^{b} \lim_{n \to \infty} f_{n}
  = \lim_{n \to \infty} \int_{a}^{b} f_{n}.
$$

*If each $f_{n}$ is differentiable, if the sequence of derivatives $(f_{n}')$ converges uniformly to some function $g$ in $[a, b]$, and if $(f_{n})$ converges at one point, then $(f_{n})$ converges uniformly on $[a, b]$ to a differentiable function $f$, and
$$
f'(x) = (\lim_{n \to\infty} f_{n})'
  = \lim_{n \to\infty} (f_{n}').
$$
(Note carefully the stronger hypothesis that the sequence of derivatives converges uniformly. Integration is a continuous operator: An integral of a function of small absolute value is small. Differentiation is a discontinuous operator: Loosely, a function of small absolute value can have arbitrarily wild derivative.)
A: It means that
$$
\lim_{R\to 0} \left( \sup_{z \in \Gamma_R} |f(z)| \right) = 0.
$$
This is a little more than saying, for example, that for each fixed $t$ with $0 \le t \le \pi$, $\lim_{r\to\infty} f(re^{it}) = 0$.
A: The wiki article has a good explanation of uniform convergence.
