Today my professor introduced the idea of uniform continuity. However, I had difficulty visualizing what the graphical interpretation of a function being uniformly continuous is. Is it possible to determine if a function is uniformly continuous by looking at its graph?
Can anyone give an intuitive interpretation of uniform continuity?
Generally it is impossible to judge whether a function is uniformly continuous only by looking at the graph.
Note that even differentiable uniformly continuous functions can have unbounded derivatives (namely unbounded slopes). One immediate example is $f(x)=x\sin(1/x)$ on $(0,1]$, see here for a proof.
Of course, if defined on a compact set (say a bounded closed interval) then it trivially suffices to check whether the function is continuous.
As far as I can tell, I can't find a good intuitive interpretation of uniform continuity.
It's really a bit hard to visualize in the general situation. An easier concept to visualize might be "$\varepsilon$-continuity" and "$\varepsilon$-uniform continuity". These are not standard terms, but by these I mean that we are writing out the definition of continuity and uniform continuity with $\varepsilon$ equal to a fixed constant, rather than universally quantified.
Then $\varepsilon$-continuity at $x$ means that there is a positive number $\delta$ and a rectangle centered at $(x,f(x))$ of height $2 \varepsilon$ and width $2 \delta$ such that the graph of $f$ crosses the rectangle on the sides, not on the top or bottom. A function is continuous at $x$ if it is $\varepsilon$-continuous at $x$ for every $\varepsilon > 0$.
$\varepsilon$-uniform continuity means that there is a single positive number $\delta$ such that if you draw a rectangle of height $2 \varepsilon$ and $2 \delta$ centered at $(x,f(x))$ for any $x$, then the graph of $f$ passes through the rectangle on the sides. A function is uniformly continuous if it is $\varepsilon$-uniformly continuous for every $\varepsilon > 0$. So you can imagine sliding this one rectangle along the graph of $f$, without ever allowing the graph to pass through the top or bottom. This immediately rules out, for instance, an increasing function with an unbounded derivative (why?)
Heuristically, you can make a good guess if you have a continuous function $f:[a,b] \to \mathbb{R}$ such that that $f$
is bounded on $[a,b]$ (there exists $M$, such that $|f(x)| \leq M$ for all $x \in [a,b]$)
It doesn't visually "wiggle" too much, or is smooth almost everywhere. The topologist's sine curve is a typical counter-example to boundedness implies uniform continuity, but you can see by looking that it just wiggles too much.
For intuition's sake consider the two following function: $f(x)=\frac{1}{x}$. This function is not uniformly continuous n $[0,\infty]$, but $is$ uniformly continuous on $[1,\infty]$, You can verify this result for practice, but just looking at the function, it is smooth and bounded, and will not "Wiggle" too much, meaning that you can catch each open neighborhood around $f(x)$ with a single $\delta$.
A real-valued function $f$ on a bounded subset $U$ of $\mathbf{R}^{n}$ is uniformly continuous if and only if $f$ admits a continuous extension to the closure of $U$.
Contrary to expectations related to the mean value theorem, unboundedness of the derivative is not a definitive test. For example:
Functions such as $f(x) = x\sin(1/x)$ and $g(x) = \sqrt{x}$ are uniformly continuous on $(0, 1)$ even though their derivatives are unbounded near $0$. (Each is the restriction of a continuous function on the compact set $[0, 1]$.)
The function $f(x) = x/|x|$ is not uniformly continuous on the set of non-zero reals, even though its derivative is identically zero. (The same formula does define a uniformly continuous function on every set $U$ obtained by removing from the reals an interval containing $0$.)
Yes, the slopes of the tangents are bounded. At least if you are dealing with a differentiable function, but this gives you the intuition.
