This question has three parts.
a) Difference between continuity and uniform continuity
b) Geometrical meaning of uniform continuity
c) Correct the example
Definition of Continuity of a function in an interval $I$ is quite similar to uniform continuity.
I know that continuity is a pointwise property of a function and that when a function is continuous in an interval it means that it is continuous in every point in that interval. It is also easy to unterstant what this geometrically means (i think).
The first one is clearly a continuous function while the second one is not. The same logic applies to functions with more variables.
However i don't understand what uniform continuity means geometrically and how do we show in which interval a function is uniformly continuous.
For Example: $$ f(x,y) = x^2 + xy $$
The definition of uniform continuity says:
We have two points $P=(x,y)$ and $P_0 = (x_0,y_o) $
$$ \forall ε>0 \exists d>0 / \forall P,P_0 \in I (|x-x_0|<δ, |y-y_0| < δ \Rightarrow |f(P) - f(P_0)|<ε )$$
So we take $ |x-x_0|<δ $ and $ |y-y_0|<δ $
$$|x^2 +xy -x_0^2 -x_0y_0|= | (x-x_0)^2 + (x-x_0)(y-y_0) + (2x_0 + y_0)(x-x_0) + x_0(y-y_0)|<δ^2+δ^2+|2x_0+y_0|δ + |x_0| δ = 2δ^2 + (|2x_0 +y_0|+|x_0|)δ<2δ^2+(3|x_0|+|y_0|)δ $$
Now if we choose $δ<1$
$$|f(P)-f(P_0)|< (3|x_0|+|y_0|+2)δ=ε \Rightarrow δ=\frac{ε}{3|x_0|+|y_0|+2} $$
Now is the condition for the function to be uniformly continuous, δ to be positive?
In this case $$ δ>0 \forall P_0 \in \mathbb R^2 $$ So the function is uniformly continuous in $ \mathbb R^2 $ .
However i previously chose $ δ <1 $ so $|PP_0|<1 $. Doesn't that affect the outcome?