again!
I simply need help understanding this example.
QUESTION: Show that the function
$$ g(x) =
\begin{cases}
\frac{xy^2}{x^2 +y^4}, & \text{if $(x,y) \neq 0$}\\
0, & \text{if $(x,y) = (0,0)$} \\
\end{cases} $$
has directional derivatives at $0 = (0,0)$ but is not differentiable at $0 = (0,0)$
(What does this even mean?)
In any case, the answer is given as:
ANSWER
Direct computation yields for every $v=(x,y) \in \Bbb R^2$,
$$D_v(g) = \lim_{t\to0} \frac {g(0+tv) – g(0)}{t} = \lim_{t\to0} \frac {txy^2}{x^2 + t^2y^4} = 0$$ Thus $g$ has directional derivatives in all directions at $(0,0)$
On the other hand we notice that $g(x^2,x) =\displaystyle \frac{1}{2} \not\to 0 = g(0,0)$ as $g(x^2,x) \to 0$ thus g is not continuous at $0$ and is there not differentiable.
Now I understand the first part.
But my concern is with the second one. Why are we discussing $g(x^2,x)$ in the matter of continuity? I understand the idea is to prove indifferentiablity but why $g(x^2,x)$ to be exact? I know it's simply an example. But I may be thrown off by the use of x in both vector components…
Best Answer
I take the question to be "How did someone come up with the particular choice of $x^2$ and $x$ to plug into $g$ and discover discontinuity?" Of course, not being a mind-reader, I don't know how it was actually done, but here's how it might have been done. The key observation is that, in the fraction that defines $g(x,y)$, the numerator $xy^2$ is the geometric mean of the two terms in the denominator, $x^2$ and $y^4$. (That might sound complicated, but it really just means that the exponents of $x$ and $y$ in the numerator are the averages of their exponents in the two terms of the denominator: For the exponents of $x$, $1$ is the average of $2$ and $0$, and for the exponents of $y$, $2$ is the average of $0$ and $4$.) So if we give $x$ and $y$ values that make the two denominator terms equal, then the numerator will automatically be equal to them also. So the numerator will match each of the terms in the denominator, and the fraction will be $1/2$. If we can find such values for $x$ and $y$ arbitrarily close to $0$, then that will make $g$ discontinuous at $(0,0)$. The easiest way to achieve this, meaning to make $x^2=y^4$, is to give $y$ an arbitrary value, say $t$, and to set $x=t^2$. So you set $(x,y)=(t^2,t)$ to find points, as close to $(0,0)$ as you like if $t$ is very small, where $g$ takes the value $1/2$.
Finally, if you're in a nasty mood, you re-name the variable $t$ as $x$, even though it serves as the value to substitute for $y$, just to confuse readers.
P.S. If you know how to use a program like Mathematica, you can have it plot the graph of $g$, and you'll probably be able to see a sort of a ridge in the graph, over the parabola $x=y^2$, at height $z=1/2$. So this might be another way to "guess" the substitution that proves discontinuity of $g$ at $(0,0)$.