One more exercise I stepped at while strolling through papers and journals for my preparation on the semester exams for multivariable calculus.
Let $D=\{(x,y): x^2 + y^2 \leq 1\}$ A function $f:D \to \mathbb R$ is called harmonic if it's continuous, has continuous partial derivatives (maximum of second grade) in the inside of $D$ and moreover : $f_{xx}(x,y) + f_{yy}(x,y) = 0 $ $\forall (x,y)$ inside $D$. Show that such a function $f$ does not have local extrema in the inside of D. Also show that $f$ gets a maximum and minimum value in the circle $C=\{(x,y): x^2 + y^2 =1\}$.
For the second part, we will use the entry which states that a continuous and real function defined in a closed and bounded subspace of $\mathbb R^2$ gets a maximum and minimum value. Thus, $f$ gets a maximum and minimum value in $D$ which is the set of points of $\mathbb R^2$ that are inside of circle (we can see that by the definition. So, $f$ gets maximum and minimum value in the circle $C=\{(x,y): x^2 + y^2 =1\}$.
My question is regarding the first part. How do we prove that it does NOT have local extrema ? My attempt was the following :
$f_{xx}(x,y) + f_{yy}(x,y) = 0 \Leftrightarrow (f_x)_x(x,y) + (f_y)_y(x,y) = 0 \Leftrightarrow f'(x,y) = c(y-x)$ as a solution to the First order ODE.
Now, the derivative of $f$ : $f'(x,y) = c(y-x) = 0 \Leftrightarrow y=x$ and that's true $\forall x,y\in \{x^2 + y^2 < 1\} : x=y $ which are infinite, thus the points on which the derivative is zero are NOT finite and thus we can conclude that $f$ does NOT have local extrema in the inside of $D$.
Is my attempt correct ? Personally I think something's off of it. Any tip/correction/help will be really appreciated.