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If f is a real valued harmonic function in $C$ we know that $f_{xx} + f_{yy} = 0$. So, $f_{xx}$ and $f_{yy}$ exists, from this it is not correct to conclude that $f_x$ and $f_y$ are continuous right? Because there are differentiable function with discontinuous partial derivatives.

But if $f_x$ and $f_y$ were continuous we would have $f_{xy}=f_{yx}$. Does f is real valued help here in any way ?

So in short my doubt is, if f is a real valued harmonic function in $C$ can we tell $f_{xy} = f_{yx}$ ?

math
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2 Answers2

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This is an excellent question. It is good to be skeptical: A function can have partial derivatives at a point while not being even continuous. See here for an example: A function whose partial derivatives exist at a point but is not continuous

But, $f_{xx}+f_{yy}=0$, with both $f_{xx}$ and $f_{yy}$ continuous, the usual definition for harmonicity, is a very strong condition. In fact, such functions are smooth, and so of course you get equality of mixed partials. There are many ways to prove this, depending on what you know.

operatorerror
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  • Is this because f is real valued ? What if f is not real valued ? – math May 13 '20 at 02:20
  • @math If $f$ is complex valued then the same thing applies to the real and imaginary parts (generally at this stage harmonic functions are taken to be real valued though). So no, not because $f$ is real valued. – operatorerror May 13 '20 at 02:25
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No (to the first question). Here's an example. Let's first start with holomorphic (analytic) functions and Cauchy-Riemann equations. Here's a standard example: $g(z)=e^{-z^{-4}}$, where we define $g(0)=0$, is clearly holomorphic outside the origin. The partial derivatives $g_x$ and $g_y$ exist at the origin and are zero so Cauchy-Riemann equations are satisfied at the origin too. But the function has an essential singularity at the origin.

Now for a solution to the Laplace equation. Let $f = \text{Re}\, g$. Then you have a function that is harmonic outside the origin, and you can check that $f_{xx}$ and $f_{yy}$ will still exist and vanish at the origin. Thus $f_{xx}+f_{yy}=0$ at all points. But the function is not harmonic, in particular it is not even continuous, let alone $C^1$ as you wanted. (Also note that $C^1$ does not imply that $f_{xy}=f_{yx}$, you need $C^2$ for that).

And this brings me to the last thing. This is why it is required in the definition of harmonic via the Laplace equation that the function is $C^2$. Just like for concluding that a function is holomorphic via the Cauchy-Riemann equations one requires $C^1$.

Well that's not quite a correct statement. One can use the Laplace equation to define harmonicity starting even with just continuous functions, if one uses the derivatives in the distribution sense, see Weyl's lemma on wikipedia. Why does this not contradict the example above? The example does not satisfy Laplace equation distributionally, it satisfies it pointwise. For distribution solutions you need the "derivatives" to be integrable, and in the example above the derivatives behave really terribly: they are not going to be integrable on any neighborhood of the origin.

The moral of all this? Pointwise solutions to differential equations can be quite terrible, even for nice differential equations. You generally need something extra such as either continuity or at least integrability of those derivatives.

Jiri Lebl
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