Can Schwarz's Theorem be used "Backwards" to prove continuity?

[Feymar]

TL;DR Summary

If the second partial (mixed) derivatives are equivalent, can it be used to prove that the function itself is continuous?

Hey folks, long story short, the Schwarz Theorem, that says that a continuous function has equivalent second partial (mixed) derivatives, can it be used "backwards", i.e. if I can show that the mixed partials are equivalent, that it proves that the function itself is continuous?

And if not, why not? I feel like I am still stuck in single-variable calculus in my intuition, and try to apply it all the time to multivariable mixed derivatives (I do have the correct intuition for first order partials and why they don't imply the same things as first order derivatives in single-variable problems). I seem to either lack some key piece that would let it all fall in place, or have a deep misunderstanding on what limits and higher order (mixed) partial derivatives don't do in multivariable calculus, but which limits and first order derivatives do in single-variable, and why that is so.... at least if the above in the original question cannot be done.

Edit: Typo.

 

[mathwonk]

you do not say whether you are asking about continuity at one point or on an open set, but the case of one point has a counterexample here.
https://math.stackexchange.com/questions/3033548/does-existence-of-mixed-derivative-imply-continuity

 

[WWGD]

All I can think of here is that if all partials exist and are continuous, then f is differentiable and thus continuous at a given point. Don't know if this applies just at a point or if it applies in an open set.

 

[Feymar]

you do not say whether you are asking about continuity at one point or on an open set, but the case of one point has a counterexample here.
https://math.stackexchange.com/questions/3033548/does-existence-of-mixed-derivative-imply-continuity

I am aware of the examples provided, however, they talk about the mere existence of some or all partial derivatives, I am specifically asking if the equivalence of second partial mixed derivatives guarantees me, rigorously, that the function is continuous, in reference to the part of Schwarz's Theorem that states that continuous functions will have equivalent second partial mixed derivatives.

All I can think of here is that if all partials exist and are continuous, then f is differentiable and thus continuous at a given point. Don't know if this applies just at a point or if it applies in an open set.

Yes, but there you need to check if the partials are continuous, which is sometimes harder than simply proving that the second oder mixed partial derivatives are simply equivalent. I am asking if it is possible to work Schwarz's Theorem backwards, i.e. does the statement that "a function that is continuous, will have equivalent second order mixed partial derivatives", also mean: "equivalent second order mixed partial derivatives prove a function continuous", or is this a case of "sufficient but not necessary" ?

 

[Feymar]

Apparently, according to this post:

https://math.stackexchange.com/questions/2095484/clairauts-theorem-is-reversible?noredirect=1&lq=1

Clairaut-Schwarz Theorem is not reversible, i.e. equivalence of mixed partials does not prove a function to be continuous and an example was provided. What a shame. I just hate doing continuity proofs hah.

 

[mathwonk]

@Feymar: I am confused as to whether you want to know about the continuity of the original function f, or the continuity of its second mixed partials. The example you cite is one where the mixed second partials are not continuous. The author states specifically that the function f he gives is considered to be extended continuously to the whole plane.

So I am a little confused about your actual question. I.e. Clairaut's theorem assumes the continuity of the second partials, not of the original function. So if your question really is whether Schwartz/Clairaut's theorem is reversible, then you seem to have a counterexample as you cited.

If instead your question is what you actually asked, namely "i.e. if I can show that [if] the mixed partials are equivalent, that it proves that the function itself is continuous?", then I do not know.

I.e. there seem to be two questions:
1) if a function has equal mixed partials, must those partials be continuous? (actual converse of Schwartz).
2) if a function has equal mixed partials, must the original function be continuous? (your actual question.)

The example you cite says the answer to the first question is "no". I do not know the answer to the second question, although there do exist functions f that possess at least first partials everywhere but such that f itself is not even continuous say at the origin, e.g. f(x,y) = (xy)/(x^2+y^2). It follows that the first partials cannot be continuous, but it does not follow that the first partials cannot themselves have partials, i.e. it is conceivable that second mixed partials exist, maybe even equal, for this discontinuous function f. I have not checked it. ok I checked it. this function does not have mixed partials at (0,0).

I apologize also for the example I cited earlier, which I did not check. Its author states incorrectly that the second partials of his discontinuous function all exist and equal zero, which would of course make the mixed partials equal to each other, (since they would all equal zero). If that were so, it would be a counterexample to the second question, but he is wrong, since in his example the first partials do not exist on a neighborhood of (0,0), since they do not exist at any point of the line x=y except (0,0). So his function is not continuous at (0,0), and does have first partials at (0,0), but not on any neighborhood of (0,0). In particular the second partials do not exist at all at (0,0).

 

[Feymar]

@Feymar: I am confused as to whether you want to know about the continuity of the original function f, or the continuity of its second mixed partials. The example you cite is one where the mixed second partials are not continuous. The author states specifically that the function f he gives is considered to be extended continuously to the whole plane.

So I am a little confused about your actual question. I.e. Clairaut's theorem assumes the continuity of the second partials, not of the original function. So if your question really is whether Schwartz/Clairaut's theorem is reversible, then you seem to have a counterexample as you cited.

If instead your question is what you actually asked, namely "i.e. if I can show that [if] the mixed partials are equivalent, that it proves that the function itself is continuous?", then I do not know.

I.e. there seem to be two questions:
1) if a function has equal mixed partials, must those partials be continuous? (actual converse of Schwartz).
2) if a function has equal mixed partials, must the original function be continuous? (your actual question.)

The example you cite says the answer to the first question is "no". I do not know the answer to the second question, although there do exist functions f that possess at least first partials everywhere but such that f itself is not even continuous say at the origin, e.g. f(x,y) = (xy)/(x^2+y^2). It follows that the first partials cannot be continuous, but it does not follow that the first partials cannot themselves have partials, i.e. it is conceivable that second mixed partials exist, maybe even equal, for this discontinuous function f. I have not checked it. ok I checked it. this function does not have mixed partials at (0,0).

I apologize also for the example I cited earlier, which I did not check. Its author states incorrectly that the second partials of his discontinuous function all exist and equal zero, which would of course make the mixed partials equal to each other, (since they would all equal zero). If that were so, it would be a counterexample to the second question, but he is wrong, since in his example the first partials do not exist on a neighborhood of (0,0), since they do not exist at any point of the line x=y except (0,0). So his function is not continuous at (0,0), and does have first partials at (0,0), but not on any neighborhood of (0,0). In particular the second partials do not exist at all at (0,0).

So here was my thought process:

If second mixed partials are continuous, then they are equivalent.
If second mixed partials are continuous, then the original function is C², and therefore continuous.
Therefore, if the second mixed partials are equivalent, the function itself is continuous.

Stated in another manner, I guess, is: Is there a noncontinuous function whose second partials are equivalent?

As I said, the motivation behind this thought process was, that to me it seems easier to calculate the possible equivalence of second order partials, than to test continuity with delta-epsilon or such.

I do realize that this is a very naive way to do maths, but for many cases, such logic seemed to be true more often than not in single-variable calculus, but in multivariable, the simple existence of derivatives seems to be holding a lot less information about the original function than in single-variable calculus, and I can intuitively see why that would be for non-mixed derivatives, but I do not "see" it intuitively why mixed partials still hold less information about the original function, than derivatives in single-variable calculus.

If it helps, I am in Engineering, and Calculus 1 was very practically minded at my University, while Calculus 2 (MV Analysis) is the same that Math or Physics majors have, so I am constantly lost.

 

[mathwonk]

I do not see your argument. e.g.= you say:
"If second mixed partials are continuous, then they are equivalent. " correct, this is Schwartz theorem.

"If second mixed partials are continuous, then the original function is C², and therefore continuous." I do not see this. Just because Fxy and Fyx are continuous, I do not see why the function is C^2, nor even C^1, since there is no guarantee that Fx and Fy are continuous, much less that Fxx and Fyy even exist or if so, are continuous.

For F to be C^2, you need Fx, Fy, Fxx, Fyy, Fxy, Fyx, all to exist and be continuous, just assuming the two mixed ones exist and are continuous does not see to imply this, without some argument.


"Therefore, if the second mixed partials are equivalent, the function itself is continuous." I do not see this of course since it requires the previous statement.

To succeed in analysis it is crucial to be careful about what exactly is being assumed in every statement. To say Fxy (exists and) is continuous, means that Fx also exists, but does not seem to imply that Fx is continuous. I.e. a discontinuous function can have partials. Fx would be continuous if both Fxx and Fxy were continuous. Thus by your argument, F would be C^2, and hence continuous, if the 4 second derivatives Fxx, Fxy, Fyy, and Fyx were all continuous.

 

[WWGD]

Well, it may not help intuition, but it may help to know that Cauchy-Riemann is sufficient for differentiability.

 

[WWGD]

I do not see your argument. e.g.= you say:
"If second mixed partials are continuous, then they are equivalent. " correct, this is Schwartz theorem.

"If second mixed partials are continuous, then the original function is C², and therefore continuous." I do not see this. Just because Fxy and Fyx are continuous, I do not see why the function is C^2, nor even C^1, since there is no guarantee that Fx and Fy are continuous, much less that Fxx and Fyy even exist or if so, are continuous.

For F to be C^2, you need Fx, Fy, Fxx, Fyy, Fxy, Fyx, all to exist and be continuous, just assuming the two mixed ones exist and are continuous does not see to imply this, without some argument.


"Therefore, if the second mixed partials are equivalent, the function itself is continuous." I do not see this of course since it requires the previous statement.

To succeed in analysis it is crucial to be careful about what exactly is being assumed in every statement. To say Fxy (exists and) is continuous, means that Fx also exists, but does not seem to imply that Fx is continuous. I.e. a discontinuous function can have partials. Fx would be continuous if both Fxx and Fxy were continuous. Thus by your argument, F would be C^2, and hence continuous, if the 4 second derivatives Fxx, Fxy, Fyy, and Fyx were all continuous.

Can either of Fxy, Fyx , Fxx , Fyy exist and be continuous if Fx, Fy are not? Fxx is the partial of Fx with respect to x. If Fx is not continuous, it's not differentiable. How would we then define Fxx?

 

[mathwonk]

this stuff is tricky and somewhat unintuitive.

Fxx exists if Fx behaves well just along the horizontal line through our point. F does not have to be differentiable or even continuous for this.

amazingly, there do exist discontinuous functions which still have both first partials (necessarily also discontinuous). Such functions, although they have first partials, are not differentiable. I.e. they are not well approximated by a linear function.
An example is the function cited above, F(x,y) = xy/(x^2+y^2), F(0,0) = 0. This function is very nice away from (0,0), and at (0,0) it is not even continuous. But since it is equal to zero on both axes, and since the x and y partials are only allowed to look at those two axes, it does have first partials also at (0,0), and both partials are zero there.
Hence both partials exist everywhere, but the function is differentiable only away from (0,0), where it is not even continuous, since it equals 1/2 on the punctured line x=y, and x≠0.

The first partials are not continuous at (0,0), since they both equal zero there, but both approach infinity , when (x,y) approaches (0,0) along each axis.

Thus Fx can exist and can even have partials, without being continuous itself. E.g. conceivably, Fx can exist and also Fxy can exist without Fxx existing. Similarly, conceivably Fx and Fy and Fxy and Fyx could all exist without any other partials existing, even while F is discontinuous. I do not have examples of this, but none of the theorems I know rule it out.

I.e. the only theorems I know are;
1) if both Fx and Fy exist and both are continuous, then F is differentiable, hence continuous.
Hence if Fx, Fxy, Fxx, all exist, and Fxx and Fxy are both continuous, then Fx is also continuous, since differentiable.
Hence if Fx,Fy,Fxx,Fxy, Fyy, Fyx all exist, and the four second partials are all continuous, (not just the mixed ones), then F is differentiable and continuous.

2) If Fx and Fy exist, and if both Fxy and Fyx exist and both are continuous, then they are equal, even if no other second partials exist, and even if F,Fx, and Fy, are not continuous.

I should check #2 to make sure I have it right. Any calculus book should suffice, I have Courant.

 

[WWGD]

this stuff is tricky and somewhat unintuitive.

Fxx exists if Fx behaves well just along the horizontal line through our point. F does not have to be differentiable or even continuous for this.

amazingly, there do exist discontinuous functions which still have both first partials (necessarily also discontinuous). Such functions, although they have first partials, are not differentiable. I.e. they are not well approximated by a linear function.
An example is the function cited above, F(x,y) = xy/(x^2+y^2), F(0,0) = 0. This function is very nice away from (0,0), and at (0,0) it is not even continuous. But since it is equal to zero on both axes, and since the x and y partials are only allowed to look at those two axes, it does have first partials also at (0,0), and both partials are zero there.
Hence both partials exist everywhere, but the function is differentiable only away from (0,0), where it is not even continuous, since it equals 1/2 on the punctured line x=y, and x≠0.

The first partials are not continuous at (0,0), since they both equal zero there, but both approach infinity , when (x,y) approaches (0,0) along each axis.

Thus Fx can exist and can even have partials, without being continuous itself. E.g. conceivably, Fx can exist and also Fxy can exist without Fxx existing. Similarly, conceivably Fx and Fy and Fxy and Fyx could all exist without any other partials existing, even while F is discontinuous. I do not have examples of this, but none of the theorems I know rule it out.

I.e. the only theorems I know are;
1) if both Fx and Fy exist and both are continuous, then F is differentiable, hence continuous.
Hence if Fx, Fxy, Fxx, all exist, and Fxx and Fxy are both continuous, then Fx is also continuous, since differentiable.
Hence if Fx,Fy,Fxx,Fxy, Fyy, Fyx all exist, and the four second partials are all continuous, (not just the mixed ones), then F is differentiable and continuous.

2) If Fx and Fy exist, and if both Fxy and Fyx exist and both are continuous, then they are equal, even if no other second partials exist, and even if F,Fx, and Fy, are not continuous.

I should check #2 to make sure I have it right. Any calculus book should suffice, I have Courant.

Ah, good point; partials consider the limit along a single direction; along the axes, while differentiability requires the limit exist along a plane-ful of directions.

 

[mathwonk]

well as usual I learn something from Courant (vol.2, pages 50-60). The theorems 1 and 2 there are stronger than what I said.

1') If the first partials Fx and Fy exist and are only bounded (not necessarily continuous) on a region R, then F itself is continuous on R (but not necessarily differentiable).

2') If Fx and Fy and Fxy exist, and Fxy is continuous, then also Fyx exists and is equal to Fxy; in particular then Fyx is also continuous since Fxy is. (but, as far as I know, F may still be neither differentiable nor continuous.)

"always read the masters."

 

[Feymar]

For F to be C^2, you need Fx, Fy, Fxx, Fyy, Fxy, Fyx, all to exist and be continuous, just assuming the two mixed ones exist and are continuous does not see to imply this, without some argument.

I understand here your counter argument, but since we (engineers in this program), were mainly thought to do second order partials from the first order partials, i.e. for Fxy and Fyx to exist, there must exist Fx and Fy and be differentiable, and since I thought that the existence Fxy and Fyx was a "more powerful" statement about the function than Fxx or Fyy (because it considered more directions at once), I thought all of the required conditions were met for C² , IF Fxy and Fyx were equivalent, because of Schwartz.

Of course, along the same lines, I did think as what you are saying in the case that Fxy and Fyx WERE NOT equivalent. That seemed to be obvious, but that essentially there is nothing more gained about the original function if Fxy and Fyx happen to be equivalent, i.e. that Schwartz cannot be taken in reverse, is wild to me.

What I am essentially trying to understand is, how or why there is so little information gained about the original function in MV Analysis through existence or equivalence of mixed partial derivatives, since "mixed" for me in my head, essentially is associated with "taking care of all directions at once"; why essentially only (total = of all) continuity of derivatives proves continuity of the function itself (if one wants or needs to prove continuity through derivatives instead of other means of course); and the question if one can "calculate" rather than "prove" continuity, if one calculates enough derivations?

Again, I am completely aware, that this is a silly and naive approach to mathematics, and that this is more to satisfy some childish predisposition of mine to think of mathematics as an engineering tool rather than in its own terms. I am aware of that and I am also aware that maths can and should stand on its own merit.

Thus, as you've said, for F to be continous, all 4 second order partials would need to exist and be continous. But Schwarzt says that a continous function has equivalent second order mixed partials, so I thought maybe that, in reverse, that would be enough. i.e. if second order mixed partials are equivalent, that that would guarantee the existence of Fxx and Fyy and their continuity too, because a continous function would have them too, as well as, per Schwartz, equivalent Fxy = Fyx.

I do see the error now and were my assumptions break down. I also see that it is fairly "easy" to prove a discontinous function as such with derivatives, but that the opposite, proving that it is continous by simple existence or equivalence of any order or number of derivatives, is a fool's errand. Thank you.

That being said, is there a way to "calculate", isntead of "prove", a functions continuity, where I would put epsilon-delta in the "prove" category?

Like, if I show that a functions is essentially a polynomial, or a rational function with a domain not containing the root of the denominator, and such? Are these valid ways to show a function is continous without invoking delta-epsilon proofs?

Edit: None of my above reasoning should be taken as arguments, I just wanted to explain why my thinking was erroneous in this particular way.

 

[mathwonk]

there is no free lunch. You have not spent the time and effort to master these precise statements, and it does not suffice to substitute vague impressions of what they mean.


I have given you already a criterion implying continuity:
1') If the first partials Fx and Fy exist and are only bounded (not necessarily continuous) on a region R, then F itself is continuous on R (but not necessarily differentiable).

 

[Feymar]

Indeed. Thank you so much for you patience and time.

 

[mathwonk]

forgive me for fussing at you. let me try to give a little intuition for derivatives. the first point is that possessing "partial derivatives" does not imply the function is "differentiable", far from it, as we have seen it may not even be continuous. this is because the partial derivatives are derivatives of the one variable functions obtained by restricting the function of x and y only to the two lines parallel to the axes.

A function F(x,y) has a graph which is a surface. To possess partial derivatives Fx, and Fy, (at the origin) just means geometrically that this surface has tangent lines at least in the two directions parallel to the axes. But it might not have a tangent line in any other direction. I.e. it only has (one variable) derivatives in two of the infinitely many possible directions at (0,0). you could start with just the two axes as part of this surface, and no matter how you fill in the rest of the surface, e.g. it could go off to infinity near the origin, you are still guaranteed to have both partials at (0,0).

Partials tell you nothing about the function except directly over the two axes. Even so, if partials just exist, you can already try to take second partials, and these can also exist but without the original function being at all nice. I.e. existence of second partials does, as you said, assume existence of first partials, but does not assume actual differentiability, which would require the function's graph to have a tangent plane, not just tangent lines.

A function that has derivatives in all directions (from (0,0)) is said to have (all) directional derivatives there. But it still might not be differentiable!

I.e. if F has directional derivatives in all directions, then the graph surface has a tangent line in every direction from (0,0), but these lines might wave around and make many different angles. If the function is actually differentiable, then not only do tangent lines exist in every direction, but they all lie in a plane, the tangent plane at (0,0).


If the function is nice and differentiable the tangent plane is the best flat approximation to the graph, and if it has nice second derivatives also, the second partials describe the shape of the quadratic surface that fits closest to the graph surface.

Since you are an applied mathematician, or engineer, I suggest you not worry much about exotic, badly behaved functions, since I presume you will be mostly concerned with very nice, continuous, differentiable, even infinitely differentiable functions. But I could be wrong. e.g. all polynomials, rational functions, trig functions, exponential functions, log functions, and all compositions of them, are infinitely differentiable everywhere they are defined, (i.e. where they do not involve division by zero).
good luck and best wishes.

 

[Feymar]

forgive me for fussing at you. [...]

[...] if F has directional derivatives in all directions, then the graph surface has a tangent line in every direction from (0,0), but these lines might wave around and make many different angles. If the function is actually differentiable, then not only do tangent lines exist in every direction, but they all lie in a plane, the tangent plane at (0,0).
[...]
If the function is nice and differentiable the tangent plane is the best flat approximation to the graph, and if it has nice second derivatives also, the second partials describe the shape of the quadratic surface that fits closest to the graph surface.

My vocabulary is infinitely limited to express the gratitude I feel for your and the work you put in educating me and everybody else here. Thank you, and there is nothing to apologize, since you were completely right in the previous post, and I took it as constructive criticism, and actually sat down and relearned some of the specific (other courses are demanding my attention to, unfortunately). It's just that I thought, that maybe I don't understand them well enough, I thought that there was more, because in single-variable calculus, derivatives can tell us so much more about a function than in multivariable, so I convinced myself that I am missing something that actually was not there and tried to shoehorn it.

I thought that maybe if I differentiate enough or if I took enough mixed ones to take care of the various directions in the multiple dimension or if the derivatives themselves not only exist but have some other nice properties, like them being equivalent or such, that that was the "holy grail" I was looking for. But thanks to your patience and these kind explanations, I saw the error of my approach, and found it easier to accept that which I am learning at the moment.

You know, naturally, there is a lot of relevant math to a subject at hand, that is left to be thought at a later date, when the student has matured, and I thought maybe here was something, that would make the ordeal more easier, if I learned it beforehand, but that was just not the case, and that is alright. I will approach the definitions I have been given to work with, with more rigor and care.

And also your explanation with actually using tangent lines, and them being in a plane (or not - in the other case) helped me more to intuitively understand these concepts, since I can imagine it somewhat accurately and work with that far more than with concepts like "... if it the linear mapping is a good approximation then we..." and so on so forth, which I can work through intellectually, but ultimately tells me nothing about the physical representation in space of what I am doing, and it makes it hard for me to apply to my engineering problems. Again, I accept that math is not and should not be bounded by that, but I also found it funny that there was taken great here in Math 1, that it is relevant to our profession, while Math 2 (same professor) goes so much into proves and exotic functions and abstractions. Which is alright, there is no useless knowledge, at least not in this sense we are talking here, and I really don't want to let down my professor or myself with all of this, so thank you from the bottom of my heart once again.
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Only one more thing on what I have said, I would more pointers to, and I will take care of properly learning them. You've said:

"if F has directional derivatives in all directions..." that would mean that there is a gradient ∇f without critical points? i.e. non-zero or non-undefined partial derivatives in all directions exist.

" then you have said: "f the function is actually differentiable, then not only do tangent lines exist in every direction, but they all lie in a plane,"

So would it make sense or be possible at least, to use the gradient ∇f, in some fashion, to test if the tangents are in the same plane, and thus prove differentiability?

 

[mathwonk]

unfortunately, although the "gradient" has the partials as entries <Fx,Fy>, and thus appears to exist if those partials exist, in fact the gradient does not "exist" in any useful sense unless the function its actually differentiable.

So the most common and useful situation is to compute the partials, and then check whether they are continuous. If so, then the function is differentiable, the tangent lines all lie in a plane, and the gradient is a good linear approximation to the function. Moreover then the directional derivatives exist in all directions and are computed by dotting with the gradient.

Although the definition of being differentiable means having a certain precise approximation by a linear function, one seldom actually checks it that way. I.e. the first job is to find that linear function. IF the function is differentiable, i.e. if it has a good linear approximation, then the only possible candidate for that linear approximation is the gadget whose entries are the partials.

Hence in practice, if you want to know whether a function F(x,y) is differentiable, first see if it has both partials. If not, then it is not differentiable, stop. If so, and both Fx and Fy exist, it still may not be differentiable. You have to check whether the linear function defined by dotting with the covector <Fx,Fy> actually approximates the function to within "little oh". This technical effort is tricky and seldom done directly.

More usual, one checks whether the partials exist, not only at your point, but also on an open neighborhood of the point, and then whether also they are continuous at your point. If so, then it is guaranteed that the "gradient" <Fx,Fy> is a good approximation to your function, i.e. your function is indeed differentiable.

There do exist functions that are differentiable at only one point, and whose partials are not continuous, so this method of checking differentiability does not work on them, but I don't know any examples offhand, and I am a professional mathematician. I don't know those exotic examples because I don't need them, except maybe when teaching a course showing how all the properties are related. (I can almost picture in my mind an exotic shaped surface with a tangent plane at one point, but not nicely shaped nearby, whose slices are curves that form angles gradually becoming 180 degrees at our point. e.g. a parabola is a limit of slightly angled curves.). But these functions do not come up in practice.

 

[mathwonk]

I got hung up on your use of the term gradient when the function may not be differentiable. I think I understand your question better now.
"So would it make sense or be possible at least, to use the gradient ∇f, in some fashion, to test if the tangents are in the same plane, and thus prove differentiability?"
Yes in some sense. I.e. if the directional derivatives all lie in a plane, then. it is the plane whose graph is the linear function defined by dotting with the "gradient" <Fx,Fy>. Thus one can indeed use the partials to test whether the directional derivatives all lie in a plane. E.g. one can ask whether in every direction, the directional derivative is equal to the result of dotting with the "gradient". I do not know how easy this is to carry out, and I do not know however whether that proves this plane is actually the tangent plane to the graph. I.e. it does contain the tangent lines in all directions, but I am not sure that guarantees the linear function it represents is really a "little-oh" approximation to the original function. I.e. the desired approximation only holds if we allow the point to approach our given point in all ways, not just straight in along one direction line.

This is an insightful question of yours, but I suggest you will be benefited if you stick to the traditional test for differentiability, i.e. existence and continuity of the partials. That also seems much easier to check than existence of all directional derivatives and agreement with dotting the gradient, even if that did suffice. But it is a nice idea.

 

[mathwonk]

ok here is an example of a function F that has directional derivatives at (0,0) in all directions, all tangent lines there lie in a plane, i.e. the directional derivative in every direction is obtained by dotting with the "gradient" <Fx,Fy>, but the function is not differentiable, not even continuous, at (0,0).

Imagine the parabola y=x^2 in the x,y plane, and shade in all points on and above it in the plane. Then reflect this region in the x axis, so that now one has shaded all points (x,y) with |y| ≥ x^2. Define F to equal zero on this region and also on the whole x axis. Notice that every line in the plane through (0,0) meets this shaded region in an interval of positive length centered at (0,0).
Hence, no matter how we define F outside this region, it will have directional derivative zero in every direction at (0,0). So just define F to be equal to 1 outside this region. Then F is not even continuous at (0,0), since the shaded region doe not contain any open disc centered at (0,0).

The "gradient" is <0,0> and hence dots to give every directional derivative at (0,0).

I think I could make this example continuous as well, and still not differentiable. I.e. I would make it slope sharply up along the curve y=x^4, so as to not have derivative zero along a curved direction, even though it does along all straight directions.

The moral is that basically none of these useful properties are invertible. The easiest way to check differentiability at (0,0) is still hopefully to have existence of partials on a neighborhood, and continuity at (0,0).

 

[Feymar]

Sorry for the late reply. I went to the library and also some more YT courses on the topic, that explain it more vividly thanks to animation, than a professor could. Of course, some things were just analogies to the abstractions that were happening. And I invested a lot into actually coming to terms what was outlined for me in the study materials, most of which I knew also before creating this post in the first place, but just had a gut feeling that I myself am missing something... and I did, but in the wrong way.
Essentially, I thought that I was missing something because I thought that one could do more with mixed partials, that without proving continuity, somehow we could say a lot about the original function, if the gradient was just right, or something was equivalent or we went high enough with the derivations, and that I didn't know. I thought that partials in some form were this extremely powerful too (which they are in engineering), I was just missing something to make them like that in rigorous Maths.

This was not the case. My ignorance stemmed entirely from understanding some definitions in unprecize terms, thinking that there was hiding something "between the lines". Not as a conspiracy, but as a doubt in myself, like surely this cannot be all, surely I was too dumb to understand it, so I made things up, and in that very process I actually made myself dumb. Irony.

Once again, I cannot thank you enough for your time and your patience with me and all of this. If I ever graduate, I will mention you in my speech. The entirety of this thread and your efforts play out a crucial role in my life and you made a difference for me. Thank you.