Hi! So I am learning multivariable calculus, and I have a doubt. In one of my assignments, I was asked to prove that if a multivariable function had continuous first partial derivatives on a domain $D$, then that function is continuous as well.
Now, when I tried working out the problem, I noticed that continuity of first partial derivatives wasn't a necessary condition. With only the condition that the partial derivatives are defined everywhere on the domain $D$, I could prove that function was continuous on the domain $D$.
So my doubt is whether only the existence of first partial derivatives everywhere in the domain $D$ guarantee the continuity of the the function? (I think yes)
@TedShifrin Continuous partials do imply differentiability of the original function. But I only want to prove the continuity of the original function, so I guess even the existence of partials everywhere would do the work.
@TedShifrin Yeah, because the derivatives along the axes exist, but the function is discontinuous at the origin when approached along $x = y$ and $x = -y$ direction.
@aderchox a single eyeball works by projecting 3D objects onto a 2D screen. We get a 3D effect by having two eyeballs, so we implicitly measure a difference between two 2D images.
Jacobson 2:18-22 18 But someone will say, “You do algebra, and I do analysis.” Show me algebra without analysis, and I will show you my algebra by way of analysis. 19 You believe that there is one Field. You do well. Even the mathematicians believe—and tremble! 20 But do you want to know, O foolish graduate student, that algebra without analysis is dead? 21 Was not Cauchy our father justified by analysis when he offered geometry his son on the altar? 22 Do you see that algebra was working together with analysis, and by analysis, algebra was made perfect?
user486313
@anakhro may be true for functional analysis ...
user486313
nice quote, btw ...
user486313
nobody in my department spends any time on algebra. everyone's doing Analysis and PDEs.
Hi all anyone know a link which proves that a transport plan between measures $\mu_1$ and $\mu_2$ has to be the product measure if either $\mu_1$ or $\mu_2$ is a dirac?
The Matrix is everywhere. It is all around us. Even now, in this very room. You can see it when you look out your window or when you turn on your television.
