For a function, if at a point $a$, the function has directional derivatives along some lines, but the function is discontinuous at $a$, does that mean along those lines, the function is continuous, but along some other directions the function is not? What does the graph of such a function look like? Continuous in some direction but discontinuous in others?
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$\begingroup$ I'm not sure the question is clear. In your first sentence, you say that the function is continuous -- did you actually mean to say discontinuous? $\endgroup$– AllawonderCommented Sep 9, 2019 at 1:06
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$\begingroup$ @Allawonder. Yes, should be discontinuous. $\endgroup$– user533661Commented Sep 9, 2019 at 1:17
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$\begingroup$ @saulspatz why in every direction? Seems undefined in directions of rational slope? $\endgroup$– user7530Commented Sep 9, 2019 at 2:56
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$\begingroup$ @user7530 You're right. Sloppy thinking. I'll try to fix it. Trouble is, I'm about to fall asleep. $\endgroup$– saulspatzCommented Sep 9, 2019 at 2:59
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$\begingroup$ @user7530 I deleted the comment. I don't know what I was thinking. $\endgroup$– saulspatzCommented Sep 9, 2019 at 13:20
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Sure. The directional derivative $D_v f$ looks at the restriction of the function to the line in the $v$ direction; it's quite possible for this restriction to be continuous whereas the overall function is discontinuous. An easy example is $$f(x,y) = \begin{cases}1, &y \geq 0\\ 0, &y < 0\end{cases}$$ which looks continuous in the $x$ direction, but not in any other direction.
As you may know, existence of directional derivatives is much weaker than differentiability of $f$ (which guarantees that the directional derivative at a point is linear as a function of the direction $v$). Perhaps counterintuitively, you can have that the directional derivative of $f$ at a point exists in all directions, but $f$ is still not continuous at that point. See this writeup for instance which proves that $$f(x,y) = \begin{cases}\frac{x^2 y}{x^2+y^4}, & (x,y)\neq (0,0)\\ 0, & x=y=0\end{cases}$$ is such a counterexample (you can create many others of the same family).
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$\begingroup$ Thanks, I can see the example is discontinuous yet all directional derivative exist, but I can not see what "continuous" along a direction means intuitively, on the graph the discontinuous point is a hole, how can the function "continuous" along a particular direction on a hole? $\endgroup$– user533661Commented Sep 9, 2019 at 4:19
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$\begingroup$ @Cathy there shouldn't be a hole (put in the missing point). $\endgroup$– user7530Commented Sep 9, 2019 at 4:21
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$\begingroup$ Then what the missing point looks like? $\endgroup$– user533661Commented Sep 9, 2019 at 4:45