f(x,y) is continuous if the partial derivatives of f(x,y) exist and one of them is bounded below? I had a test and I couldn't solve this problem:
f is continuous if the partial derivatives of f(x,y) exist and one of them is bounded below.
I know that if f is differentiable then f must be continuous. But this isn't a necessary condition. How do you do that?
 A: Suppose that $f$ is not continuous at the origin and $f(0,0)=0,f_{x}$ is bounded below.Then there is an $\varepsilon_{0}>0,P_{n}=(x_{n},y_{n})\rightarrow0$ as $n\rightarrow\infty$ so that $|f(P_{n})|\geq\varepsilon_{0}.$
If $\{P_{n}\}$ is in the first quadrant and suppose that $f(P_{n})\geq0$, according to the existence of $f_{x}$, we know that there is a $\delta_{1}>0$ so that when $x\in(0,\delta_{1}),|f(x,0)|<\frac{\varepsilon_{0}}{4}.$
Similarly, $\exists\delta_{2}>0$ so that when $y\in(0,\delta_{2}),|f(x,y)-f(x,0)|<\frac{\varepsilon_{0}}{4}$ where $x\in(0,\delta_{1}).$
Hence $$\begin{aligned}
|f(x,y)|&=|f(x,y)-f(x,0)+f(x,0)|\\
&\leq|f(x,y)-f(x,0)|+|f(x,0)|\\
&<\dfrac{\varepsilon_{0}}{2}.
\end{aligned}$$
Since $P_{n}=(x_{n},y_{n})\rightarrow0$ as $n\rightarrow\infty$,when $n$ large enough,there is $|P_{n}|<\min\{\delta,\delta_{2}\}.$ Thus we have $|f(x,y_{n})|<\frac{\varepsilon_{0}}{2}.$
Using the Lagrange mean value theorem,we know there exists $\xi_{n}$ between $x$ and $x_{n}$ so that $$f_{x}(\xi_{n},y_{n})=\dfrac{f(x,y_{n})-f(x_{n},y_{n})}{x-x_{n}}<-\dfrac{\varepsilon_{0}}{2x}.$$
Let $x\rightarrow0$ we can get $f_{x}(\xi_{n},y_{n})\rightarrow-\infty$ which is contradict to $f_{x}$ is bounded below.
When $\{P_{n}\}$ is in other quadrant we can discuss similarly to conclude the contradiction,and in general,we can just take a subsequence $\{P_{n_{k}}\}$ which in some quadrant into consideration.
Above all,$f$ is continuous.
