Prove that $f(x,y)$ is continuous if for every constant $y_0$, $f(x,y_0)$ continuous and the partial derivative by $y$ is bounded. I had a test and I couldn't solve this problem:
Given $f: \mathbb R^2 \rightarrow \mathbb R$.For every constant $y_0$, $f(x,y_0)$ is known to be continuous.Also, $\frac{\partial f}{\partial y}(x,y)$ is defined and bounded for all $(x,y)$.
I needed to prove that $f$ is continuous for all $\mathbb R^2$. How do you do that?
 A: Let $(x_0,y_0) \in \mathbb R^2$. For $(x,y) \in \mathbb R^2$ we have
$|f(x,y)-f(x_0,y_0)|=$
$|f(x,y)-f(x,y_0)+f(x,y_0)-f(x_0,y_0)| \le |f(x,y)-f(x,y_0)|+|f(x,y_0)-f(x_0,y_0)| $
There is $L \ge 0$ such that $|\frac{\partial f}{\partial y}(x,y)| \le L$ for all $(x,y) \in \mathbb R^2$. By the Mean value theorem, there is $t$ between $y$ and $y_0$ such that
$|f(x,y)-f(x,y_0)|=|\frac{\partial f}{\partial y}(x,t)(y-y_0)| \le L|y-y_0|$.
Hence
$|f(x,y)-f(x_0,y_0)| \le L|y-y_0|+|f(x,y_0)-f(x_0,y_0)| $.
Now it can be seen, that $f(x,y) \to f(x_0,y_0)$ for $(x,y) \to (x_0,y_0)$ .
A: Hint:
For the continuity you have to consider $|f(x_0,y_0)-f(x,y)|$. The information that $x\mapsto f(x,y_0)$ is continuous says how $f$ behaves in the $x$-direction while $\partial_yf(x,y)$ is bounded says how it behaves in the $y$-direction. Therefore you have to write your term such that you can use this information. It means, you write
\begin{align}
|f(x_0,y_0)-f(x,y)|&=|f(x_0,y_0)-f(x,y_0)+f(x,y_0)-f(x,y)|\\
&\leq|f(x_0,y_0)-f(x,y_0)|+|f(x,y_0)-f(x,y)| 
\end{align}
For each term you have to use one of the information you have.
Proof:

 Let be $(x_0,y_0)\in\mathbb R^2$ and $\varepsilon>0$. Since $x\mapsto f(x,y_0)$ is continuous there exists $\delta_1>0$ such that for $x\in\mathbb R$ with $|x_0-x|<\delta_1$ holds $$|f(x_0,y_0)-f(x,y_0)|<\frac\varepsilon2.$$ Futher exists $C>0$ such that $|\partial_yf(x,y)|<C$ for all $(x,y)\in\mathbb R^2$ by assumption. The MVT yields $$|f(x,y_0)-f(x,y)|<C|y_0-y|.$$ Define $\delta<\min\left\{\delta_1,\frac{\varepsilon}{2C}\right\}$. For $(x,y)\in\mathbb R^2$ with $\|(x_0,y_0)-(x,y)\|_\infty<\delta$ holds $$|x_0-x|<\delta<\delta_1\text{ and }|y_0-y|<\delta<\frac\varepsilon{2C}$$ and therefore\begin{align}|f(x_0,y_0)-f(x,y)|&=|f(x_0,y_0)-f(x,y_0)+f(x,y_0)-f(x,y)|\\&\leq|f(x_0,y_0)-f(x,y_0)|+|f(x,y_0)-f(x,y)|\\&<\frac{\varepsilon}2+C|y_0-y|\\&<\varepsilon\end{align} 

A: f(x,y)-f($x_0$,$y_0$) = [f(x,y)-f(x,$y_0$) ] + [ f(x,$y_0$) - f($x_0$,$y_0$) ] .
On the first bracketed term use the mean value theorem and the boundedness of the partial derivative : on the second use the given continuity .  
