How to prove that F is continuous?

Question

I have the function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ which is continuous and i know that it I can differentiate it to the first variable and that $D_1f : \mathbb{R}^2\rightarrow\mathbb{R}$ is continuous. How do i prove continuity for the following function? $$ F(x) = \int_a^xf(x,y)dy, x\in\mathbb{R}$$

Answer 1

Fix $\epsilon>0$. By the continuity property of $f$, it follows there exists $\delta>0$ such that \begin{align} |f(x+h, y)-f(x, y)|<\epsilon \ \ \Rightarrow \ \ |f(x+h, y)| <\epsilon+|f(x, y)| \end{align} whenever $|h|<\delta$. Also, by the mean value theorem, we know there exists $ \xi \in (x, x+h)$ such that \begin{align} f(x+h, y)-f(x, y) = D_1f(\xi, y)h. \end{align} Combining the above two facts yield the estimate \begin{align} |F(x+h) - F(x)| =&\ \left| \int^{x+h}_a f(x+h, y)\ dy - \int^x_a f(x, y)\ dy \right|\\ \leq&\ \int^{x+h}_x|f(x+h, y)|\ dy+ \int^x_a|f(x+h, y)-f(x, y)|\ dy\\ \leq&\ \int^{x+h}_x [|f(x, y)|+\epsilon]\ dy +\int^x_a |D_1 f(\xi, y)||h|\ dy \\ \leq&\ [|f(x, y)|+\epsilon]||h| + \sup_{\xi\in [x, x+h]}|D_1f(\xi, y)||h||x-a|. \end{align} Hence the problem can be finished using the usual $\epsilon-\delta$ argument which I will leave as an exercise for the reader.