PROBLEM:
Let $K:[a,b]\times [a,b]\to\mathbb{R}$ y $f:[a,b]\to\mathbb{R}$ continuous functions. Let $\hat{f}:[a,b]\to\mathbb{R}$ given by: \begin{equation*} \widehat{f}(x):=\int_{a}^{b}K(x,y)f(y)dy \end{equation*} Prove or disprove the continuity of $\widehat{f}(x)$.
My ideas: I know that K is uniformly continuous because $[a,b]$ is compact and also $f$ is uniformly continuous.
Also if I proving that $\int_{a}^{b}K(x,y)f(y)dy$ is differentiable, then $\widehat{f}(x)$ would be continuous but I don't know how to find the right path. Any idea would be nice ♥
Yes, it is true, and here's an elementary way of proving it. Let $M= \sup\limits_{y\in [a,b]}|f(y)|$; this is finite because $f$ is continuous and $[a,b]$ is compact. Now, for any $x_1,x_2\in [a,b]$, we have \begin{align} \left|\hat{f}(x_1) - \hat{f}(x_2)\right| &= \left|\int_a^b K(x_1,y) f(y)\, dy - \int_a^b K(x_2,y) f(y)\, dy\right|\\ &\leq \int_a^b \left|K(x_1,y) - K(x_2, y)\right| \cdot \left|f(y)\right|\, dy \\ &= M \cdot \int_a^b\left|K(x_1,y) - K(x_2, y)\right| \, dy \end{align} Can you now see how to use uniform continuity of $K$ to complete the proof?