What bounds does an approximate identity give: $f_{\epsilon} = \int f(x-y)\frac{1}{\epsilon}\phi(\frac{y}{\epsilon})dy$ for the following inequalities?
$\sup_{x}|f_{\epsilon}(x)|\leq ?\sup_{x}|f(x)|$
$\sup_{x}|f_{\epsilon}'(x)|\leq ?\sup_{x}|f(x)|$
$\sup_{x}|f_{\epsilon}''(x)|\leq ?\sup_{x}|f(x)|$
I think the first one is bounded by $\frac{1}{\epsilon}$ and then you just continually take derivatives because the function $\phi$ is smooth but I'm pretty apprehensive that it is that simple.
We have $$\sup_x|f_\epsilon|\leq\sup_x|f(x)|\int\frac1\epsilon\phi(\frac y\epsilon)dy=\sup_x|f(x)|\int\phi(y) dy=\sup_x|f(x)|$$ where I'm assuming $\phi$ is positive, has integral 1, and we are in 1 dimension. For the others, the only additional thing to notice is that we can commute the convolution so that the derivative falls on the mollifier. $$\sup_x|f^{(n)}(x)|=\sup_x|\int f(y)\frac1{\epsilon^{1+n}}\phi^{(n)}(\frac{x-y}\epsilon)dy|\leq\sup_x|f(x)|\frac1{\epsilon^n}\int|\phi^{(n)}(y)|dy$$ which is the best possible bound.