I use the standard mollifier: $ \varphi(x) = c \cdot \exp \left( \frac{-1}{1-x^2} \right) $ for $x \in [-1,1]$, and $0$ elsewhere. Where $c$ is a constant to mormalize the integral.
I use this mollifier for 2 functions:
1) For $f = x \cdot 1_{x \geq 0}$. I saw this in an article, they claimed that $ 0 \leq f_\lambda(x) \leq f(x) + \lambda$ for $|x| < \lambda$, and $f_\lambda(x) = f(x)$ for $|x| \geq \lambda$. How can I show that?
2) For the function $g(x) = -x log(x)$ for $x \in [0,1]$, and $0$ elsewhere. How can I bound $| g_\lambda(x) - g(x) |$ ?
In both cases: $$f_\lambda(x) = f(x) \ast\varphi_\lambda(x) = \int_{\mathbb{R}^n} f(x-y) \varphi_\lambda(y) dy $$ That is the convolution of the two functions, and: $\varphi_\lambda(x) = \lambda^{-1} \varphi(x/\lambda)$
Is there a generic rule for bounding such differences?
I came up with a solution to part 1 only, here it is:
Let $|x| \geq \lambda$.
If $x \geq \lambda$, then $f(x-t) = x-t$ for all $t \in [-\lambda, \lambda]$, so: \begin{align*} f_\lambda(x) & = \int_{-\lambda}^{\lambda} f(x-t) \varphi_\lambda(t) dt = \int_{-\lambda}^{\lambda} (x-t) \varphi_\lambda(t) dt \\ & = x \cdot \int_{-\lambda}^{\lambda} \varphi_\lambda(t) dt - \int_{-\lambda}^{\lambda} t \varphi_\lambda(t) dt \\ & = x \cdot 1 - 0 = x = f(x) \end{align*}
If $x \leq -\lambda$, then $x-t \leq 0$ for all $t \in [-\lambda, \lambda]$, so: \begin{equation*} f_\lambda(x) = \int_{-\lambda}^{\lambda} f(x-t) \varphi_\lambda(t) dt = \int_{-\lambda}^{\lambda} 0 \cdot \varphi_\lambda(t) dt = 0 = f(x) \end{equation*}
Now, let $|x| < \lambda$, then $f(x-t) = 0$ for $x \leq t \leq \lambda$, so: \begin{align*} f_\lambda(x) & = \int_{-\lambda}^{\lambda} f(x-t) \varphi_\lambda(t) dt = \int_{-\lambda}^{x} (x-t) \varphi_\lambda(t) dt \end{align*} We divide to two cases: