Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Construct a set of functions $\{g_\epsilon(x) \}$, such that for every $\epsilon > 0, \; g_\epsilon(x)$ is infinitely differentiable and $$ g_\epsilon \rightarrow f,$$ where $f(x) = |x|,$ in the sup norm as $\epsilon \downarrow 0$.

share|cite|improve this question

Try $f_\varepsilon(x):=\sqrt{x^2+\varepsilon}$.

share|cite|improve this answer

Let the sequence $\{\phi_\epsilon\}$ be a mollifier. If $f$ is any continuous function, then the convolutions $g_\epsilon(x) := (f * \phi_\epsilon) (x) := \int f(y) \phi_\epsilon(x-y) dy$ converge uniformly on compact sets to $f$ as $\epsilon \rightarrow 0$.

In fact, we can choose $\phi_\epsilon=\epsilon^{-1} \phi(\epsilon^{-1}x)$, where $\phi$ has integral $\int_\mathbb{R} \phi(y)dy =1$. Then we have $$g_\epsilon (x) =\int_{-\epsilon}^\epsilon f(y) \phi_\epsilon(x-y) dy = \int_{-1}^1 f(x-\epsilon y) \phi(y) dy $$ and $$|g_\epsilon(x)-f(x)| \leq |\int_{-1}^1 f(x-\epsilon y) \phi(y) dy - f(x)| =\int_{-1}^1 \phi(y)|f(x-\epsilon y) - f(x)| dy$$ For $x \in K$, a compact set, $f$ is uniformly continuous, taking the limit as $\epsilon \rightarrow 0$ shows that $g_\epsilon \rightarrow f$ on $K$.

share|cite|improve this answer
    
... but $\mathbb{R}$ is not compact (I assume the OP wants $\lim_{\epsilon \to 0} \sup |f(x) - g_\epsilon(x)| = 0$). In this case however we can explicitly choose a mollifier such that $g_\epsilon \equiv f$ outside $B_{\epsilon}$. – Willie Wong Nov 20 '12 at 12:47
    
@WillieWong, you are right. I guess the mollifier approach only gives convergence on compact sets. Am I right? – Cantor Nov 20 '12 at 13:15
    
Well... it allows you to directly conclude (without needing to argue it yourself) that we have convergence on compact sets. But in our case $f$ is much better than just continuous: it is globally Lipschitz... – Willie Wong Nov 20 '12 at 13:19
    
@WillieWong: If we take $g_epsilon = f$ outside $B_\epsilon$ will $g_\epsilon$ still be smooth? what happens at points $\{-\varepsilon, \varepsilon \}$? – Arunangshu Biswas Nov 22 '12 at 13:02
    
@ArunangshuBiswas There is nothing that prevents $g_\epsilon$ to be smooth even it it equals $f$ outside $B_\epsilon$, so long as $f$ itself is smooth outside $B_\epsilon$. For example, if you take the mollifier $\phi_\epsilon$ to be an even function, you can manifestly check that in this case ($f(x) = |x|$) outside of $B_\epsilon$ we do have $g_\epsilon \equiv f$. – Willie Wong Nov 22 '12 at 13:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.