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Is there a single, continuously differentiable function $g(x,k)$ that approximates the following:

$f(x)= \begin{cases} 0 & x<0 \\ x & 0 \le x \le 1 \\ 1 & x>1\end{cases}$

$k$ is a real parameter such that $g \rightarrow f$ as $k \gg 1$

Edit: 1. I apologise for my lack of rigour; I'm the worst kind of mathematician: an engineer ;)

  1. I realise that my question doesn't imply that $g(x,k)$ is bounded by [0,1] for all $x$. So that's an additional constraint. Thus I'm not sure a Fourier series will be satisfactory in this case.

  2. What I mean by $g \rightarrow f$ is that $\left| g(x,k) - f(x) \right| < \epsilon$ for any given $x$ and a small positive number $\epsilon$ for a sufficiently large $k$.

CjS
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  • Let's assume that you have a function $g_{0}(x,k)$ which satisfies your assumptions. The function $g_{1} (x,k)$ defined as $g_{1} (x,k) = g_{0} (x, 2 k)$ is different from $g_{0}$ but still satisfies your assumptions, doesn't it ? – jibe Jul 01 '14 at 16:08
  • First, assume there exists a function $\theta_{k} \in \mathcal{C}^{\infty}(\mathbb{R})$, such that $\theta_{k}(x) = 0$ when $x\leq0$, and $\theta_{k}(x) = 1$ when $x \geq \frac{1}{k}$. I suggest you to derive your result with this hypothesis. Then, try to prove this hypothesis. Note that asking $g \to f$ doesn't make any sense without precising the underlying topology : are you talking about pointwise convergence, uniform convergence, an other type of convergence ? – Tlön Uqbar Orbis Tertius Jul 01 '14 at 16:12
  • Edit : you can even make the weaker hypothesis $\theta_{k} \in \mathcal{C}^{1}(\mathbb{R})$, it'll be easier to prove .But first, try to derive your result assuming $\theta_{k}$ exists for every $k$ as explained in the preceding post. – Tlön Uqbar Orbis Tertius Jul 01 '14 at 16:22
  • Do you know about Fourier series? Incidentally, you should explain what do you mean by convergence of a sequence of functions since there are many different notions. – Moishe Kohan Jul 01 '14 at 16:28

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