0
$\begingroup$

The function $f$ is defined : $\begin{cases} C_{[0,1]}\to\mathbb{R} \\f\to f(0)\end{cases}$ And $C_{[0,1]}$ is the metric space of continuous functions with the integral metric $d$. We define a function $g$ that lies on the neighbourhood of $f$:

\begin{align}g:[0,1]&\to\mathbb{R}\\ x&\rightarrow\begin{cases}f(\epsilon^´)\frac{x}{\epsilon´}&\text{if }x<\epsilon´\\ f(x)&\text{otherwise}\end{cases}\end{align}


My question is about the integral. How can $\int_0^{\epsilon´}\left|f-g\right|\leqslant 2M\epsilon$ be true. Is this right? I got it from a textbook exercise solution. $M=sup\left|f\right|$.Thanks for reading!

\begin{align}d(f,g_n)&=\int_0^1\left|f-g\right|\\ &\ =\int_0^{\epsilon´}\left|f-g\right|\\ &\leqslant 2M\epsilon´ \end{align}

$\endgroup$
  • $\begingroup$ Presumably $M$ is a bound for $|f|$ in which case the functions $f,g$ differ only for $x \in [0,\epsilon']$ and then $2M $ is a conservative upper bound for $|f(x)-g(x)|$ there. $\endgroup$ – copper.hat Mar 20 '17 at 15:53
  • $\begingroup$ I have thought of that but when you compute the integral you get $M\epsilon´-M\epsilon´=0$ instead of $2M\epsilon$, right? $\endgroup$ – Pedro Gomes Mar 20 '17 at 15:59
1
$\begingroup$

Note that $g(x) = f(x)$ for $x > \epsilon'$.

$\int_0^1 |f-g| = \int_0^{\epsilon'} |f(x)-g(x)| dx \le \int_0^{\epsilon'} (|f(x)|+|g(x)|)dx \le \int_0^{\epsilon'} 2M dx = 2M \epsilon'$.

$\endgroup$
  • $\begingroup$ Thank you I was doing$\int_0^{\epsilon'} (|f(x)|-|g(x)|)dx $ which was wrong. $\endgroup$ – Pedro Gomes Mar 20 '17 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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