0
$\begingroup$

I'm working on problem 3 from Exercises 2E of Axler's "Measure, Integration and Real Analysis" which says: enter image description here

I thought of something like $f_n (x) = \frac{1}{x + \frac{1}{n}}$ , but then this converges to $f(x) = 1/x$, which isn't defined at $x = 0$ so it is not a function $f: [0,1] \rightarrow \mathbf{R}$. When I attempt to fix this I lose either the continuity of $f_n$ or the convergence to $f$.

I'd appreciate any insight

$\endgroup$
0
3
$\begingroup$

Define

$$f_n(x) = \begin{cases} n^2 x &0 \leq x \leq \frac 1n \\ \frac 1x &\frac 1n \lt x \leq 1. \end{cases}$$

Then $f_n$ converges pointwise to $\frac 1x$ for $x \in (0, 1]$ and $\forall n~f_n(0)=0.$

$\endgroup$

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.