Consider a sequence $f_n:R\to R,$ $f_n(x)=e^{-nx^2}$.
This sequence converges pointwise to $$ f(x) = \left\{\begin{array}{cl}1 & \quad \mbox{if } x = 0 \\ 0 & \quad\mbox{if } x \neq 0 \end{array} \right. $$
Now $\|f_n-f\|_\infty=\sup|e^{-nx^2}-f(x)|=$ ?
What I don't understand: Apparently it equals $1$, but I don't see why this is the case.
So because $1$ doesn't converge to $0$ as $n\to \infty$, $f_n$ doesn't converge uniformly.
I need help on why the equality above equals $1$.
Thanks in advance!
We have $0 \leqslant e^{-nx^2} \leqslant 1$ for all $n \geqslant 0$ and $x\in \mathbb{R}$, hence $\lVert f_n - f\rVert_\infty \leqslant 1$. Since $f_n$ is continuous with $f_n(0) = 1$, we have
$$\lim_{x\to 0} f_n(x) = 1.$$
But $f$ is discontinuous in $0$, and we have
$$\lim_{\substack{x\to 0\\x\neq 0}} f(x) = 0,$$
and therefore
$$\lim_{\substack{x\to 0\\x\neq 0}} \left[f_n(x) - f(x)\right] = 1,$$
which implies $\lVert f_n - f\rVert_\infty \geqslant 1$. Together, it follows that $\lVert f_n - f\rVert_\infty = 1$ for all $n \geqslant 0$.