In many typical analysis books, when authors introduce "uniform convergence", they would introduce "space of bounded functions" with supremum norm.
For example, let $X$ be an arbitrary set, $E$ be a complete metric space. Let $B(X,E)$ be the space of bounded functions. Also, let $f_n$ be a sequence of functions and $f$ a function in $B(X,E)$. It is easy to see the following three conditions are equivalent.
It is easy to see the following three conditions are equivalent:
$f_n \to f (uni)$
$f_n \to f $ in $B(X,E)$
$||f_n -f||_{\infty} \to 0$ in $\mathbb{R}$.
However in Amann's Analysis I. The authors say that $f_n$ and $f$ needn't be in $B(X,E)$. And they replace the above requirement 2 with 2', as follows:
$f_n \to f (uni)$
' $f_n - f \to 0$ in $B(X,E)$ (This means $f_n - f$ is in $B(X,E)$, that is the difference of two functions is in $B(X,E)$)
$||f_n -f||_{\infty} \to 0$ in $\mathbb{R}$.
My question is: Is Amann's generalization(The second list of statements) true? To me, it is quite possible that $f_n -f$ is not bounded for some $n$, that is to say: $f_n - f \notin B(X,E)$ for some $n$. I know the basic fact that "In an arbitrary metric space, every convergent sequence is bounded". But when it comes to function spaces, it's not so convinceable to me.
For example, I wonder if there is the possibility that, for every $\epsilon$, there is an integer $N$ s.t. $||f_n - f - 0|| < \epsilon$ for $n > N$. However, there is some $n_0 \leq N$ s.t. $||f_n - f - 0|| = \infty$
Surely possible. Let $X=\mathbb R$, $f_1(x)=x, f_n(x)=0$ and $f(x)=0$ for all $n \geq 2$ for all $x$. Then $\|f_n-f\|=0<\epsilon$ for all $n >1$ wheras $\|f_1-f\|=\infty$.