Q:

Subset closed with respect to the norm $\|\cdot\|_{\infty}$ but not with respect to $\|\cdot\|_1$? 

Let $E = C([0,1],\mathbb{R})$ be the vector space of the continuous functions $[0,1] \rightarrow \mathbb{R}$.
For all $f \in E$, we denote $\|f\|_{\infty} = \sup_{t \in [0,1]} |f(t)|$ and $\|f(t)\|_1 = \int_0^1 |f(t)| dt$.
We consider now $A = \{f \in E \mid f(0)=0 \text{ and } \int_0^1 f(t) \, dt \geq 1 \}$. I have to show that $A$ is a closed subset of $E$ with $\|\cdot\|_{\infty}$ but is not a closed subset of $E$ with $\|\cdot\|_1$. Do you have an idea of how to do it ?  Thank you in advance. 

A:

Here's a short outline for the first part: take a sequence $(f_n)$ in $A$ that converges in norm $\lVert \cdot \rVert_\infty$ to some $f \in X$. In other words, $f_n$ converges uniformly to $f$. That $f(0) =0$, should be easy enough to prove.  For the second condition consider
$$
\int_0^1 f(t) dt = \int_0^1 f_n(t) dt + \int_0^1 [ f(t) -f_n(t) ]dt. 
$$
The first term is always bigger than $1$ and the second term can be made arbitrarily small using the uniform convergence. 
For the second part, you need to find a sequence $(f_n)$ in $A$ that converges to $f$ in $\lVert \cdot \rVert_1$ but with $f(0) \neq 0$. You don't have to look far for examples.  For example take $f=1$, the constant function and $f_n$ to be linear close to $t=0$ but becomes a constant $a_n$ after some $t_n$. As $n$ becomes larger, $t_n$ should go to zero and $a_n$ 
to $1$.

