Let $X$ be the space of real sequences having finitely many nonzero terms with $||\ ||_p , 1\leq p \leq \infty$ . Define $f: X\to\mathbb{R}$ by $f(x)=\sum_{j=1}^{\infty} x_j $ for $x=(x_j) \in X$. Then
$(a)$ $f$ is continous only for $p=1$.
$(b)$ $f$ is continous only for $p=2$.
$(c)$ $f$ is continous only for $p=\infty$.
$(d)$ $f$ is not continous only for any $ p , 1\leq p \leq \infty$.
My attempt:
$X$ is just $c_{00}$ with $||\ ||_p$
For $(c)$ if I take $z_n=(1,1,1,...1,0,0,0,...)$ then $||(z_n)||_\infty=1$ but $f((z_n))=n$ that means norm of $f$ is unbounded means $f$ is not continuous.
For $(a)$ if I take $z_n=(0,0,0,...,1,0,0,0,...)$ $ 1$ at $n^{th}$ place. Then $||(z_n)||_1=1$ but $f((z_n))=\sum_{j=1}^{\infty} 1 =\infty$ that is norm of $f$ is unbounded. Therefore $f$ can not be continuous. But I don't know how to proceed for $(b)$ and $(d)$. Could you please help me. Thank you very much.
The corrrect answer is a): $f$ is continuous only for $p=1$. For $p=1$ we have $|f(x)| \leq \sum |x_i|=\|x\|$ so $f$ is bounded. For $ 1<p<\infty$ consider $x=(1,1,\cdots ,1,0,0...)$ where there are $n$ ones. Then $f(x)=n$. If $|f(x) \leq C\|x\|$ we get $n \leq Cn^{1/p}$ for all $n$ and we get a contradiction by dividing by $n^{1/p}$ and letting $n \to \infty$. For $p=\infty$ your example works.