Does $\|f\|_K = \sup_{z \in K} |f(z)|$ define a norm on $C(\mathbb{C})$ and $H(\mathbb{C})$ for compact $K$? 
Let $C(\mathbb{C})$ denote the vector space of continuous complex valued functions on $\mathbb{C}$ and $H(\mathbb{C})$ denote vector space of entire functions. For any function $f$ in $C(\mathbb{C})$ and $H(\mathbb{C})$ and for any compact subset $K$ define $\|f\|_K = \sup_{z\in K} |f(z)|$.
  
  
*
  
*Is $\|\cdot\|_K$ a norm on $C(\Bbb C)$ for every compact $K\subset \Bbb C$?
  
*Is $\|\cdot\|_K$ a norm on $H(\Bbb C)$ for every compact $K\subset \Bbb C$?
  

I can't figure out how should I proceed with these problems.
It will be great if someone could figure out what to do here.
 A: First notice that for every continuous function $f \colon \mathbb{C} \to \mathbb{C}$ we have $\|f\|_\emptyset = -\infty$, so we will only look at non-empty compact sets.

Now let $K \subseteq \mathbb{C}$ is some non-empty compact subset. If $f \colon \mathbb{C} \to \mathbb{C}$ is continus then the restriction $f|_K$ is also continuous with $|f(x)| \geq 0$ for all $x \in K$. By the compactness of $K$ we find that
$$
 0 \leq \sup_{x \in K} |f(x)| < \infty.
$$
So $\|f\|_K$ is well defined. For all continuous functions $f,g \colon \mathbb{C} \to \mathbb{C}$ and $\alpha \in \mathbb{C}$ we now have
$$
 \| \alpha f \|
 = \sup_{x \in K} |\alpha f(x)|
 = |\alpha| \sup_{x \in K} |f(x)|
 = |\alpha| \|f\|_K
$$
and
$$
 \|f+g\|_K
 = \sup_{x \in K} |f(x)+g(x)|
 \leq \sup_{x \in K} |f(x)| + \sup_{y \in K} |g(y)|
 = \|f\|_K + \|g\|_K,
$$
so $\|\cdot\|_K$ is absolutely homogeneous and satisfies the triangle inequality. It thus is a seminorm on $C(\mathbb{C})$. Now if we have $\|f\|_K = 0$ then $f|_K = 0$. But this is not enough for $\|\cdot\|_K$ to be a norm on $C(\mathbb{C})$! For if we take a continus function $f \colon \mathbb{C} \to \mathbb{C}$ with $f|_K = 0$ but $f \neq 0$, for example the distance function
$$
 f_K(x) = \inf_{y \in K} |x-y|,
$$
then $\|f_K\|_K = 0$, but $f_K \neq 0$. (Notice that $f(x) \neq x$ for all $x \notin K$ because $K$ is closed, and such $x$ exist because $K$ is bounded.) So $\|\cdot\|_K$ is only a seminorm, but never a norm on $C(\mathbb{C})$.

Remark: There is a standard trick to actually get a metric out of this seminorms: Take a compact exhaustion of $\mathbb{C}$, i.e. a sequence $(K_n)_{n \in \mathbb{N}}$ of compact subsets $K_n \subseteq \mathbb{C}$ which is increasing, i.e. $K_0 \subseteq K_1 \subseteq K_2 \subseteq \dotsb$ and exhausts $\mathbb{C}$ i.e. $\bigcup_{n \in \mathbb{N}} K_n = \mathbb{C}$. Then we can define
$$
 d(f,g) = \sum_{n \in \mathbb{N}} 2^{-n} \frac{\|f-g\|_{K_n}}{1+\|f-g\|_{K_n}}
 \quad
 \text{for all $f,g \in C(\mathbb{C})$}.
$$
Because each $\|\cdot\|_{K_i}$ is a seminorm it follows that $d$ is symmetric and satisfies the triangle inequality. But if $d(f,g) = 0$ then $\|f-g\|_{K_i} = 0$ for every $i \in \mathbb{N}$, and because $(K_n)_{n \in \mathbb{N}}$ exhausts $\mathbb{C}$ it follows that $f-g = 0$ and thus $f = g$. So $d$ is a metric on $C(\mathbb{C})$ (and thus by restricton also on the subspace $H(\mathbb{C})$.)
It can also be shown that for a sequence $(f_n)_{n \in \mathbb{N}}$ in $C(\mathbb{C})$ and $f \in C(\mathbb{C})$ one has $d(f_n,f) \to 0$ if and only if $\|f - f_n\|_K \to 0$ for every compact subset $K \subseteq \mathbb{C}$. So the topology generated by $d$ does actually not dependent on the exhaustion.

Now what about $H(\mathbb{C})$? Because this is a subspace of $C(\mathbb{C})$ we find that $\|\cdot\|_K$ still defines a seminorm on $H(\mathbb{C})$ for every compact subset $K \subseteq \mathbb{C}$.
Now suppose that $K \subseteq \mathbb{C}$ is a non-discrete compact subset (and thus in particular non-empty), i.e. has an accumulation point. Then if $f \colon \mathbb{C} \to \mathbb{C}$ is holomorphic with $f(x) = 0$ for all $x \in K$, then by the identity theorem we already have $f = 0$ . So in this case it follows from $\|f\|_K = 0$ that $f = 0$. So if $K \subseteq \mathbb{C}$ is a non-discrete compact subset then $\|\cdot\|_K$ is actually a norm on $H(\mathbb{C})$.
If on the other hand $K \subseteq \mathbb{C}$ is non-empty and compact, but discrete, then this does not necessarily hold: Take for example any finite non-empty subset $K \subseteq \mathbb{C}$. Then for the polynomial $f(z) = \prod_{\xi \in K} (z-\xi)$ we have $\|f\|_K = 0$ but $f$ is non-zero outside of $K$. Because every discrete compact subset $K \subseteq \mathbb{C}$ is already finite, it follows that if $K \subseteq \mathbb{C}$ is non-empty, compact and discrete, then $\|\cdot\|_K$ still only is a seminorm, but a not a norm. (To see that $K$ is finite notice that every $x \in K$ has an open neighborhood $U_x \subseteq \mathbb{C}$ with $U_x \cap K = \{x\}$. Because $K$ is compact it can be covered by finitely many $U_{x_i}$. So
$$
 K
 = K \cap (U_{x_1} \cup \dotsb \cup U_{x_n})
 = \{x_1, \dotsc, x_n\}
$$
for some finitely many $x_1, \dotsc, x_n \in K$.)
