Problem with $F$ bounded $\rightarrow$ $F$ continuous I have to proof: $F$ bounded in the ball $\rightarrow$ $F$ continuous, where $F: H \rightarrow \mathbb{R}$ and $H$ a real Hilbert space and $F$ linear.
Suppose (absurde) exists a sequence $\{ x_n \}$ with $\left \| x_n \right \|=1$ $\forall n \in N$ such that $\frac {F(x_n)}{\left \| F(x_n) \right \|} \xrightarrow{n \to \infty} + \infty$.
So:
$$\frac{x_n}{F(x_n)}= \frac{x_n}{\left \| F(x_n) \right \| } \cdot \frac{\left \| F(x_n) \right \|}{F(x_n)} \xrightarrow{n \to \infty} 0,$$ but $F \left ( \frac{x_n}{F(x_n)} \right )=1 $ (why?)
Is it make sense?.. Maybe I don't copy well from the blackboard!
Thanks
 A: There isn't much to prove. Linear operators are continuous if and only if they are bounded, therefore you must prove that $F$ is bounded. But what does boundedness mean? It means several equivalent things, the most useful of which is the following description: $F$ is bounded if and only if $\| F \| < \infty$, where
$$ \| F \| = \sup \limits _{\| x \| \le 1} \frac {|F(x)|} {\| x \|} .$$
Since $F$ is linear, the above fraction is in fact $\left| F \left(\dfrac x {\| x \|} \right) \right|$, which is bounded by hypothesis (because $\dfrac x {\| x \|}$ is an element of the sphere of radius $1$, where $F$ is asumend to be bounded), and this is it. The trick was to think of boundedness, not of continuity, because they are equivalent (but boundedness is easier to work with, because it can be expressed by a single number, $\| F \|$, so you don't have to work with sequences or nets).
Concerning the proof that you give, it doesn't make sense: since $F$ takes values in $\Bbb R$, and the norm in $\Bbb R$ is the usual modulus, then $\| F(x) \| = |F(x)|$, so $\left| \dfrac {F(x_n)} {\| F(x_n) \|} \right| = \dfrac {| F(x_n) |} {| F(x_n) |} = 1$, which can never tend to $\infty$. Therefore, you simply may not assume that $\left| \dfrac {F(x_n)} {\| F(x_n) \|} \right| \to \infty$.
What your professor probably does is to use the fact that an operator must be bounded on the sphere of radius $1$ in order to be continuous. Therefore he assumes $F$ unbounded on the sphere, so there exist a sequence $x_n$ on the sphere (which means $\| x_n \| = 1$) such that $F(x_n)$ is an unbounded sequence, which means $| F(x_n) | \to \infty$. This implies that $\dfrac {| F(x_n) |} {\| x_n \|} \to \infty$, but on the other hand you have that $\dfrac {| F(x_n) |} {\| x_n \|} = \left| F \left( \dfrac {x_n} {\| x_n \|} \right) \right| < \infty$ by the linearity of $F$ and by the boundedness of $F$ on the ball (and therefore on the sphere), and a bounded equence can not tend to $\infty$, hence the original assumption must be false, therefore $F$ must be bounded on the sphere, therefore continuous.
A: question: $F \left ( \frac{x_n}{F(x_n)} \right )=1$ why?
You know
$$
F(x_n) = F(x_n)
$$
right?  Next, multiply by a scalar,
$$
\frac{1}{F(x_n)}\;F(x_n) = 1
$$
Then from linearity, put the scalar inside
$$
F\left(\frac{x_n}{F(x_n)}\right) = 1
$$
