To expand on my comment:$\DeclareMathOperator{\Hom}{Hom}$
There are many ways to equip $X \times Y$ with a norm, the most natural ones are $\|(x,y)\| = \max{\{\|x\|,\|y\|\}}$ and $\|(x,y)\| = \|x\| + \|y\|$ since they correspond to the categorical product and coproduct operations (in the category of Banach or normed linear spaces and linear maps of norm $\leq 1$). Be that as it may, it is a good exercise to check that all the $p$-norms $\|(x,y)\|_{p} = \left(\|x\|^{p} + \|y\|^{p}\right)^{1/p}$ on $X \times Y$ are equivalent, thus a linear map $T: X \times Y \to Z$ is continuous if and only if it is bounded with respect to any of the norms with which you can equip the space $X \times Y$.
Note that linear means $T(\lambda x, \lambda y) = \lambda T(x,y)$ and $T(x+x',y+y') = T(x,y) + T(x',y')$ for all $(x,y), (x',y') \in X \times Y$ and all $\lambda \in \mathbb{R}$.
The multiplication is not linear in this sense, but it is bilinear $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$. As such, it is obviously continuous and it is bounded with respect to the norm on the bilinear maps $B: X \times Y \to Z$ given by
$$\|B\| = \sup_{\|x\|,\|y\| \leq 1} \|B(x,y)\|_{Z}$$
and it is a good exercise to check:
- A bilinear map $B$ is continuous if and only if it is bounded with respect to the norm above. Note that this simply means that $\|B(x,y)\|_{Z} \leq \|B\| \,\|x\|_{X} \, \|y\|_{Z}$ by bilinearity.
- If $Z$ is a Banach space then the space of bilinear maps $X \times Y \to Z$ is complete with respect to that norm.
Added: In fact, this idea naturally leads to the notion of the projective tensor norm.
Recall that a bilinear map $B: X \times Y \to Z$ corresponds to a linear map $b: X \otimes Y \to Z$. An element of the tensor product can be written as a finite sum $\sum x_i \otimes y_i$ and since $b$ is linear we have $b(\sum x_i \otimes y_i) = \sum b(x_i \otimes y_i) = \sum B(x_i,y_i)$. Now we want a norm on $X \otimes Y$ such that $b$ is bounded if and only if $B$ is bounded. Now $B$ is bounded if and only if $B$ is bounded on the elementary tensors, for which we have $\|B(x,y)\| \leq \|B\|\,\|x\|\,\|y\|$ so the norm of $\sum x_i \otimes y_i$ on $X \otimes Y$ had better be comparable to $\sum \|x_i\| \, \|y_i\|$. Now the problem is that this is not well defined because an element of $X \otimes Y$ has many decompositions into sums of elementary tensors. It turns out that the correct definition for the norm of $\omega \in X \otimes Y$ is
$$\|\omega \|_{\pi} = \inf\left\{ \sum \|x_i\|_X \, \|y_i\|_{Y} \,:\, \omega = \sum x_i \otimes y_{i}\right\}$$
where the infimum is taken over all (finite) representations $\omega = \sum x_i \otimes y_i$ as sum of elementary tensors. It is quite obvious that $\|\cdot\|_{\pi}$ is a semi-norm on $X \otimes Y$ satisfying $\|x \otimes y\|_{\pi} \leq \|x\|_{X} \|y\|_{Y}$. A bit more work shows that actually $\|x \otimes y\|_{\pi} = \|x\|_{X} \|y\|_{Y}$ and that $\| \cdot \|_{\pi}$ is a norm. Moreover, one can check that $\|B\| = \|b\|$, when the latter is computed as operator norm on $X \otimes Y$ with respect to $\|\cdot \|_{\pi}$. This gives us a bijection
$$\text{Bil} (X,Y;Z) = \text{Hom}(X \otimes Y, Z)$$
between the spaces of bounded bilinear maps $X \times Y \to Z$ and bounded linear maps $X \otimes Y \to Z$. Now if $X,Y$ happen to be Banach spaces then $X \otimes Y$ no longer is a Banach space in general, so we may simply complete it and we write $X \widehat{\otimes} Y$ for this completion. Combining this with the observation made by Mark in his answer, we get the (isometric) correspondences
$$\Hom{(X \widehat{\otimes} Y, Z)} = \text{Bil}(X,Y;Z) = \Hom(X,\Hom(Y,Z))$$
which we know well from linear algebra. We equip all $\Hom$-spaces with their natural operator norms and the space $\text{Bil}$ with the norm I defined above.
Let $A \in L(X,Y)$, $\tilde y \in Y^*$, $x \in X$. Then from $\left(A^*(\tilde y)\right)(x) := \tilde y (A(x))$ you get
$$|A^*(\tilde y)(x)|= |\tilde y (A(x))|≤\|\tilde y\|_{Y^*} \|A(x)\|_Y ≤ \|\tilde y\|_{Y^*} \|A\|_{L(X,Y)} \|x\|_X$$
For $A^*$ to be bounded you must consider the following expression
$$\|A^*\|_{L(Y^*,X^*)}=\sup_{\tilde y \in Y^*, \ \|\tilde y\|_{Y^*}≤1} \left\{\|A^*(\tilde y)\|_{X^*}\right\}$$
Note that you have the following inequality, which follows from the first one
$$\|A^*(\tilde y)\|_{X^*}=\sup_{x \in X, \ \|x\|_X≤1}\{|A^*(\tilde y)(x)|\}≤\sup_{x \in X, \ \|x\|_X≤1}\{\|\tilde y\|_{Y^*} \|A\|_{L(X,Y)}\ \|x\|_X \}$$
Putting it together gives you
$$\|A^*\|_{L(Y^*,X^*)}≤\|A\|_{L(X,Y)}$$
Note further that the construction gives you $\|A^{**}\|_{L(X^{**},Y^{**})}≤\|A^*\|_{L(Y^*,X^*)}≤\|A\|_{L(X,Y)}$. If you restrict $A^{**}$ to the subspace given by the embedding of $X$ into $X^{**}$ you get precisely $A$ again (composed with the embedding of $Y$ into $Y^{**}$). So $\|A^{**}\|_{L(X^{**},Y^{**})}≥\|A\|_{L(X,Y)}$ also holds, and the two inequalities are equalities, so also $\|A^*\|=\|A\|$.
Best Answer
Then $\dfrac{\|Lx\|}{\|x\|}=\dfrac{\|Lx\|}{\alpha}\leq\sup_{\|y\|\leq 1}\|Ly\|$ and hence $\|L\|\leq\sup_{\|y\|\leq 1}\|Ly\|$.
Now for $\|y\|\leq 1$, wrtie $\|Ly\|=\dfrac{\|Ly\|}{\|y\|}\|y\|\leq\dfrac{\|Ly\|}{\|y\|}\leq\|L\|$, of course this inequality is trivial for $y=0$, then $\sup_{\|y\|\leq 1}\|Ly\|\leq\|L\|$.