Finding the norm of linear operator $A:X\to X$ as $Ax=f(x)y$ where $y\in X$ and $f\in X^*$ Let $X$ be a linear normed space, $y\in X$ and $f\in X^*$. Define $A:X\to X$ as $Ax=f(x)y$. Prove that $A\in \mathcal{L}(X)$ and find its norm.
Hint: when finding $||A||$, you may use the corollary of Hahn-Banach theorem to show that $\exists x\in X$ s.t. $||x||=1$ and $|f(x)|=||f||$.
Okay so showing that $A\in \mathcal{L}(X)$ was easy. Then my candidate for the norm is $||A||=||f||||y||$. I showed that $||A||\leq||f||||y||$ but I am stuck with the part which uses the hint. I know the corollary I am supposed to use is: For every nonzero $x\in X$ there exists a functional $g\in X^*$ s.t. $g(x)=||x||$ and $||g||=1$. I am not sure if I am supposed to explicitly come up with some $x$ then use the corollary on it and end up with the hint somehow. I tried using the definition of $A$ since $Ax\in X$ but I got nothing. I know how to proceed if I just figure out how to prove the hint.
 A: You have been mislead. The hint is not valid. (It is valid in reflexive spaces, not in general normed linear spaces). For example, let $f(x)=\sum (1-\frac  1 n) x_n$ on $\ell^{1}$. Then $\|f\|=1$ and there is no $x$ such that $\|x\|=1$ and $f(x)=1$. 
However, you can prove the result using the following: given $\epsilon >0$ there exists a vector  $x$ of norm $1$ such that $f(x) >\|f\|-\epsilon$. 
A: $$\|A\|=\sup\limits_{x\ne0}\frac{\|f(x)y\|}{\|x\|}=\sup\limits_{x\ne0}\frac{|f(x)|\cdot\|y\|}{\|x\|}=\|y\|\sup\limits_{x\ne0}\frac{|f(x)|}{\|x\|}=\|y\|\cdot\|f\|$$
A: Remember that if $f\in X^{*}$, then $\|f\|=\sup\limits_{x\in X\smallsetminus{\{0\}}}\dfrac{|f(x)|}{\|x\|}$. We consider $z=(yf(y))/(\|y\|\|f(y)\|)$ and note that $\|z\|=1$, then $|f(z)|=\left |f\left(\dfrac{yf(y)}{\|y\|\|f(y)\|}\right)\right|=\dfrac{\|f(y)\| |f(y)|}{\|y\|\|f(y)\|}$ (since $f(y)\in\mathbb{K}$ so $ |f(yf(y))|=|f(y)f(y)|=\|f(y)\||f(y)|$), hence, $|f(z)|=\dfrac{|f(y)|}{\|y\|}\leq\sup\limits_{x\in X\smallsetminus{\{0\}}}\dfrac{|f(x)|}{\|x\|}=\|f\|$ $\therefore$ $|f(z)|\leq\|f\|$ $(1)$.
Now, as $\dfrac{|f(y)|}{\|y\|}=|f(z)|$, then $\dfrac{|f(y)|}{\|y\|}\leq|f(z)|$ thus $\sup\limits_{x\in X\smallsetminus{\{0\}}}\dfrac{|f(x)|}{\|x\|}=\|f\|\leq |f(z)|$ $(2)$. By $(1)$ and $(2)$ it follow that $|f(z)|=\|f\|$, and therefore $|f(z)|\|y\|=\|f\|\|y\|$, i.e. $\|A_{z}\|=\|f\|\|y\|$. 
I think that this is the you need.
