Let $E$ and $F$ be normed spaces. Is it true that $\mathcal{L}(E,F)$ contains an isometric copy of $E'$? 
Let $E$ and $F$ be normed spaces. Is it true that $\mathcal{L}(E,F)$
contains an isometric copy of $E'$?

I want to use this idea in my proof, but I don't know if it's true. How could I define a function to check the isometric isomorphism between $E'$ and $\mathcal{L}(E,F)$.
Note: $E'$ is the topological dual of $E$.
 A: I am just adding the details on the comment of Giuseppe Negro:
Let $X,Y$ be normed spaces. Fix $y_0\in Y$ with $\|y_0\|=1$. Denote by $X^*$ the topological dual of $X$. For $f\in X^*$, define $T_f:X\to Y$ by $T_f(x)=f(x)\cdot y_0$. Note that $T_f\in L(X,Y)$ since $\|T_f(x)\|=\|f(x)y_0\|=|f(x)|\cdot\|y_0\|=|f(x)|\le\|f\|\cdot\|x\|$, so $\|T_f\|\le\|f\|$, since this is true for all $x\in X$.
Define a map $\Phi:X^*\to L(X,Y)$ by $\Phi(f)=T_f$. Then $\Phi$ is a well-defined linear map that satisfies $\|\Phi(f)\|=\|T_f\|\le\|f\|$ for all $f\in X^*$. We show that $\Phi$ is actually isometric: Indeed, let $f\in X^*$ and $\varepsilon>0$. Since $\|f\|=\sup_{x\in X,\|x\|\le1}|f(x)|$, find $x\in X$ with $\|x\|\le1$ such that $\|f\|\le|f(x)|+\varepsilon$. Then, since $|f(x)|=|f(x)|\cdot\|y_0\|=\|f(x)y_0\|=\|T_f(x)\|$, we actually have that $\|f\|\le \|T_f(x)\|+\varepsilon$. But $\|T_f(x)\|\le\|T_f\|\cdot\|x\|\le\|T_f\|$, so $\|f\|\le\|T_f\|+\varepsilon$. Letting $\varepsilon\to0^+$, yields $\|f\|\le\|T_f\|=\|\Phi(f)\|$, proving the other inequality as we wanted.
