$X,Y$ are normed spaces. For a linear map $T:X\rightarrow Y$ define $T^{*}:Y^*\rightarrow X^*$ by $T^*(f)=f\circ T$.
If $T$ is bounded I have to show that $\|{T}\|=\|{T^*}\|$. Showing $\|T^*\|\leq\|T\|$ was easy. For the converse I tried this :
We have $\|T^{**}\|\leq\|T^*\|$. If I can show $\|T\|\leq\|T^{**}\|$ then I am done. Now $\|T^{**}(f)\|=\|f\circ T^*\|$. Then $\|T^{**}(J_x)\|=\|J_x\circ T^*\|$ where $J$ is the natural inclusion of $X$ into $X^{**}$. $$|J_x\circ T^*(f)|=|J_x(T^*(f))|=|J_x(f\circ T)|=|f\circ T(x)|$$ Now for each $x\in X$ $\exists$ a continuous linear functional $f$ on $Y$ such that $\|f\|=1$ and $f(Tx)=\|Tx\|$ (Hahn-Banach Theorem). For such an $f$ the last expression becomes $\|Tx\|$. So that $$\|J_x\circ T^*\|\geq\|Tx\|$$ which means $$\|T^{**}\|=\sup\limits_{\|g\|=1}\|T^{**}(g)\|\geq\sup\limits_{\|x\|=1}\|T^{**}(J_x)\|\geq\sup\limits_{\|x\|=1}\|Tx\|=\|T\|$$ since $\|J_x\|=\|x\|$.
Is there something wrong? Is there a simpler proof? Any help is much appreciated.
Here is an argument without involving the double dual.
Given $x\in X$, one of the most common application of the Hahn-Banach Theorem implies there exists $g\in Y^*$ such that $\|g\|=1$ and $$ \|Tx\|=g(Tx). $$ So $$ \|Tx\|=g(Tx)=(T^*g)x\leq\|T^*g\|\,\|x\|\leq \|T^*\|\,\|x\|. $$ Thus $\|T\|\leq\|T^*\|$.