Show that this function is a contraction Let $H$ denote a Hilbert space and $\mathcal{B}_1(H)$ denote the space of trace class operators on $H$. I believe that the following map is an isometry but I'm stuck on the proof that it's a contraction.
$$\Phi :H^* \widehat{\otimes} H \rightarrow \mathcal{B}_1(H), ~~\varphi \otimes f\mapsto\varphi f.$$
Here $\varphi f : H \rightarrow H$ is that trace class operator mapping $h \in H$ to $\varphi(h)f \in H$.
Edit: the norm on $H^* \otimes H$ is the projective one.
 A: (Initially, I didn't know your norm was the projective one; maybe you could edit that into the question) 

First, $\Phi$ is injective. Indeed, if $\sum_{j=1}^n\varphi_j f_j=0$, we can do the following. Let $z_1,\ldots,z_m$ be a basis of $\text{span}\,\{f_1,\ldots,f_n\}$. So 
$$f_j=\sum_{k=1}^m\alpha_{jk}z_k,
%\ \ \ \ z_k=\sum_{j=1}^n\beta_{kj}f_j
$$
for appropriate coefficients. Then
$$
0=\sum_{j=1}^n\varphi_jf_j=\sum_{k=1}^m\left(\sum_{j=1}^n\alpha_{jk}\varphi_j\right)\,z_k.
$$
As the $\{z_k\}$ are linearly independent, we get for each $k$
$$
\sum_{j=1}^n\alpha_{jk}\varphi_j=0.
$$
Then
$$
\sum_{j=1}^n\varphi_j\otimes f_j=\sum_{k=1}^m\left(\sum_{j=1}^n\alpha_{jk}\varphi_j\right)\otimes z_k=0.
$$

Now consider the norm, on $H^*\otimes H$, given by 
$$
\|x\|_0:=\|\Phi(x)\|_1.
$$
The linearity and injectivity of $\Phi$ make $\|\cdot\|_0$ a norm. And it is a cross norm: it is not hard to check that the one-norm of $\varphi f$ is $\|\varphi\|\,\|f\|$. Indeed, by the Riesz Representation Theorem, the functional $\varphi$ is implemented by a vector $\varphi\in H$, i.e. $f\longmapsto \langle f,\varphi\rangle$. Let $T=\varphi f$. Then 
$$
\langle T^*h,k\rangle=\langle h,Tk\rangle=\overline{\langle k,\varphi\rangle}\,\langle h,f\rangle=\langle\,\langle h,f\rangle\,\varphi,k\rangle.
$$
So $T^*h=\langle h,f\rangle\,\varphi$ (that is, $(\varphi f)^*=f\varphi$; this is of course confunsing, and one usually uses the notation $f\otimes \varphi$ for this; but of course this notation would really cloud you question). Thus
$$
T^*Th=\langle Th,f\rangle\,\varphi=\|f\|^2\,\langle h,\varphi\rangle\,\varphi.
$$
One can check that $h\longmapsto \frac1{\|\varphi\|^2}\langle h,\varphi\rangle\,\varphi$ is a projection (i.e., selfadjoint and equal to its square). It follows that a (so "the") positive square root of $T^*T$ is 
$$
(T^*T)^{1/2}=\frac{\|f\|}{\|\varphi\|}\,\langle\cdot,\varphi\rangle\,\varphi. 
$$
So
$$
\|\varphi f\|_1=\text{Tr}((T^*T)^{1/2})=\|f\|\,\|\varphi\|\,\text{Tr}\,\left( \frac1{\|\varphi\|^2}\,\langle\cdot,\varphi\rangle\,\varphi\right)=\|f\|\,\|\varphi\|
$$
(a rank-one projection has trace one: this can be seen by using an orthonormal basis that includes $\varphi/\|\varphi\|$).

Now, as the projective norm is the largest cross norm on $H^*\otimes H$, we get 
$$
\|\phi(x)\|=\|x\|_0\leq \|x\|.
$$
