# Calculating the norm of $S\otimes T$ for bounded linear $S,T$.

Let $X,Y,W,Z$ be Banach Spaces. Let $X\otimes_{\pi}Y$ denote the tensor product endowed with the projective norm.

If I have $S\in B(X,Z)$, $T\in B(Y,W)$, it is straightforward to show that $S\otimes T\in B(X\otimes_{\pi} Y, Z\otimes_{\pi}W)$ with $\|S\otimes T\|\leq \|S\|\cdot\|T\|$.

According to Raymond Ray in Introduction to Tensor Products of Banach Spaces, the reverse inequality "... follows easily from $S\otimes T(x\otimes y) = Sx\otimes Ty$."

I cannot see how to get it.

For fixed $\epsilon > 0$, I tried choosing elements $x\in B_{X}$ and $y\in B_{Y}$ such that $$\|Sx\| > \|S\| - \epsilon\text{ and }\|Ty\| > \|T\| - \epsilon$$ which gives $$\|S\otimes T(x\otimes y)\| > \|T\|\cdot \|S\| - \epsilon (\|S\| + \|T\|) + \epsilon$$

But this turns out to be fruitless, because even though I can take $\epsilon\to 0$, $x\otimes y$ need not be in the unit ball of $X\otimes_{\pi} Y$.

Can anyone give me a hint (the vaguer the better)?

$\bf{\text{Available Fact}}$:

The closed unit ball $B_{X\otimes_{\pi} Y}$ is equal to the convex hull of $B_{X}\otimes B_{Y}$.

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If the closed unit ball is the convex hull of $B_X \otimes B_Y$, aren't you done because $x \otimes y \in B_X \otimes B_Y \subset B_{X \otimes Y}$? – Daniel Fischer Jul 7 '13 at 0:46
The projective norm is a cross norm (the largest one). – 1015 Jul 7 '13 at 0:54
Right. I was looking at things totally incorrectly. Thanks! facepalm – roo Jul 7 '13 at 1:07

Since projective tensor norm is a cross norm we have $\Vert x\otimes y\Vert=\Vert x\Vert\Vert y\Vert\leq 1$. Then $$\Vert S\otimes T\Vert \geq\Vert S\otimes T\Vert\Vert x\otimes y\Vert \geq\Vert(S\otimes T)(x\otimes y)\Vert =\Vert S(x)\otimes T(y)\Vert =\Vert S(x)\Vert\Vert T(y)\Vert \geq(\Vert S\Vert-\varepsilon)(\Vert T\Vert-\varepsilon)$$ Taking the limit $\varepsilon\to 0$ we get $\Vert S\otimes T\Vert\geq \Vert S\Vert\Vert T\Vert$.