There are two exercises about projective tensor products.
1) If either $X$ or $Y$ is finite dimensional, then the projective tensor product $X\otimes Y$ is complete.
2) If both $X$ and $Y$ are infinite dimensional, then $X\otimes Y$ is never complete. (As a result, $X\otimes Y\ne X\otimes_{\pi}Y$)
If both $X$ and $Y$ are finite dimensional, then $\dim X\otimes Y=(\dim X)(\dim Y)$ is finite, and hence the norm space $X\otimes Y$ is complete and is just $X\otimes_{\pi}Y$ itself.
But I can't make it for only just either of $X$ or $Y$ is finite dimensional.
Any idea?
Note that here both $X$ and $Y$ are Banach spaces, and the projective tensor product norm defined on $X\otimes Y$ is \begin{align*} \pi(u)=\inf\left\{\sum_{i=1}^{n}\|x_{i}\|\|y_{i}\|: u=\sum_{i=1}^{n}x_{i}\otimes y_{i}\right\}, \end{align*} where $u=\displaystyle\sum_{i=1}^{n}x_{i}\otimes y_{i}$ is any representation of $u$ in $X\otimes Y$. The space $X\otimes_{\pi}Y$ is the completion of $X\otimes Y$ under this norm.
Maybe the first one can be dealt in this way:
I suspect that $(X\otimes Y)/Y\cong X$ where $Y$ is assumed to be finite dimensional, so $(X\otimes Y)/X$ is complete, as $X$ is also complete, then $X\otimes Y$ is complete by the three-space property of Banach spaces.
It is not necessarily true that $(X \otimes Y)/Y$ is isomorphic to $X$; it may help verify that even in the finite-dimensional case, the dimensions of these spaces do not match.
I think that the easiest approach for the case where $Y$ is finite dimensional is to note that if $Y$ has dimension $n$ (so that $Y$ is isomorphic to $\Bbb C^n$ under some norm), we have $$ X \otimes Y \cong X \otimes \Bbb C^n \cong \overbrace{X \oplus \cdots \oplus X}^{n \text{ times}}. $$ In the infinite-dimensional case, suppose that $\{x_i\} \subset X$ and $\{y_i\} \subset Y$ are linearly independent sequences with $\|x_i\| = \|y_i\| = 1$. It suffices to verify that the sequence defined by $$ z_n = \sum_{i=1}^n \frac{1}{2^i} x_i \otimes y_i $$ is Cauchy, but fails to converge in $X \otimes Y$ (since the infinite sum cannot be expressed as a finite sum of tensor products).