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Finding a counterexample to “operator norm of linear transformation $f$ = $\sup \{ \|f(x)\| \mid \|x\| = 1\}$”
I want to show that the notion of operator norm is not commutative.
Assuming the definition is right, the notion of operator norm is essentially the same as the notion of $\ell_2$ norm except the $\sup$ operator. So my impression is that, among all $\ell_2$ norms, the operator norm corresponds to the $\ell_\infty$ norm. I can’t really show that any $2 \times 2$ matrix doesn’t satisfy the relation, that is the operator norm of any linear transformation is the supremum of the operator norms of linear transformations with respect to the $\ell_2$ norm, although intuitively I know that that is not true. Please help me find an easy counter example. Thanks a lot.
A:
As you say the notion of operator norm is the $\ell_\infty$ norm of
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inplace of:
c2_a2.sendto(pack(c3_a1.recv()),”127.0.0.1″, 3000)
You’re not actually using the recv() function, which is why you only get junk sent to the other end. You can make a few changes to the sendto function to make it more useful:
def sendto_net(s, addr, len):
s.settimeout(4.5)
s.sendto(pack(bytes(s, addr, len)), addr)
And then use it like:
sendto_net(c2_a2,c3_a1,4)
You still haven’t used the recv() function, which you should call recv(4) to make sure you can actually get 4 bytes.
But I also have a more intuitive way to do this:
From this answer:
>>> from socket import *
>>> s = socket(AF_INET, SOCK_DGRAM)
>>> s.sendto(bytes(‘I am sending my address’, ‘ascii’), (‘127.0.
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