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BruceW

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Hi everyone!

I've been trying to get my head around the Gateaux differential. (or derivative, however it is called). So anyway, I've seen it defined like this:

[tex]\lim_{\tau \rightarrow 0} \frac{F(u + \tau \psi) - F(u)}{\tau} [/tex]

where the function ##F## takes ##X## to ##Y## (which are both locally convex topological vector spaces). Also, ##u## is an element of some open subset of ##X## and ##\psi## is any element of ##X##. i.e. the Gateaux differential only exists at ##u## if this limit is the same for any choice of ##\psi##. Now, I'm not a mathematician. So I'm not used to this kind of stuff, so please say if I've done something stupid. But I decided to try a really simple example, just have ##X## and ##Y## the real numbers. And, to keep it super-simple, use the function ##F(x)=x##. So, now check if the Gateaux differential exists:

[tex]\lim_{\tau \rightarrow 0} \frac{u + \tau \psi - u}{\tau} [/tex]

Which is then equal to:

[tex]\lim_{\tau \rightarrow 0} \frac{\tau \psi}{\tau} [/tex]

Which, after cancelling the ##\tau## simply gives:

[tex]\psi[/tex]

But clearly, this is not the same for any choice of ##\psi##. So does that mean the Gateaux differential does not exist for this simple example? Or is ##\psi## meant to have unit magnitude, in some sense? hmm. the fact that ##X## is a locally convex topological vector space means that it must be a vector space with a family of seminorms. So it would make sense if ##\psi## must be defined to have a set norm or seminorm. For my super-simple example, we could define the norm as the absolute value, and if we say that ##\psi## must have a set norm, then the Gateaux differential would exist.

So, is that the answer? ##\psi## is not any element of ##X##, but its norm must be set, and then if the limit exists (for any ##\psi## with the same norm), then the Gateaux differential exists?

And so, in the context of Lagrangian mechanics, when we calculate the functional derivative of the Action, we vary the test function, but we cannot vary the norm of the test function. So, say if we chose ##L^2## as our space of test functions, then for our Gateaux differential to exist, we require that we get the same answer with any ##L^2## function that has the same value for its square integral. Also, I realize that to calculate the functional derivative of the action, we don't really need to care about keeping the norm of the test function the same, because we can just say that far away from the minimized path, the test function takes whatever values it needs so that the norm of the test function stays the same. And I think we can usually do this with nice enough functions, so I won't worry too much about this when I'm doing Lagrangian mechanics.

Anyway, I hope someone can verify some of this, or has some insight on this topic. It's pretty interesting, I think.

edit: in fact, the use of the Gateaux differential in extremizing the action is kinda weird, because when the Euler-Lagrange equations are not satisfied, the Gateaux differential does not exist. So it's true we are choosing the action to be stationary, but it is not like the functional derivative of the action is non-zero on nearby paths. The functional derivative does not exist on nearby paths! Although, I guess we don't care, since if we have found a stationary action, that is all we need.

p.s. I've been assuming that we want to use the Gateaux differential as the functional derivative simply because (I think) the Gateaux differential is the most general mathematical definition for the functional derivative.

I've been trying to get my head around the Gateaux differential. (or derivative, however it is called). So anyway, I've seen it defined like this:

[tex]\lim_{\tau \rightarrow 0} \frac{F(u + \tau \psi) - F(u)}{\tau} [/tex]

where the function ##F## takes ##X## to ##Y## (which are both locally convex topological vector spaces). Also, ##u## is an element of some open subset of ##X## and ##\psi## is any element of ##X##. i.e. the Gateaux differential only exists at ##u## if this limit is the same for any choice of ##\psi##. Now, I'm not a mathematician. So I'm not used to this kind of stuff, so please say if I've done something stupid. But I decided to try a really simple example, just have ##X## and ##Y## the real numbers. And, to keep it super-simple, use the function ##F(x)=x##. So, now check if the Gateaux differential exists:

[tex]\lim_{\tau \rightarrow 0} \frac{u + \tau \psi - u}{\tau} [/tex]

Which is then equal to:

[tex]\lim_{\tau \rightarrow 0} \frac{\tau \psi}{\tau} [/tex]

Which, after cancelling the ##\tau## simply gives:

[tex]\psi[/tex]

But clearly, this is not the same for any choice of ##\psi##. So does that mean the Gateaux differential does not exist for this simple example? Or is ##\psi## meant to have unit magnitude, in some sense? hmm. the fact that ##X## is a locally convex topological vector space means that it must be a vector space with a family of seminorms. So it would make sense if ##\psi## must be defined to have a set norm or seminorm. For my super-simple example, we could define the norm as the absolute value, and if we say that ##\psi## must have a set norm, then the Gateaux differential would exist.

So, is that the answer? ##\psi## is not any element of ##X##, but its norm must be set, and then if the limit exists (for any ##\psi## with the same norm), then the Gateaux differential exists?

And so, in the context of Lagrangian mechanics, when we calculate the functional derivative of the Action, we vary the test function, but we cannot vary the norm of the test function. So, say if we chose ##L^2## as our space of test functions, then for our Gateaux differential to exist, we require that we get the same answer with any ##L^2## function that has the same value for its square integral. Also, I realize that to calculate the functional derivative of the action, we don't really need to care about keeping the norm of the test function the same, because we can just say that far away from the minimized path, the test function takes whatever values it needs so that the norm of the test function stays the same. And I think we can usually do this with nice enough functions, so I won't worry too much about this when I'm doing Lagrangian mechanics.

Anyway, I hope someone can verify some of this, or has some insight on this topic. It's pretty interesting, I think.

edit: in fact, the use of the Gateaux differential in extremizing the action is kinda weird, because when the Euler-Lagrange equations are not satisfied, the Gateaux differential does not exist. So it's true we are choosing the action to be stationary, but it is not like the functional derivative of the action is non-zero on nearby paths. The functional derivative does not exist on nearby paths! Although, I guess we don't care, since if we have found a stationary action, that is all we need.

p.s. I've been assuming that we want to use the Gateaux differential as the functional derivative simply because (I think) the Gateaux differential is the most general mathematical definition for the functional derivative.

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