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I'm trying to prove whether the following statement is true or false:

Let $Y$ be a normed space and $x\in K$. $K$ is one dimensional. A function $f:K\rightarrow Y$ is Gateaux Differentiable at $x$ if and only if it is Frechet Differentiable at $x$.

I can't decide whether it is true or false. My initial instinct was that it was false, but I can't find an example to disprove it, so I'm not sure.

Any help would be appreciate

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    Hint: It is false. – 2017-02-12
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    It is wrong on any not 1 dimensional space. – 2017-02-12
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    So, if K is 1 dimensional and Y has any dimension, then it's true? – 2017-02-12
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    Duplicate: http://math.stackexchange.com/questions/2137197/gateaux-and-frechet-derivatives-on-vector-valued-functions/2137700#2137700 – 2017-02-12

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It is known that Frechet differentiability implies Gateaux differentiability.

Now if $M=\langle e \rangle$ is a one-dimensional normed space, $f : M \to N$ is a map into another normed space, and if, given $x_0 \in M,$ the Gateaux derivative $$ \delta f(h)=\lim_{\lambda \to 0} \frac{f(x_0+\lambda h)-f(x_0)}{\lambda} $$ at $x_0$ exists for all $h \in M,$ then $$ \delta f(\mu e)= \mu d=\mu \delta f(e) $$ where $\mu \in \mathbf R$ and where $d=\delta f(e) \in N$ is the vector $$ d=\lim_{\lambda \to 0} \frac{f(x_0+\lambda e)-f(x_0)}\lambda $$ (there are two cases to consider: $\mu \ne 0$ and $\mu=0$). We can simplify the argument by letting $M$ be equal to $\mathbf R.$

We see that the map $\delta f$ is linear and continuous, and it can be shown that it is Frechet derivative of $f$ at $x_0$ (just verify the definition with $\delta f$ as the potential Frechet derivative of $f$ at $x_0$). So yes, if $M$ is one-dimensional the statement is true, although it is not true in general.

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    Is the statement true when the domain is one dimensional, and the range can be of any dimension though? – 2017-02-12
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    Yes, the range can be of any dimension. – 2017-02-12