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Let $E=\mathcal{C}[0,1]$ provide with $\|\cdot\|_\infty$ norm. Let $f:E\to \mathbb{R}$, given by $f(p)=e^{\int_0^1 p}$. I need to prove that $f$ is differentiable.

My approach: Let $p,q\in\mathcal{C}[0,1]$ then, $$f(p+tq)-f(p)=e^{\int_0^1 p+tq}-e^{\int_0^1 p}=e^{\int_0^1 p}\left( e^{t\int_{0}^1 q}-1\right)=f(p)\left(1+t\int_{0}^1q+t\ o(1)-1 \right)$$ The last equality is true, just applying Taylor's Formula. Then we have $$\lim_{t\to 0}\frac{f(p+tq)-f(p)}{t}= f'_{q}(p)=f(p)\int_{0}^1 q$$ So I deduce that if $f$ is differentiable, then $Df(p)=f(p)T$, where $T(q)=\int_{0}^1q$. The operator $f(p)T$ is linear because $T$ is linear, and is continuous because: $\left|f(p)\int_{0}^1 q \right|\leq f(p)\|q\|_{\infty}$, from this I know that $\left(\int_{0}^1q\right)^2\leq \|q\|_{\infty}^2$.

Maybe my question is silly, but how can I prove that: $$f(p+q)-f(p)=f(p)T(q)+\|q\|_{\infty}\ o(1)$$

I tried for longer but finally I'm stuck! Thanks in advance.

Valent
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1 Answers1

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We are given the formula $Df_p(q) = f(p) \cdot \int_0^1 q$, and we need to prove that this is true.

To prove this, just realize that $$\big| f(p+q)-f(p)-Df_p(q) \big| = \bigg|e^{\int_0^1 p+q} -e^{\int_0^1 p} - e^{\int_0^1p} \int_0^1 q \bigg|$$ $$= \big| e^{\int_0^1 p} \big| \cdot \bigg| e^{\int_0^1 q}-1 - \int_0^1 q \bigg|$$

For small values of $\alpha$, we know by Taylor expansion that $e^{\alpha}-1-\alpha = O(\alpha^2)$. In particular, by applying this to $\alpha = \int_0^1 q$, we see that $\bigg| e^{\int_0^1 q}-1 - \int_0^1 q \bigg| = O\big( \big[\int_0^1 q \big]^2\big)$.

But since $\big| \int_0^1 q \big| \leq \|q\|_{\infty}$, it follows that $\big| e^{\int_0^1 q}-1 - \int_0^1 q \big| \leq O\big(\|q\|_{\infty}^2\big)$.

Hence $\big| f(p+q)-f(p)-Df_p(q) \big| \leq O\big(\|q\|_{\infty}^2\big)=\|q\|_{\infty} \; o(1)$, and the result follows.

shalop
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