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$\phi: (C([0,1];\Bbb R ),||\cdot||_\infty )\to (\Bbb R, |\cdot | ); \: \: \: \: \: \: \phi(u):=\int_0^1 u^2(t) dt $

Is this function continuous or even uniformly continuous? (I know that the function $g: M\to \Bbb R , g(x) := x^2$ is continuous but not uniformly)

Also, is there a non-empty subset of $ C([0,1];\Bbb R )$ so that $\phi$ is Lipschitz-continuous on that set?

Thanks in advance!

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$\phi$ will be continuous. Indeed, fix any $u\in C_{[0,1],\mathbb{R}}$, and let $\delta > 0$ such that $\delta(2\lVert u\rVert_\infty + \delta) \leq \varepsilon$.

Fix any $\varepsilon > 0$. If $v$ is such that $\lVert u - v\rVert_\infty \leq \delta$, then $$\begin{align} \lvert \phi(u)-\phi(v) \rvert &= \left\lvert \int_{[0,1]} u^2-v^2\right\rvert \leq \int_{[0,1]} \left\lvert u^2-v^2\right\rvert \\ &= \int_{[0,1]} \underbrace{\left\lvert u-v\right\rvert}_{\leq \lVert u - v\rVert_\infty} \cdot \underbrace{\left\lvert u+v\right\rvert}_{\leq 2\lVert u\rVert_\infty+\delta} \leq \delta(2\lVert u\rVert_\infty + \delta) \leq \varepsilon \end{align}$$ i.e. $\phi$ is continuous.

It is easy to adapt the above to show that $\phi$ will be Lipschitz (and therefore uniformly continuous) on any subset of any $C_{[0,1],[-a,a]}\subseteq C_{[0,1],\mathbb{R}}$ (i.e., for functions bounded by some universal constant). The Lipschitz constant will then be $2a$ (replacing the $2\lVert u\rVert_\infty + \delta$ term).

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  • $\begingroup$ Thanks a lot! Is it enough to say that the function $x^2$ is not uniformly continuous to say that $\phi$ isn't either? $\endgroup$ – DeltaChief May 26 '16 at 16:14
  • $\begingroup$ Not quite sure right now, but maybe (considering the constant functions $f_x\colon t\in [0,1]\to x$) -- I'll think about it once back. $\endgroup$ – Clement C. May 26 '16 at 16:15
  • $\begingroup$ Yes, the same proof/argument will go through. (To show non uniform continuity.) $\endgroup$ – Clement C. May 26 '16 at 16:23
  • $\begingroup$ Oh, my bad, your proof is already the proof for uniform continouity. $\endgroup$ – DeltaChief May 26 '16 at 16:25
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    $\begingroup$ No, i prove continuity: and uniform continuity (via Lipschitz) over any subset of functions with bounded range. $\endgroup$ – Clement C. May 26 '16 at 16:27

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