Let \(I=[0,1]\) and write \[ \|f\|_p=\left(\int_I|f|^p\right)^{1/p} \quad(1\le p<\infty), \qquad \|f\|_\infty=\operatorname*{ess\,sup}_I|f|. \]
Show that \(C(I)\) with the norm \(\|\cdot\|_p\), \(1\le p<\infty\), is not Banach.
Prove that \(L^p(I)\), meaning measurable finite-\(p\)-norm functions modulo equality almost everywhere, is Banach for \(1\le p\le\infty\).
Prove that there is no smooth function \(h\in C^\infty(\mathbb R)\) such that \[ (f*h)(x):=\int_0^1 f(y)h(x-y)\,dy=f(x) \qquad\text{for every }f\in L^1([0,1])\text{ and almost every }x\in[0,1]. \]
Prove Hölder’s inequality: if \(1/p+1/q=1\), then \[ \int_I|fg|\le \|f\|_p\|g\|_q. \]
Deduce Young’s convolution inequality on \(\mathbb R\): if \[ 1\le p,q,r\le\infty, \qquad \frac1p+\frac1q=1+\frac1r, \] then \[ \|f*g\|_r\le\|f\|_p\|g\|_q. \]
Solution
1. \(C(I)\) is not complete in the \(L^p\) norm for finite \(p\).
Let \[ f=\mathbf1_{[0,1/2]}. \] Choose continuous functions \(f_n\) which equal \(1\) on \([0,1/2-1/n]\), equal \(0\) on \([1/2+1/n,1]\), and interpolate linearly in between. Then \[ \|f_n-f\|_p^p\le \frac{2}{n}\longrightarrow0. \] Thus \((f_n)\) is Cauchy in \(C(I)\) with the \(L^p\) norm. If it converged in that norm to some \(g\in C(I)\), uniqueness of \(L^p\) limits would give \(g=f\) almost everywhere. Continuity would then force \(g=1\) on \([0,1/2)\) and \(g=0\) on \((1/2,1]\), impossible at \(1/2\). Hence \(C(I)\) is not complete.
2. \(L^p(I)\) is complete.
First let \(1\le p<\infty\) and let \((f_n)\) be Cauchy in \(L^p\). Choose a subsequence \((f_{n_k})\) such that \[ \|f_{n_{k+1}}-f_{n_k}\|_p\le2^{-k}. \] Set \[ g_m=|f_{n_1}|+\sum_{k=1}^{m}|f_{n_{k+1}}-f_{n_k}|. \] By Minkowski, \[ \|g_m\|_p \le \|f_{n_1}\|_p+\sum_{k=1}^{m}2^{-k}, \] so by monotone convergence the pointwise limit \(g=\lim_m g_m\) belongs to \(L^p\). Therefore \[ \sum_{k=1}^\infty |f_{n_{k+1}}(x)-f_{n_k}(x)|<\infty \] for almost every \(x\). The subsequence consequently converges pointwise a.e. to some measurable \(f\).
Moreover, for every \(k\), \[ |f-f_{n_k}| \le \sum_{j=k}^\infty |f_{n_{j+1}}-f_{n_j}|, \] and Minkowski gives \[ \|f-f_{n_k}\|_p\le\sum_{j=k}^\infty2^{-j}\longrightarrow0. \] Since the original sequence is Cauchy, it follows that \(f_n\to f\) in \(L^p\).
For \(p=\infty\), if \((f_n)\) is Cauchy in essential-supremum norm, choose representatives after discarding the countable union of null exceptional sets in the Cauchy estimates. Off that null set, \((f_n(x))\) is uniformly Cauchy, hence converges uniformly to a bounded measurable \(f\), and \(\|f_n-f\|_\infty\to0\). Thus \(L^\infty(I)\) is complete as well.
3. There is no smooth convolution identity on \(L^1([0,1])\).
Suppose \(h\in C^\infty(\mathbb R)\) satisfied the stated identity for every \(f\in L^1([0,1])\). Fix such an \(f\). The displayed convolution is finite for every \(x\in[0,1]\), since \(h\) is bounded on \([-1,1]\), and is unchanged by replacing \(f\) on a null set. Define \[ \omega_h(\delta)=\sup\{|h(s)-h(t)|:s,t\in[-1,1],\ |s-t|\leq\delta\}. \] Uniform continuity on this compact interval gives \(\omega_h(\delta)\to0\) as \(\delta\to0\). For \(x,x'\in[0,1]\), \[ \begin{aligned} |(f*h)(x')-(f*h)(x)| &\le \int_0^1 |f(y)|\,|h(x'-y)-h(x-y)|\,dy\\ &\le \|f\|_1\omega_h(|x'-x|). \end{aligned} \] Here \(x-y\) and \(x'-y\) both lie in \([-1,1]\) for \(0\leq y\leq1\). The bound therefore proves that \(f*h\) is continuous on \([0,1]\).
Now take \[ f=\mathbf1_{[0,1/2]}. \] This belongs to \(L^1([0,1])\). The assumed identity would give \[ f(x)=(f*h)(x) \qquad\text{for almost every }x\in[0,1], \] while the right-hand side is continuous. Any continuous function equal to this indicator almost everywhere must equal one throughout \((0,1/2)\) and zero throughout \((1/2,1)\): a different value would persist on an interval of positive measure. These two values contradict continuity at \(1/2\). Hence no smooth \(h\) can act as the identity convolution kernel on all of \(L^1([0,1])\), even with equality required only almost everywhere.
4. Hölder’s inequality.
The endpoint cases \(p=1,q=\infty\) and \(p=\infty,q=1\) are immediate. Assume \(1<p,q<\infty\) and \(1/p+1/q=1\). If either norm vanishes there is nothing to prove, so normalize \[ F=\frac{|f|}{\|f\|_p}, \qquad G=\frac{|g|}{\|g\|_q}. \] Young’s scalar inequality gives \[ F(x)G(x)\le \frac{F(x)^p}{p}+\frac{G(x)^q}{q}. \] Integrating, \[ \int_I FG \le \frac1p\int_I F^p+\frac1q\int_I G^q =\frac1p+\frac1q=1. \] Multiplying back by the norms yields \[ \int_I|fg|\le\|f\|_p\|g\|_q. \]
5. Young’s convolution inequality.
If \(r=\infty\), then \(1/p+1/q=1\), so Hölder gives directly \[ |(f*g)(x)|\le\|f\|_p\|g\|_q. \]
Assume \(r<\infty\). The relation implies \(r\ge p,q\). For fixed \(x\), write \[ |f(y)g(x-y)| = \bigl(|f(y)|^p|g(x-y)|^q\bigr)^{1/r} |f(y)|^{1-p/r}|g(x-y)|^{1-q/r}. \] Apply generalized Hölder in \(y\) with reciprocal exponents \[ \frac1r, \qquad \frac1p-\frac1r, \qquad \frac1q-\frac1r, \] whose sum is \(1\). This gives \[ |f*g(x)| \le \left(\int |f(y)|^p|g(x-y)|^q\,dy\right)^{1/r} \|f\|_p^{1-p/r}\|g\|_q^{1-q/r}. \] Raise to the \(r\)th power and integrate in \(x\). Tonelli and translation invariance give \[ \begin{aligned} \|f*g\|_r^r &\le \|f\|_p^{r-p}\|g\|_q^{r-q} \iint |f(y)|^p|g(x-y)|^q\,dy\,dx\\ &= \|f\|_p^{r-p}\|g\|_q^{r-q} \|f\|_p^p\|g\|_q^q\\ &=\|f\|_p^r\|g\|_q^r. \end{aligned} \] Taking \(r\)th roots proves \[ \|f*g\|_r\le\|f\|_p\|g\|_q. \]